Solved Examples on Probability

Genotype of Man = AS
The man has a sickle cell trait, SCT (because of the S)
Genotype of Woman = AS
The woman also has a sickle cell trait, SCT (because of the S)
Assume they intend to have four children
Let us draw a Punnett Square to illustrate the likely genotypes of their four children

Woman → Man ↓	$A$	$S$
$A$	$AA$	$AS$
$S$	$AS$	$SS$

If they have four children, it is likely (not certain) that:
1 child will have genotype AA
This means that one child is likely to be typical (no blood disorder).

2 children will have genotype AS
This means that two children are likely to inherit the sickle cell traits.

1 child will have genotype SS
This means that one child is likely to have the sickle cell disease.
The issue is with the SS genotype. The child with that genotype will likely need a lot of medical care.
Why bring a child into this world to suffer the child?
Do not recommend the marriage.

$ n(SS) = 1 \\[3ex] n(S) = 4 \\[3ex] P(SS) = \dfrac{n(SS)}{n(S)} \\[5ex] P(SS) = \dfrac{1}{4} = 0.25 = 25\% \\[3ex] $ If they have four children, there is a 25% probability that one of the children will have the sickle cell disease.

Genotype of Man = $AS$
Genotype of Woman = $AS$
Let us draw a Punnett Square to illustrate the likely genotypes of their four children

Woman → Man ↓	$A$	$S$
$A$	$AA$	$AS$
$S$	$AS$	$SS$

If they have four children, it is likely (not certain) that:
1 child will have genotype AA
2 children will have genotype AS
1 child will have genotype SS

The issue is with the SS genotype. The child with that genotype will likely need a lot of medical care.
Why bring a child into this world to suffer the child?
Do not recommend the marriage.

$ n(SS) = 1 \\[3ex] n(S) = 4 \\[3ex] P(SS) = \dfrac{n(SS)}{n(S)} \\[5ex] P(SS) = \dfrac{1}{4} = 0.25 = 25\% \\[3ex] $ If they have four children, there is a 25% probability that one of the children will have the sickle cell disease.

$ S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] n(S) = 12 \\[3ex] E = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] n(E) = 9 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{9}{12} \\[5ex] P(E) = \dfrac{3}{4} $

$ S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] n(S) = 12 \\[3ex] E = \{ \} \\[3ex] n(E) = 0 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{0}{12} \\[5ex] P(E) = 0 \\[3ex] $ This is an impossible event.
It is impossible to obtain a number greater than $12$ when a $12-sided$ doe is rolled one time.

$ (a.) \\[3ex] A = \{Ruth\:\:Kings,\:\:Esther\:\:Proverbs\} \\[3ex] (b.) \\[3ex] n(A) = 2 \\[3ex] n(S) = 9 \\[3ex] P(A) = \dfrac{n(A)}{n(S)} \\[5ex] P(A) = \dfrac{2}{9} \\[5ex] (c.) \\[3ex] A' = \{Joshua\:\:Judges,\:\:Samuel\:\:Daniel,\:\:Nehemiah\:\:Job,\:\:Isaiah\:\:Psalms,\:\:Jeremiah\:\:Amos,\:\:Ezekiel\:\:Joel,\:\:Micah\:\:Nahum\} $

(a.) The sequence of digits is: 3, 9, 3, 2, 9, 1, 7, 7, 0, 5

(b.) The sequence of letters is: H, T, H, H, T, H, T, T, H, T

(c.) The longest streak of heads in the list is 2 heads (3rd and fourth letters)

(d.) The percentage of flips that were heads:

$ number\;\;of\;\;flips = 10 \\[3ex] number\;\;of\;\;heads = 5 \\[3ex] \%\;\;of\;\;flips\;\;that\;\;were\;\;heads \\[3ex] = \dfrac{number\;\;of\;\;heads}{number\;\;of\;\;flips} * 100 \\[5ex] = \dfrac{5}{10} * 100 \\[5ex] = 50\% $

(a.) The sequence of digits is: 2, 6, 6, 1, 3, 0, 4, 5, 6, 6, 2, 4, 9, 8, 6, 7, 8, 4, 9, 8

(b.) The sequence of letters is: C, C, C, P, P, C, C, P, C, C, C, C, P, C, C, P, C, C, P, C

(c.) The percentage of the 20 participants that were assigned to the computer group:

$ number\;\;of\;\;participants = 20 \\[3ex] number\;\;in\;\;computer\;\;group = 14 \\[3ex] \%\;\;of\;\;participants\;\;that\;\;were\;\;computer\;\;group \\[3ex] = \dfrac{number\;\;of\;\;computer\;\;group}{number\;\;of\;\;participants} * 100 \\[5ex] = \dfrac{14}{20} * 100 \\[5ex] = 70\% \\[3ex] $ (d.) Other ways the Random Number Table could have been used to randomly assign participants to one of the two groups include:

(i) The odd dogits could have assigned participants to the computer group, and the even diigits could have assigned participants to the pen and paper group.

(ii) The digits: 0, 1, 2, 3, 4 could have assigned participants to the pen and paper group, and the digits: 5, 6, 7, 8, 9 could have assigned participants to the computer group.

(iii) The digits: 0, 1, 2, 3, 4 could have assigned participants to the computer group, and the digits: 5, 6, 7, 8, 9 could have assigned participants to the pen and paper group.

We can solve this question in two ways
Use any method you prefer

Let the number of additional red marbles = $r$

$ \underline{First\:\:Approach} \\[3ex] \underline{Initial} \\[3ex] n(S) = 42 \\[3ex] n(Red) = 16 \\[3ex] \underline{Updated} \\[3ex] n(Red) = 16 + r \\[3ex] n(S) = 42 + r \\[3ex] P(Red) = \dfrac{16 + r}{42 + r} \\[5ex] P(Red) = \dfrac{3}{5} \\[5ex] \therefore \dfrac{16 + r}{42 + r} = \dfrac{3}{5} \\[5ex] Cross\:\: Multiply\:\: method \\[3ex] 5(16 + r) = 3(42 + r) \\[3ex] 80 + 5r = 126 + 3r \\[3ex] 5r - 3r = 126 - 80 \\[3ex] 2r = 46 \\[3ex] r = \dfrac{46}{2} \\[5ex] r = 23 \\[5ex] \underline{Second\:\:Approach} \\[3ex] P(Red) = \dfrac{3}{5} \\[5ex] P(Red) + P(Red') = 1 ...Complementary\:\:Rule \\[3ex] P(Red') = 1 - P(Red) \\[3ex] P(Red') = 1 - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5}{5} - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5 - 3}{5} \\[5ex] P(Red') = \dfrac{2}{5} \\[5ex] n(Red') = n(Yellow) + n(Green) \\[3ex] n(Red') = 7 + 19 = 26 \\[3ex] P(Red') = \dfrac{n(Red')}{n(S)} \\[5ex] n(S) * P(Red') = n(Red') \\[3ex] n(S) = \dfrac{n(Red')}{P(Red')} \\[5ex] n(S) = n(Red') \div P(Red') \\[3ex] n(S) = 26 \div \dfrac{2}{5} \\[5ex] n(S) = 26 * \dfrac{5}{2} \\[5ex] n(S) = 13(5) \\[3ex] n(S) = 65 \\[3ex] Additional\:\:red\:\:balls = 65 - (16 + 7 + 19) \\[3ex] Additional\:\:red\:\:balls = 65 - 42 \\[3ex] Additional\:\:red\:\:balls = 23 \\[3ex] $ $23$ red marbles need to be added so that the probability of drawing a red marble is $\dfrac{3}{5}$

Because the options/answers are in percentages, we shall just work with percents

$ P(winning) + P(losing) = 100\% ...Complementary\:\: Rule \\[3ex] \underline{Saturday} \\[3ex] P(winning) = 60\% \\[3ex] P(losing) = 100\% - 60\% = 40\% \\[3ex] \underline{Sunday} \\[3ex] P(winning) = 35\% \\[3ex] P(losing) = 100\% - 35\% = 65\% \\[3ex] P(losing\:\: both\:\: games) = 40\% * 65\% ...Multiplication\:\: Rule \\[3ex] = \dfrac{40}{100} * \dfrac{65}{100} \\[5ex] = \dfrac{40 * 65}{100 * 100} \\[5ex] = \dfrac{2600}{1000} \\[5ex] = 0.26 \\[3ex] = 26\% $

It is much better to represent this information in a table.
Because the question asked for the probability of the "sum", we shall be adding the elements.

$(+)$	$2$	$3$	$4$
$1$	$3$	$\color{red}{4}$	$\color{red}{5}$
$3$	$\color{red}{5}$	$\color{red}{6}$	$7$
$5$	$7$	$8$	$9$

Numbers in red are greater than $3$ and less than $7$
Let $x$ be those numbers

$ n(S) = 9 \\[3ex] n(3 \lt x \lt 7) = 4 \\[3ex] P(3 \lt x \lt 7) = \dfrac{n(3 \lt x \lt 7)}{n(S)} \\[5ex] P(3 \lt x \lt 7) = \dfrac{4}{9} $

$ Let: \\[3ex] red = R \\[3ex] green = G \\[3ex] yellow = Y \\[3ex] blue = B \\[3ex] n(R) = 3 \\[3ex] n(G) = 2 \\[3ex] n(Y) = 1 \\[3ex] n(B) = 4 \\[3ex] n(S) = 3 + 2 + 1 + 4 = 10 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] \underline{Without\:\:Replacement\:\:condition} \\[3ex] blue-green-blue...in\:\:that\:\:order \\[3ex] P(BGB) = \dfrac{4}{10} * \dfrac{2}{9} * \dfrac{3}{8} ...Option\:D \\[5ex] P(BGB) = \dfrac{1 * 1 * 1}{10 * 3 * 1} \\[5ex] P(BGB) = \dfrac{1}{30} $

$ Let\:\:Samuel = A \\[3ex] Dominic = D \\[3ex] Chukwuemeka = C \\[3ex] (a.)\:\:S = \{AD, AC, DC\} \\[3ex] n(S) = 3 \\[3ex] (b.)\:\:n(DC) = 1 \\[3ex] P(DC) = \dfrac{n(DC)}{n(S)} = \dfrac{1}{3} \\[5ex] $ If Chukwuemeka stays home, this means that Samuel and Dominic will attend

$ n(AD) = 1 \\[3ex] (c.)\:\:P(AD) = \dfrac{n(AD)}{n(S)} = \dfrac{1}{3} $

$ Let\:\: Member = M \\[3ex] Let\:\:Officer = F \\[3ex] n(S) = 16 \\[3ex] n(M) = 13 \\[3ex] n(F) = 3 \\[3ex] n(Adrian) = 1 \\[3ex] Adrian\:\: \epsilon \:\:M \\[3ex] P(Adrian) = \dfrac{n(Adrian)}{n(M)} \\[5ex] P(Adrian) = \dfrac{1}{13d} $

$ Let\:\: grape = G \\[3ex] n(G) = 2 \\[3ex] n(S) = 10 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} = \dfrac{2}{10} = \dfrac{1}{5} \\[5ex] P(G') = 1 - P(G) \\[3ex] ... Complementary\;\;Rule...Complementary\;\;Events \\[3ex] P(G') = 1 - \dfrac{1}{5} \\[5ex] = \dfrac{5}{5} - \dfrac{1}{5} \\[5ex] = \dfrac{5 - 1}{5} \\[5ex] = \dfrac{4}{5} $

$ Let \\[3ex] red\:\:jelly\:\:bean = R \\[3ex] green\:\:jelly\:\:bean = G \\[3ex] white\:\:jelly\:\:bean = W \\[3ex] S = \{4R, 5G, 3W\} \\[3ex] n(S) = 4 + 5 + 3 = 12 \\[3ex] n(G) = 5 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} \\[5ex] P(G) = \dfrac{5}{12} $

$ Let\:\: Member = M \\[3ex] Let\:\:Officer = F \\[3ex] n(S) = 13 \\[3ex] n(M) = 10 \\[3ex] n(F) = 3 \\[3ex] n(Samara) = 1 \\[3ex] Samara\:\: \epsilon \:\:M \\[3ex] P(Samara) = \dfrac{n(Samara)}{n(M)} \\[5ex] P(Samara) = \dfrac{1}{10} $

The number of times the die was thrown is the total frequency
The total frequency is the cardinality of the sample space

$ n(S) = 25 + 30 + x + 28 + 40 + 32 \\[3ex] n(S) = 155 + x \\[3ex] P(3) = 0.225 \\[3ex] P(3) = \dfrac{x}{155 + x} \\[5ex] \rightarrow \dfrac{x}{155 + x} = 0.225 \\[5ex] x = 0.225(155 + x) \\[3ex] x = 34.875 + 0.225x \\[3ex] x - 0.225x = 34.875 \\[3ex] 0.775x = 34.875 \\[3ex] x = \dfrac{34.875}{0.775} \\[5ex] x = 45 \\[3ex] n(S) = 155 + x \\[3ex] n(S) = 155 + 45 \\[3ex] (a.)\:\: n(S) = 200 \\[3ex] $ Let the event of getting an even number be $E$
Let the event of getting a prime number be $P$

$ E = \{2, 4, 6\} \\[3ex] n(E) = 3 \\[3ex] P = \{2, 3, 5\} \\[3ex] n(P) = 3 \\[3ex] E\:\:\:AND\:\:\:P = \{2\} \\[3ex] n(E\:\:\:AND\:\:\:P) = 1 \\[3ex] P(E\:\:\:OR\:\:\:P) = P(E) + P(P) - P(E\:\:\:AND\:\:\:P) ...Addition\:\:Rule \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{3}{45} \\[5ex] P(P) = \dfrac{n(P)}{n(S)} = \dfrac{3}{45} \\[5ex] P(E\:\:\:AND\:\:\:P) = \dfrac{n(E\:\:\:AND\:\:\:P)}{n(S)} = \dfrac{1}{45} \\[5ex] \rightarrow P(E\:\:\:OR\:\:\:P) = \dfrac{3}{45} + \dfrac{3}{45} - \dfrac{1}{45} \\[5ex] P(E\:\:\:OR\:\:\:P) = \dfrac{3 + 3 - 1}{45} = \dfrac{5}{45} = \dfrac{1}{9} \\[5ex] (b.)\:\: P(E\:\:\:OR\:\:\:P) = \dfrac{1}{9} $

$ n(S) = 24 \\[3ex] n(A's) = 8 \\[3ex] P(A's) = \dfrac{n(A's)}{n(S)} = \dfrac{8}{24} = \dfrac{1}{3} $

$ n(S) = 1890 \\[3ex] n(Math) = 232 + 442 + 59 = 733 \\[3ex] n(Art) = 1036 + 68 + 53 = 1157 \\[3ex] n(10^{th}) = 232 + 1036 = 1268 \\[3ex] n(11^{th}) = 442 + 68 = 510 \\[3ex] n(12^{th}) = 59 + 53 = 112 \\[3ex] (a.) \\[3ex] P(Math) = \dfrac{n(Math)}{n(S)} \\[5ex] P(Math) = \dfrac{733}{1890} \\[5ex] (b.) \\[3ex] n(11^{th}) = 510 \\[3ex] P(11^{th}) = \dfrac{n(11^{th})}{n(S)} \\[5ex] = \dfrac{510}{1890} \\[5ex] = \dfrac{17}{63} \\[5ex] (c.) \\[3ex] n(12^{th}) = 112 \\[3ex] P(12^{th}) = \dfrac{n(12^{th})}{n(S)} \\[5ex] = \dfrac{112}{1890} \\[5ex] n(11^{th}\:\:AND\:\:12^{th}) = 0 \\[3ex] P(11^{th}\:\:\:OR\:\:\:12^{th}) = P(11^{th}) + P(12^{th}) - P(11^{th}\:\:\:AND\:\:\:12^{th}) ... Addition\:\:Rule \\[3ex] P(11^{th}\:\:\:OR\:\:\:12^{th}) = \dfrac{510}{1890} + \dfrac{112}{1890} - 0 \\[5ex] = \dfrac{510 + 112}{1890} \\[5ex] = \dfrac{622}{1890} \\[5ex] = \dfrac{311}{945} \\[5ex] (d.) \\[3ex] n(11^{th}\:\:\:AND\:\:\:Math) = 442 \\[3ex] P(11^{th}\:\:\:AND\:\:\:Math) = \dfrac{n(11^{th}\:\:\:AND\:\:\:Math)}{n(S)} \\[5ex] = \dfrac{442}{1890} \\[5ex] = \dfrac{221}{945} \\[5ex] (e.) \\[3ex] P(11^{th}\:\:\:OR\:\:\:Math) = P(11^{th}) + P(Math) - P(11^{th}\:\:\:AND\:\:\:Math) ... Addition\:\:Rule \\[3ex] P(11^{th}\:\:\:OR\:\:\:Math) = \dfrac{510}{1890} + \dfrac{733}{1890} - \dfrac{442}{1890} \\[5ex] = \dfrac{510 + 733 - 442}{1890} \\[5ex] = \dfrac{267}{630} \\[5ex] = \dfrac{89}{210} \\[5ex] (f.) \\[3ex] P(Math\:\:from\:\:11^{th}) = \dfrac{n(11^{th}\:\:\:AND\:\:\:Math)}{n(11^{th})} \\[5ex] = \dfrac{442}{510} \\[5ex] = \dfrac{13}{15} \\[5ex] (g.) \\[3ex] n(12^{th}\:\:\:AND\:\:\:Math) = 59 \\[3ex] P(Math\:\:from\:\:12^{th}) = \dfrac{n(12^{th}\:\:\:AND\:\:\:Math)}{n(12^{th})} \\[5ex] = \dfrac{59}{112} \\[5ex] (h.) \\[3ex] n(10^{th}) = 1268 \\[3ex] n(10^{th}\:\:\:AND\:\:\:Math) = 232 \\[3ex] P(Math\:\:from\:\:10^{th}) = \dfrac{n(10^{th}\:\:\:AND\:\:\:Math)}{n(10^{th})} \\[5ex] = \dfrac{232}{1268} \\[5ex] = \dfrac{58}{317} \\[5ex] (i.) \\[3ex] P(Two\:\:Math\:\:from\:\:11^{th}) = P(Math\:\:from\:\:11^{th}) * P(Math\:\:from\:\:11^{th}) ... Multiplication\:\:Rule \\[5ex] P(Two\:\:Math\:\:from\:\:11^{th}) = \dfrac{13}{15} * \dfrac{13}{15} \\[5ex] = \dfrac{13 * 13}{15 * 15} \\[5ex] = \dfrac{169}{225} \\[5ex] (j.) \\[3ex] P(Two\:\:Math\:\:from\:\:10^{th}) = P(Math\:\:from\:\:10^{th}) * P(Math\:\:from\:\:10^{th}) ... Multiplication\:\:Rule \\[5ex] P(Two\:\:Math\:\:from\:\:10^{th}) = \dfrac{58}{317} * \dfrac{58}{317} \\[5ex] = \dfrac{58 * 58}{317 * 317} \\[5ex] = \dfrac{3364}{100489} \\[5ex] (k.) \\[3ex] P(12^{th}\:\:\:OR\:\:\:Math) = P(12^{th}) + P(Math) - P(12^{th}\:\:\:AND\:\:\:Math) ... Addition\:\:Rule \\[3ex] P(12^{th}\:\:\:OR\:\:\:Math) = \dfrac{112}{1890} + \dfrac{733}{1890} - \dfrac{59}{1890} \\[5ex] = \dfrac{112 + 733 - 59}{1890} \\[5ex] = \dfrac{786}{1890} \\[5ex] = \dfrac{131}{315} $

Let the event of that a randomly selected chip from this batch of $100$ is Type $I$ and manufactured by Company $B$ = $E$

$ n(S) = 100 \\[3ex] n(E) = 30 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{30}{100}...Option\:\:B \\[5ex] = \dfrac{3}{10} $

$ P(smile) = 12\% = \dfrac{12}{100} = \dfrac{3}{25} \\[5ex] P(smile') = 1 - P(smile) ... Complementary\:\:Rule \\[3ex] P(smile') = 1 - \dfrac{3}{25} = \dfrac{25}{25} - \dfrac{3}{25} = \dfrac{25 - 3}{25} = \dfrac{22}{25} \\[5ex] (a.)\:\:P(both\:\:smile) = \dfrac{3}{25} * \dfrac{3}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] = \dfrac{3 * 3}{25 * 25} = \dfrac{9}{625} \\[5ex] (b.)\:\:P(both\:\:do\:\:not\:\:smile) = \dfrac{22}{25} * \dfrac{22}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] = \dfrac{22 * 22}{25 * 25} = \dfrac{484}{625} \\[5ex] (c.)\:\:P(only\:\:one\:\:smiles) \\[3ex] = P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile\:\:OR\:\:1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) \\[3ex] = P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile) + P(1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) ...Addition\:\:Rule \\[3ex] P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile) = \dfrac{3}{25} * \dfrac{22}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[5ex] P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile) = \dfrac{3 * 22}{25 * 25} = \dfrac{66}{625} \\[5ex] P(1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) = \dfrac{22}{25} * \dfrac{3}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[5ex] P(1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) = \dfrac{22 * 3}{25 * 25} = \dfrac{66}{625} \\[5ex] = \dfrac{66}{625} + \dfrac{66}{625} \\[5ex] = \dfrac{66 + 66}{625} \\[5ex] = \dfrac{132}{625} \\[5ex] (d.)\:\:can\:\:be\:\:done\:\:in\:\:two\:\:ways \\[3ex] \underline{First\:\:Method - Less\:\:Work} \\[3ex] P(at\:\:least\:\:one\:\:smiles) = 1 - P(both\:\:do\:\:not\:\:smile) ... Complementary\:\:Rule - Complementary\:\: Events \\[3ex] = 1 - \dfrac{484}{625} \\[5ex] = \dfrac{625}{625} - \dfrac{484}{625} \\[5ex] = \dfrac{625 - 484}{625} \\[5ex] = \dfrac{141}{625} \\[5ex] \underline{Second\:\:Method - More\:\:Work} \\[3ex] P(at\:\:least\:\:one\:\:smiles) = P(only\:\:one\:\:smiles) + P(both\:\:smile) \\[3ex] = \dfrac{132}{625} + \dfrac{9}{625} \\[5ex] = \dfrac{132 + 9}{625} \\[5ex] = \dfrac{141}{625} $

$ Let\:\: Defective = D \\[3ex] Non-defective = ND \\[3ex] n(S) = 50 \\[3ex] n(D) = 11 \\[3ex] n(ND) = 50 - 11 = 39 \\[3ex] P(ND) = \dfrac{n(ND)}{n(S)} = \dfrac{39}{50} $

$ P(P\:\: wins) = P(P) = 0.75 \\[3ex] P(P\:\: loses) = (P') = 1 - 0.75 = 0.25 \\[3ex] P(Q\:\: wins) = P(Q) = 0.60 \\[3ex] P(Q\:\: loses) = P(Q') = 1 - 0.60 = 0.40 \\[3ex] (i)\:\: P(both\:\: win) \\[3ex] = P(PQ) \\[3ex] = P(P) * P(Q) \\[3ex] = (0.75)(0.6) \\[3ex] = 0.45 \\[5ex] (ii)\:\: P(none\:\:wins) = P(both\:\:lose) \\[3ex] = P(P'Q') \\[3ex] = P(P') * P(Q') \\[3ex] = (0.25)(0.4) \\[3ex] = 0.1 \\[5ex] (iii)\:\: We\:\:can\:\:solve\:\:in\:\:two\:\:ways \\[3ex] \underline{First\:\:Method:\:\: Quantitative\:\:Reasoning} \\[3ex] P(at\:\:least\:\:one\:\:wins) \\[3ex] = P(P\:\:wins\:\:AND\:\:Q\:\:loses) \:\:OR\:\: P(P\:\:loses\:\:AND\:\:Q\:\:wins) \:\:OR\:\: P(P\:\:wins\:\:AND\:\:Q\:\:wins) \\[3ex] = (0.75)(0.4) + (0.25)(0.6) + (0.75)(0.6) \\[3ex] = 0.3 + 0.15 + 0.45 \\[3ex] = 0.9 \\[3ex] \underline{Second\:\:Method:\:\: Complementary\:\: Events} \\[3ex] P(at\:\:least\:\:one\:\:wins) = 1 - P(both\:\: lose) ... Complementary\:\: Rule \\[3ex] = 1 - 0.1 \\[3ex] = 0.9 $

$ S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex] n(S) = 10 \\[3ex] E = \{3, 10\} \\[3ex] n(E) = 2 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{2}{10} = \dfrac{1}{5} $

Recall:
In testing for marijuana and for most applicable tests:

A False Positive means that the applicant did not use it but tested positive for it.

A False Negative means that the applicant used it but tested negative for it.

A True Positive means that the applicant used it and tested positive for it.

A True Negative means that the applicant did not use it and tested negative for it.

$ Let\:\: Positive = P \\[3ex] True\:\:Positive = TP \\[3ex] False\:\:Positive = FP \\[3ex] Negative = N \\[3ex] True\:\:Negative = TN \\[3ex] False\:\:Negative = FN \\[3ex] n(P) = 146 \\[3ex] n(FP) = 26 \\[3ex] n(TP) = 146 - 26 = 120 \\[3ex] n(N) = 160 \\[3ex] n(FN) = 3 \\[3ex] n(TN) = 160 - 3 = 157 \\[3ex] $

(a.) Marijuana Use

	True	False
Positive Result	$120$	$26$
Negative Result	$157$	$3$

$ (b.)\:\:n(S) = n(P) + n(N) \\[3ex] = 146 + 160 \\[3ex] = 306\:\:applicants \\[3ex] (c.)\:\:n(No\:\:Marijuana) = n(TN) + n(FP) \\[3ex] = 157 + 26 \\[3ex] = 183\:\:applicants \\[3ex] (d.)\:\:n(Yes\:\:Marijuana) = n(TP) + n(FN) \\[3ex] = 120 + 3 \\[3ex] = 123\:\:applicants \\[3ex] (e.)\:\:P(No\:\:Marijuana) = \dfrac{n(No\:\:Marijuana)}{n(S)} \\[5ex] = \dfrac{183}{306} \\[5ex] = \dfrac{61}{102} \\[5ex] (f.)\:\:P(Yes\:\:Marijuana) = \dfrac{n(Yes\:\:Marijuana)}{n(S)} \\[5ex] = \dfrac{123}{306} \\[5ex] = \dfrac{41}{102} \\[5ex] (g.)\:\:P(N \cup NM) = P(N) + P(NM) - P(N \cap NM) ... Addition\:\:Rule \\[3ex] P(N) = \dfrac{n(N)}{n(S)} = \dfrac{160}{306} \\[5ex] P(NM) = \dfrac{n(NM)}{n(S)} = \dfrac{183}{306} \\[5ex] P(N \cap NM) = \dfrac{n(N \cap NM)}{n(S)} = \dfrac{n(TN)}{n(S)} = \dfrac{157}{306} \\[5ex] \therefore P(N \cup NM) = \dfrac{160}{306} + \dfrac{183}{306} - \dfrac{157}{306} \\[5ex] = \dfrac{160 + 183 - 157}{306} \\[5ex] = \dfrac{186}{306} \\[5ex] = \dfrac{31}{51} $

Let the event of finding a citizen who gets at least six hours of sleep a night = $E$

$ n(S) = 11 + 31 + 74 + 89 + 81 + 10 + 4 = 300\:\:milion \\[3ex] At\:\:least\:\:6\:\:means\:\: \gt 6 \\[3ex] n(E) = 74 + 89 + 81 + 10 + 4 = 258\:\: million \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{258}{300} = \dfrac{86}{100} = \dfrac{43}{50} $

We can solve this in at least two ways
Choose whatever approach you prefer

$ Let: \\[3ex] red = R \\[3ex] blue = B \\[3ex] white = W \\[3ex] S = \{8R, 6B, 6W\} \\[3ex] n(S) = 8 + 6 + 6 = 20 \\[5ex] \underline{First\:\: Method: Quantitative\:\: Reasoning} \\[3ex] NOT\:\:White \implies R\:\:AND\:\:B \\[3ex] n(R) = 8 \\[3ex] n(B) = 6 \\[3ex] n(R\:\:AND\:\:B) = 8 + 6 = 14 \\[3ex] P(NOT\:\:White) = P(R\:\:AND\:\:B) = \dfrac{n(R\:\:AND\:\:B)}{n(S)} \\[5ex] P(NOT\:\:White) = \dfrac{14}{20} \\[5ex] P(NOT\:\:White) = \dfrac{7}{10} \\[5ex] \underline{Second\:\: Method: Complementary\:\:Rule} \\[3ex] NOT\:\:white = W' \\[3ex] S = \{8R, 6B, 6W\} \\[3ex] n(S) = 8 + 6 + 6 = 20 \\[3ex] P(White) + P(NOT\:\:White) = 1 ... Complementary\:\: Rule \\[3ex] P(W) + P(W') = 1 \\[3ex] n(W) = 6 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} = \dfrac{6}{20} \\[5ex] P(W') = 1 - P(W) \\[3ex] P(W') = 1 - \dfrac{6}{20} \\[5ex] P(W') = \dfrac{20}{20} - \dfrac{6}{20} \\[5ex] P(W') = \dfrac{20 - 6}{20} \\[5ex] P(W') = \dfrac{14}{20} \\[5ex] P(W') = \dfrac{7}{10} $

$ n(S) = 10 + 8 + 6 + 4 + 2 = 30 \\[3ex] Let\:\: Black = B \\[3ex] n(B) = 6 \\[3ex] Green = G \\[3ex] n(G) = 2 \\[3ex] \underline{Without \:\: Replacement \implies Dependent\:\: Events} \\[3ex] P(B \:\:AND\:\: B) = P(BG) \\[3ex] P(BG) = \dfrac{6}{30} * \dfrac{2}{29}...Multiplication\:\: Rule \\[5ex] P(BG) = \dfrac{1}{5} * \dfrac{2}{29} \\[5ex] P(BG) = \dfrac{2}{145} $

$ n(S) = 17 \\[3ex] n(15\:\: rebounds) = 3 \\[3ex] P(15\:\: rebounds) = \dfrac{n(15\:\: rebounds)}{n(S)} = \dfrac{3}{17} $

$ Let\:\: red = R \\[3ex] yellow = Y \\[3ex] purple = P \\[3ex] n(S) = 100 \\[3ex] n(P) = 35 \\[3ex] P(P) = \dfrac{n(P)}{n(S)} \\[5ex] P(P) = \dfrac{35}{100} \\[5ex] P(P) = \dfrac{7}{20} $

Let us analyze each of these options.
F. At least $1$ marble is red.
At least $1$ means "$1$ or more", $\ge 1$
This is a correct option because it is possible that there are one or more red marbles in the bag.
But, let us look at the other options before making a conclusion.

G. Exactly $1$ marble is red.
We are not really sure of this option.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than only $1$ is red.
At least $1$ means $1$ or more
That also includes $1$

H. Exactly $3$ marbles are red.
We are not sure of this option either.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than only $3$ are red.
At least $1$ means $1$ or more
That also includes $3$

J. All the marbles are red.
We are not sure of this option as well.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than $all$ are red.
At least $1$ means $1$ or more
That also includes $all\:\: the\:\: marbles$

K. The bag contains more red marbles that marbles of other colors.
We are not sure of this option.
What if the bag contained only red marbles?
What if the bag contained more green marbles than red marbles and one just picked only a red marble each time because of the "with replacement" condition?
So, it is not okay to go with this option given the nature of the question.

The correct option is F. At least $1$ marble is red.

$ 1:\;\; YES \\[3ex] 1.37:\;\; NO \\[3ex] 0.26:\;\; YES \\[3ex] -0.52:\;\; NO \\[3ex] 0.05:\;\; YES \\[3ex] 0:\;\; YES \\[3ex] \dfrac{25}{12} = 2.08333:\;\; NO \\[5ex] 120\% = \dfrac{120}{100} = 1.2:\;\; NO \\[5ex] $ The probabilities of events could be: $0, 0.05, 0.26, 1$
The probability of an impossible event is $0$
The probability of an event that must occur (a surely certainty event) is $1$
The probability of any event is from $0$ through $1$

$ \Sigma Probability = 1.05 \\[3ex] \Sigma Probability \ne 1 \\[3ex] $ Therefore, it does not represent a probability model.
The event of selecting a brown color is an impossible event.

$ n(S) = 5 \\[3ex] n(correct\:\: answer) = 1 \\[3ex] n(incorrect\:\: answers) = 5 - 1 = 4 \\[3ex] P(correct\:\: answer) = \dfrac{n(correct\:\: answer)}{n(S)} = \dfrac{1}{5} \\[5ex] P(incorrect\:\: answers) = \dfrac{n(incorrect\:\: answers)}{n(S)} = \dfrac{4}{5} $

The die is not loaded (it is a fair die) because each value has an approximately equal chance of occurrence.
The frequency of the values are very close to each other.

$ Let \\[3ex] blue\:\:marble = B \\[3ex] red\:\:marble = R \\[3ex] green\:\:marble = G \\[3ex] S = \{5B, 4R, 3G\} \\[3ex] n(S) = 5 + 4 + 3 = 12 \\[3ex] n(R) = 4 \\[3ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{4}{12} \\[5ex] P(R) = \dfrac{1}{3} $

The die is loaded (it is a loaded die) because each value does not have an approximately equal chance of occurrence.
The frequency of the values of $2$ and $4$ are very high.

$ MATRICULATION \\[3ex] n(S) = 13 \\[3ex] Vowels = \{A,I,U,A,I,O\} \\[3ex] n(Vowels) = 6 \\[3ex] P(Vowels) = \dfrac{n(Vowels)}{n(S)} \\[5ex] P(Vowels) = \dfrac{6}{13} $

Based on the fact that they gave the conditional statement: If the probability that he passes at least one of the subjects is $\dfrac{7}{12}$
We have to use it.
At least one means one or more.
In this case, "at least one" means "one or both"
The probability that he passes at least one of the subjects means that:
Either he passes Mathematics AND does not pass Economics OR
He passes Economics AND does not pass Mathematics OR
He passes both subjects...he passes both Mathematics and Economics

$ Let: \\[3ex] pass\:\:Mathematics = M \\[3ex] P(M) = \dfrac{3}{4} \\[5ex] does\:\:not\:\:pass\:\:Mathematics = M' \\[3ex] P(M') = 1 - \dfrac{3}{4}...Complementary\:\:Rule \\[3ex] = \dfrac{4}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{4 - 3}{4} \\[5ex] = \dfrac{1}{4} \\[5ex] pass\:\:Economics = E \\[3ex] P(E) = \dfrac{5}{8} \\[5ex] does\:\:not\:\:pass\:\:Economics = E' \\[3ex] P(E') = 1 - \dfrac{5}{8}...Complementary\:\:Rule \\[5ex] = \dfrac{8}{8} - \dfrac{5}{8} \\[5ex] = \dfrac{8 - 5}{8} \\[5ex] = \dfrac{3}{8} \\[5ex] pass\:\:both = ME...what\:\:we\:\:need\:\:to\:\:find \\[3ex] P(ME') = P(M) * P(E')...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] = \dfrac{3}{4} * \dfrac{3}{8} \\[5ex] = \dfrac{3 * 3}{4 * 8} \\[5ex] = \dfrac{9}{32} \\[5ex] P(EM') = P(E) * P(M')...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] = \dfrac{5}{8} * \dfrac{1}{4} \\[5ex] = \dfrac{5 * 1}{8 * 4} \\[5ex] = \dfrac{5}{32} \\[5ex] P(at\:\:least\:\:one\:\:of\:\:the\:\:subjects) = \dfrac{7}{12} \\[5ex] P(at\:\:least\:\:one\:\:of\:\:the\:\:subjects) = P(ME') + P(EM') + P(ME) \\[3ex] \implies \dfrac{7}{12} = \dfrac{9}{32} + \dfrac{5}{32} + P(ME) \\[5ex] \dfrac{9}{32} + \dfrac{5}{32} + P(ME) = \dfrac{7}{12} \\[5ex] \dfrac{9 + 5}{32} + P(ME) = \dfrac{7}{12} \\[5ex] \dfrac{14}{32} + P(ME) = \dfrac{7}{12} \\[5ex] \dfrac{7}{16} + P(ME) = \dfrac{7}{12} \\[5ex] P(ME) = \dfrac{7}{12} - \dfrac{7}{16} \\[5ex] = \dfrac{28}{48} - \dfrac{21}{48} \\[5ex] = \dfrac{28 - 21}{48} \\[5ex] = \dfrac{7}{48} \\[5ex] $ The probability that he passes both subjects is $\dfrac{7}{48}$

We can solve this question in at least two ways.
It is highly recommended that you use the Second Method...using the Complementary Rule

At least one means one or more
This means that out of the 4 randomly selected CDS,
Reject the entire batch if at least one of the CDs is defective.
This means that Reject the entire batch if:
the 1st is defective AND the 2nd is not AND the 3rd is not AND the 4th is not
OR
the 1st is defective AND the 2nd is defective AND the 3rd is not AND the 4th is not
OR
the 1st is defective AND the 2nd is defective AND the 3rd is defective AND the 4th is not
OR
All four are defective
OR
the 1st is not AND the 2nd is defective AND the 3rd is not AND the 4th is not
OR
the 1st is not AND the 2nd is not AND the 3rd is defective AND the 4th is not
OR...and the possibilities keep going...
It is a Herculean task to use this method.

So, we shall use the Complementary Rule
(At least one is defective) is 1 minus (none are defective)

$ Let \\[3ex] Defective\:\:CD = D \\[3ex] Non-defective\:\:CD = D' \\[3ex] n(S) = 5000 \\[3ex] n(D) = 180 \\[3ex] n(D') = 5000 - 180 = 4820 \\[3ex] \underline{2nd\:\:Method:\:\:Complementary\:\:Rule} \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) \\[3ex] = 1 - P(none\:\:are\:\:defective) \\[3ex] = 1 - P(D'D'D'D') \\[3ex] P(D'D'D'D') = \dfrac{4820}{5000} * \dfrac{4820}{5000} * \dfrac{4820} * \dfrac{4820}{5000} \\[5ex] P(D'D'D'D') = \left(\dfrac{4820}{5000}\right)^4 \\[5ex] P(D'D'D'D') = 0.964^4 \\[3ex] P(D'D'D'D') = 0.8635910556 \\[3ex] \implies P(at\:\:least\:\:one\:\:is\:\:defective) = 1 - 0.8635910556 \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) = 0.1364089444 \\[3ex] converting\:\:to\:\:percent \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) = 0.1364089444 * 100 \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) = 13.64089444\% $

$ (a.) \\[3ex] Let \\[3ex] Window\:\:seat = W \\[3ex] Middle\:\:seat = M \\[3ex] Aisle\:\:seat = A \\[3ex] n(W) = 501 \\[3ex] n(M) = 10 \\[3ex] n(A) = 297 \\[3ex] n(S) = 501 + 10 + 297 = 808 \\[3ex] P(M) = \dfrac{n(M)}{n(S)} = \dfrac{10}{808} = 0.0123762376 \\[5ex] P(M) \approx 0.012 \\[3ex] (b.) \\[3ex] $ Rare Event Rule
Yes, it is unlikely for a passenger to prefer the middle seat because the probability that a passenger prefers the middle seat, $0.012$ is less than $0.05$
This may be due to the fact the middle seat lacks easy access to the aisle and the outside view among others.

A number is divisible by 4 if it is an even number AND if the last two digits are divisible by 4
Between 1 and 300, there are 75 numbers that are divisible by 4

Student: Samdom For Peace, how did you know?
Why not list them and count them?
Teacher: Good question.
This is a JAMB question.
You should solve this question accurately on time without a calculator
Between $1$ and $12$, list the numbers that are divisible by $4$
Student: The numbers are: $4, 8, 12$
Teacher: How many are there?
Student: Three numbers, Sir
Teacher: Correct!
Between $1$ and $20$, list the numbers that are divisible by $4$
Student: The numbers are: $4, 8, 12, 16, 20$
Teacher: How many are there?
Student: Five numbers, Sir
Teacher: Correct!
Student: Okay, I understand

$ 12 \div 4 = 3 \\[3ex] 20 \div 4 = 5 \\[3ex] 300 \div 4 = 75 \\[3ex] $ Let the set of numbers that are divisible by $4 = E$

$ n(E) = 75 \\[3ex] n(S) = 300 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{75}{300} \\[5ex] P(E) = \dfrac{1}{4} $

$ OR \implies \cup \\[3ex] AND \implies \cap \\[3ex] S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex] n(S) = 10 \\[3ex] Let\:\:A\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:prime \\[3ex] A = \{2, 3, 5, 7\} \\[3ex] n(A) = 4 \\[3ex] P(A) = \dfrac{n(A)}{n(S)} = \dfrac{4}{10} \\[5ex] Let\:\:B\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:multiple\:\:of\:\:3 \\[3ex] B = \{3, 6, 9\} \\[3ex] n(B) = 3 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} = \dfrac{3}{10} \\[5ex] A \cap B = \{3\} \\[3ex] n(A \cap B) = 1 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(S)} = \dfrac{1}{10} \\[5ex] P(A \cup B) = P(A) + P(B) - P(A \cap B)...Addition\:\:Rule \\[3ex] P(A \cup B) = \dfrac{4}{10} + \dfrac{3}{10} - \dfrac{1}{10} \\[5ex] P(A \cup B) = \dfrac{4 + 3 - 1}{10} \\[5ex] P(A \cup B) = \dfrac{6}{10} \\[5ex] P(A \cup B) = \dfrac{3}{5} \\[5ex] $ Between the numbers $1$ and $10$ inclusive, there is a $\dfrac{3}{5}$ probability of picking a prime number or a multiple of $3$

We can solve this question in two ways.
Use any method you prefer

Let the number of green marbles to put in the bag = $x$

$ Let\:\: red = R \\[3ex] n(R) = 3 \\[3ex] green = G \\[3ex] \underline{First\:\:Method} \\[3ex] n(G) = x \\[3ex] n(S) = 3 + x \\[3ex] P(R) = \dfrac{n(R)}{n(S)} = \dfrac{3}{3 + x} \\[5ex] Also\:\: P(R) = \dfrac{1}{4}...from\:\:the\:\:question \\[5ex] \rightarrow \dfrac{3}{3 + x} = \dfrac{1}{4} \\[5ex] Cross\:\:Multiply \\[3ex] 1(3 + x) = 3(4) \\[3ex] 3 + x = 12 \\[3ex] x = 12 - 3 \\[3ex] x = 9 \\[3ex] \underline{Second\:\:Method} \\[3ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{1}{4} \\[5ex] \rightarrow \dfrac{3}{n(S)} = \dfrac{1}{4} \\[5ex] Cross\:\:Multiply \\[3ex] n(S) * 1 = 3(4) \\[3ex] n(S) = 12 \\[3ex] n(G) = 12 - 3 \\[3ex] n(G) = 9 \\[3ex] $ Martin should put $9$ green marbles in the bag.

$ \underline{Check} \\[3ex] 3 + x = 3 + 9 = 12 \\[3ex] P(R) = \dfrac{3}{3 + x} = \dfrac{3}{12} = \dfrac{1}{4} $

$ Let:\:\:Class\:\:I\:\:milk = M \\[3ex] \implies NOT\:\:Class\:\:I\:\:milk = M' \\[3ex] P(M) = 88\% = \dfrac{88}{100} = 0.88 \\[5ex] P(M) + P(M') = 1 ...Complementary\:\:Events \\[3ex] P(M') = 1 - P(M) \\[3ex] P(M') = 1 - 0.88 \\[3ex] P(M') = 0.12 \\[3ex] (a.) \\[3ex] P(MM) = (0.88)(0.88) ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(MM) = 0.7744 \\[3ex] (b.) \\[3ex] P(M'M'M'M') = (0.12)(0.12)(0.12)(0.12) = 0.12^4 ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(M'M'M'M') = 0.00020736 \\[3ex] $ (a.) The probability that two containers of milk selected at random contain Class $I$ milk is $0.7744$
(b.) The probability that four containers of milk selected at random DO NOT contain Class $I$ milk is $0.00020736$

$ Sample\:\:Space \\[3ex] 1 + 3 = 4 \\[3ex] 1 + 6 = 7 \\[3ex] 3 + 6 = 9 \\[3ex] S = \{4, 7, 9\} \\[3ex] n(S) = 3 \\[3ex] NOT\:\:Odd = \{4\} \\[3ex] n(NOT\:\:Odd) = 1 \\[3ex] P(NOT\:\:Odd) = \dfrac{n(NOT\:\:Odd)}{n(S)} \\[5ex] P(NOT\:\:Odd) = \dfrac{1}{3} $

This is a case of Conditional Probability
Let us reword the question:
What is the probability of selecting a senior given that a junior was selected

$ n(juniors) = 8 \\[3ex] n(seniors) = 4 \\[3ex] n(S) = 8 + 4 = 12 \\[3ex] P(junior) = \dfrac{n(juniors)}{n(S)} = \dfrac{8}{12} = \dfrac{2}{3} \\[5ex] P(senior) = \dfrac{n(seniors)}{n(S)} = \dfrac{4}{12} = \dfrac{1}{3} \\[5ex] P(senior | junior) = \dfrac{P(senior\:\:AND\:\:junior)}{P(junior)} \\[5ex] P(senior\:\:AND\:\:junior) = P(senior) * P(junior)...Multiplication\:\:Rule\:\:for\:\:Dependent\:\:Events \\[3ex] = \dfrac{4}{12} * \dfrac{8}{11}...Without\:\:Replacement \\[5ex] = \dfrac{1}{3} * \dfrac{8}{11} \\[5ex] = \dfrac{8}{33} \\[5ex] \implies P(senior | junior) = P(senior\:\:AND\:\:junior) \div P(junior) \\[3ex] = \dfrac{8}{33} \div \dfrac{2}{3} \\[5ex] = \dfrac{8}{33} * \dfrac{3}{2} \\[5ex] = \dfrac{4}{11} $

$ Let: \\[3ex] five-dollar\:\:bill = F \\[3ex] n(F) = 2 \\[3ex] ten-dollar\:\:bill = T \\[3ex] n(T) = 9 \\[3ex] twenty-dollar\:\:bill = W \\[3ex] n(W) = 5 \\[3ex] S = \{2F, 9T, 5W\} \\[3ex] n(S) = 2 + 9 + 5 = 16 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] P(W) = \dfrac{5}{16} $

This is a case of false negative
Suspect $A$ actually murdered her daughter.
However, because the prosecution team could not provide substantial evidence that she murdered her daughter, the jury acquitted her.

NOTE: This is similar to Type $II$ error in Statistics (Errors in Hypothesis Tests): Acquitting a guilty person

This is a case of false positive
This means that he did not smoke cannabis but tested positive for it.
Augustine did not actually smoke cannabis but tested positive for it because he probably inhaled it from his friend.

This is a case of false positive
This means that he did not commit the crime but was jailed for it.

NOTE: This is similar to Type $I$ error in Statistics (Errors in Hypothesis Tests): Convicting an innocent person

This is a case of false negative
Felicty was pregnant.
However, she probably did not have a high enough blood and urine pregnancy hormone (HCG - Human Chorionic Gonadotropin) level to get a positive pregnancy test.

It is best to represent this as a table.
But, please keep in mind that this is without replacement
So, we will need to delete the sum of the repeated numbers such as:

$ (1, 1 = 1 + 1 = 2), \\[3ex] (2, 2 = 2 + 2 = 4), \\[3ex] (3, 3 = 3 + 3 = 6), \\[3ex] (4, 4 = 4 + 4 = 8), \\[3ex] (5, 5 = 5 + 5 = 10) \\[3ex] $ The highlighted numbers in the table are omitted ...because the two balls are drawn without replacement.
The numbers in red color (the $7's$) are the numbers in the event space.

$(+)$	$1$	$2$	$3$	$4$	$5$
$1$	$2$	$3$	$4$	$5$	$6$
$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$
$3$	$4$	$5$	$6$	$\color{red}{7}$	$8$
$4$	$5$	$6$	$\color{red}{7}$	$8$	$9$
$5$	$6$	$\color{red}{7}$	$8$	$9$	$10$

$ n(S) = 25 - 5 = 20 \\[3ex] n(7) = 4 \\[3ex] P(7) = \dfrac{n(7)}{n(S)} \\[5ex] P(7) = \dfrac{4}{20} \\[5ex] P(7) = \dfrac{1}{5} $

$ Let: \\[3ex] gold = G \\[3ex] silver = L \\[3ex] bronze = B \\[3ex] \underline{First\;\;Approach} \\[3ex] n(G) = 30 \\[3ex] n(L) = 22 \\[3ex] n(B) = 18 \\[3ex] n(S) = 30 + 22 + 18 = 70 \\[3ex] n(G') = 70 - 30 = 40 \\[5ex] \underline{Second\;\;Approach} \\[3ex] n(G') = 22 + 18 = 40 \\[3ex] P(G') = \dfrac{n(G')}{n(S)} \\[5ex] P(G') = \dfrac{40}{70} \\[5ex] P(G') = \dfrac{4}{7} $

$ n(S) = \Sigma F = 4 + 6 + 8 + 7 = 25 \\[3ex] n(outfielder) = 7 \\[3ex] P(outfielder) = \dfrac{n(outfielder)}{n(S)} \\[5ex] P(outfielder) = \dfrac{7}{25} \\[5ex] n(pitcher) = 8 \\[3ex] P(pitcher) = \dfrac{n(pitcher)}{n(S)} \\[5ex] P(pitcher) = \dfrac{8}{25} \\[5ex] P(outfielder\:\:OR\:\:\:pitcher) = P(outfielder) + P(pitcher) - P(outfielder-pitcher)...Multiplication\:\:Rule \\[3ex] P(outfielder-pitcher) = 0 ...Mutually\:\:Exclusive\:\:Event \\[3ex] P(outfielder\:\:OR\:\:\:pitcher) = \dfrac{7}{25} + \dfrac{8}{25} \\[5ex] = \dfrac{7 + 8}{25} \\[5ex] = \dfrac{15}{25} \\[5ex] to\:\:percent = \dfrac{15}{25} * 100 \\[5ex] = 15(4) \\[3ex] = 60\% $

$ n(S) = 360^\circ...\angle s \;\;in\;\;a\;\;circle \\[3ex] n(S) = 90 + x + 2x + (2x + 10) \\[3ex] n(S) = 90 + 3x + 2x + 10 \\[3ex] n(S) = 5x + 100 \\[3ex] n(S) = n(S) \\[3ex] \implies 5x + 100 = 360 \\[3ex] 5x = 360 - 100 \\[3ex] 5x = 260 \\[3ex] x = \dfrac{260}{5} \\[5ex] x = 52^\circ \\[3ex] n(C) = 2x + 10 \\[3ex] n(C) = 2(52) + 10 \\[3ex] n(C) = 104 + 10 \\[3ex] n(C) = 114^\circ \\[3ex] P(C) = \dfrac{n(C)}{n(S)} \\[5ex] P(C) = \dfrac{114}{360} \\[5ex] P(C) = \dfrac{19}{60} $

At least $60$ inches means $60$ inches or more

Let $E$ be the event that a randomly chosen student jumped at least $60$ inches

$ E = \{60, 61, 62, 63, 66, 70, 71, 72\} \\[3ex] n(E) = 8 \\[3ex] OR \\[3ex] n(E) = 5 + 3 = 8 \\[3ex] n(S) = 2 + 4 + 5 + 5 + 3 = 19 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{8}{19} $

The person chosen will NOT be a high school student and will NOT have replied with no opinion means that the person is a college student who approved or disapproved OR the person is a nonstudent who approved or disapproved. We shall add all of them.
Let $E$ be the event of selecting such a person.
This includes:
All college students who approved = $14$
All college students who disapproved = $10$
All nonstudents who approved = $85$
All nonstudents who disapproved = $353$

$ n(E) = 14 + 10 + 85 + 353 = 462 \\[3ex] n(S) = 560 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{462}{560} \\[5ex] P(E) = 0.825 \\[3ex] P(E) \approx 0.83 $

$ Total\:\:marks = 60 \\[3ex] Half\:\:of\:\:it = \dfrac{60}{2} = 30 \\[5ex] n(students) = 10 \\[3ex] n(students\:\:who\:\:scored\:\:less\:\:than\:\:30) = 2...the\:\:students\:\:who\:\:scored\:\:29\:\:and\:\:26 \\[3ex] P(students\:\:who\:\:scored\:\:less\:\:than\:\:30) = \dfrac{n(students\:\:who\:\:scored\:\:less\:\:than\:\:30)}{n(students)} \\[5ex] = \dfrac{2}{10} \\[5ex] = \dfrac{1}{5} $

Of the two balls chosen, one is red and the other is blue

This means it could be in any order
So, it means that the:
first ball is red AND the second ball is blue OR the
first ball is blue AND the second ball is red

$ Let: \\[3ex] yellow = Y \\[3ex] red = R \\[3ex] blue = B \\[3ex] n(R) = 3 \\[3ex] n(B) = 2 \\[3ex] n(S) = 7 + 3 + 2 = 12 \\[3ex] \underline{Without\:\:Replacement\:\:condition} \\[3ex] \underline{Multiplication\:\:Rule\:\:for\:\:Dependent\:\:Events} \\[3ex] red-blue \\[3ex] P(RB) = \dfrac{3}{12} * \dfrac{2}{11} \\[5ex] = \dfrac{1}{2} * \dfrac{1}{11} \\[5ex] = \dfrac{1 * 1}{2 * 11} \\[5ex] = \dfrac{1}{22} \\[5ex] blue-red \\[3ex] P(BR) = \dfrac{2}{12} * \dfrac{3}{11} \\[5ex] = \dfrac{1}{22} \\[5ex] red-blue\:\:OR\:\:blue-red \\[3ex] \underline{Addition\:\:Rule\:\:for\:\:Mutually-Exclusive\:\:Events} \\[3ex] P(RB\:\:OR\:\:BR) = \dfrac{1}{22} + \dfrac{1}{22} \\[5ex] = \dfrac{1 + 1}{22} \\[5ex] = \dfrac{2}{22} \\[5ex] = \dfrac{1}{11} $

$ \underline{Initial\:\:Count} \\[3ex] Let: \\[3ex] number\:\:of\:\:boys = p \\[3ex] number\:\:of\:\:girls = p + 5...(5\:\:more\:\:girls\:\:than\:\:boys) \\[3ex] \underline{Later\:\:Count...based\:\:on\:\:the\:\:condition,\:\: "if"} \\[3ex] number\:\:of\:\:boys = p + 2...(2\:\:more\:\:boys\:\:join) \\[3ex] number\:\:of\:\:girls = p + 5...(no\:\:change) \\[3ex] Ratio\:\:of\:\:girls:boys = 5:4 \\[3ex] \implies \dfrac{p + 5}{p + 2} = \dfrac{5}{4} \\[5ex] 4(p + 5) = 5(p + 2) \\[3ex] 4p + 20 = 5p + 10 \\[3ex] 20 - 10 = 5p - 4p \\[3ex] 10 = p \\[3ex] p = 10 \\[3ex] Number\:\:of\:\:boys = p = 10 \\[3ex] 10\:\:boys \\[3ex] $ Student: Excuse me Ma'am/Sir
I thought the boys would be $10 + 2 = 12$
Rather than $10$
Teacher: Yes, you have a point.
The initial number of boys is $p = 10$
However, the later count is based on a conditional statement, "if"
"If" $2$ boys join the class...then the ratio is ...
This does not imply that $2$ boys "actually" joined them
This was to assist us in determining the number of girls and boys in the class
So, we have to go by the Initial Count, rather than the "Conditional" Later Count.

$ Number\:\:of\:\:girls \\[3ex] = p + 5 \\[3ex] = 10 + 5 = 15 \\[3ex] 15\:\:girls \\[3ex] total\:\:number\:\:of\:\:pupils\:\:in\:\:the\:\:class \\[3ex] 10 + 15 = 25 \\[3ex] 15\:\:pupils \\[3ex] n(S) = 25 \\[3ex] Let\:\:E\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:boy\:\:as\:\:the\:\:class\:\:prefect \\[3ex] n(boys) = 10 \\[3ex] P(E) = \dfrac{n(boys)}{n(S)} \\[5ex] = \dfrac{10}{25} \\[5ex] = \dfrac{2}{5} \\[5ex] $ The probability of selecting a boy as the class prefect is $\dfrac{2}{5}$

For exactly one of them to pass the test means that either:
Mary passes AND Kofi does not pass OR
Mary does not pass AND Kofi passes

$ Let: \\[3ex] Mary\:\:passing\:\:the\:\:test = M \\[3ex] Kofi\:\:passing\:\:the\:\:test = K \\[3ex] P(M) = \dfrac{3}{4} \\[5ex] P(M') = 1 - \dfrac{3}{4} = \dfrac{4}{4} - \dfrac{3}{4} = \dfrac{1}{4} \\[5ex] P(K) = \dfrac{3}{5} \\[5ex] P(K') = 1 - \dfrac{3}{5} = \dfrac{5}{5} - \dfrac{3}{5} = \dfrac{2}{5} \\[5ex] P(exactly\:\:one\:\:passes) \\[3ex] = P(M) * P(K') + P(M') * P(K) \\[3ex] = \left(\dfrac{3}{4} * \dfrac{2}{5}\right) + \left(\dfrac{1}{4} * \dfrac{3}{5}\right) \\[5ex] = \left(\dfrac{3}{2} * \dfrac{1}{5}\right) + \dfrac{3}{20} \\[5ex] = \dfrac{3}{10} + \dfrac{3}{20} \\[5ex] LCD = 20 \\[3ex] = \dfrac{6}{20} + \dfrac{3}{20} \\[5ex] = \dfrac{6 + 3}{20} \\[5ex] = \dfrac{9}{20} $

$ P(pass) = \dfrac{2}{5} \\[5ex] P(fail) = 1 - P(pass) ... Complementary\:\:Rule \\[3ex] P(fail) = 1 - \dfrac{2}{5} = \dfrac{5}{5} - \dfrac{2}{5} = \dfrac{5 - 2}{5} = \dfrac{3}{5} \\[5ex] P(one\:\:passes\:\:AND\:\:one\:\:fails) \\[3ex] = P[(1st\:\:passes\:\:AND\:\:2nd\:\:fails)\:\:OR\:\:(1st\:\:fails\:\:AND\:\:2nd\:\:passes)] \\[3ex] = P(1st\:\:passes\:\:AND\:\:2nd\:\:fails) + P(1st\:\:fails\:\:AND\:\:2nd\:\:passes) ... Addition\:\:Rule \\[3ex] P(1st\:\:passes\:\:AND\:\:2nd\:\:fails) = \dfrac{2}{5} * \dfrac{3}{5} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] P(1st\:\:fails\:\:AND\:\:2nd\:\:passes) = \dfrac{3}{5} * \dfrac{2}{5} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] = \dfrac{2}{5} * \dfrac{3}{5} + \dfrac{3}{5} * \dfrac{2}{5} \\[5ex] = \dfrac{2 * 3}{5 * 5} + \dfrac{3 * 2}{5 * 5} \\[5ex] = \dfrac{6}{25} + \dfrac{6}{25} \\[5ex] = \dfrac{6 + 6}{25} \\[5ex] = \dfrac{12}{25} $

Used: Punnett Square
Sample Space for the Toss of a Fair Die and a Fair Coin

A Die in the Column and A Coin in the Row

$1\:\:Coin\:\:\rightarrow$ $1\:\:Die\:\:\downarrow$	$H$	$T$
$1$	$1H$	$1T$
$2$	$2H$	$2T$
$3$	$3H$	$3T$
$4$	$4H$	$4T$
$5$	$5H$	$5T$
$6$	$6H$	$6T$

$ S = \{ \\[3ex] 1H, 1T, \\[3ex] 2H, 2T, \\[3ex] 3H, 3T, \\[3ex] 4H, 4T, \\[3ex] 5H, 5T, \\[3ex] 6H, 6T \} \\[3ex] n(S) = 12 \\[5ex] (i.) \\[3ex] E = \{6H\} \\[3ex] n(E) = 1 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{1}{12} \\[5ex] (ii) \\[3ex] E =\{5H, 6H\} \\[3ex] n(E) = 2 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{2}{12} \\[5ex] P(E) = \dfrac{1}{6} \\[5ex] (iii) \\[3ex] E = \{2T, 4T, 6T\} \\[3ex] n(E) = 3 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{3}{12} \\[5ex] P(E) = \dfrac{1}{4} $

$ Let: \\[3ex] blue = B \\[3ex] n(B) = 5 \\[3ex] red = R \\[3ex] n(R) = 3 \\[3ex] white = W \\[3ex] n(W) = 2 \\[3ex] S = \{5B, 3R, 2W\} \\[3ex] n(S) = 5 + 3 + 2 = 10 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} \\[5ex] P(B) = \dfrac{5}{10} \\[5ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{3}{10} \\[5ex] (a.) \\[3ex] With\:\:Replacement\:\:condition \\[3ex] P(BB) = \dfrac{5}{10} * \dfrac{5}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(BB) = \dfrac{25}{100} \\[5ex] P(RR) = \dfrac{3}{10} * \dfrac{3}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(RR) = \dfrac{9}{100} \\[5ex] P(BB\:\:OR\:\:\:RR) = P(BB \cup RR) \\[3ex] P(BB \cup RR) = P(BB) + P(RR) - P(BB \cap RR) ...Addition\:\:Rule \\[3ex] P(BB \cap RR) = 0 ...Mutually\:\:Exclusive\:\:Events \\[3ex] \rightarrow P(BB \cup RR) = \dfrac{25}{100} + \dfrac{9}{100} \\[5ex] P(BB \cup RR) = \dfrac{25 + 9}{100} \\[5ex] P(BB \cup RR) = \dfrac{34}{100} \\[5ex] P(BB \cup RR) = \dfrac{17}{50} \\[5ex] (b) \\[3ex] 1\:\:Red\:\:AND\:\:1\:\:White \implies in\:\:any\:\:order \\[3ex] This\:\:means\:\:RW\:\:OR\:\:WR \\[3ex] P(RW) = \dfrac{3}{10} * \dfrac{2}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(RW) = \dfrac{6}{100} \\[5ex] P(WR) = \dfrac{2}{10} * \dfrac{3}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(WR) = \dfrac{6}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{6}{100} + \dfrac{6}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{6 + 6}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{12}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{3}{25} $

Because the bus can be early, on time, or late (3 conditions),
The probability that the bus is early plus the probability that the bus is on time plus the probability that the bus is late is $1$

$ P(early) + P(on\;\;time) + P(late) = 1 \\[3ex] 0.1 + 0.6 + P(late) = 1 \\[3ex] 0.7 + P(late) = 1 \\[3ex] P(late) = 1 - 0.7 \\[3ex] P(late) = 0.3 \\[3ex] $ The probability that the bus is late is $0.3$

Let the event of picking a blue jelly bean be $B$
and the event of picking a yellow jelly bean be $Y$

$ n(S) = 28 \\[3ex] n(B) = 8 \\[3ex] n(Y) = 4 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} = \dfrac{8}{28} \\[5ex] P(Y) = \dfrac{n(Y)}{n(S)} = \dfrac{4}{28} \\[5ex] P(B\:\:\:AND\:\:\:Y) = 0 ...Mutually\:\:Exclusive\:\:Event \\[3ex] P(B\:\:\:OR\:\:\:Y) = P(B) + P(Y) - P(B\:\:\:AND\:\:\:Y) \\[3ex] = \dfrac{8}{28} + \dfrac{4}{28} - 0 \\[5ex] = \dfrac{8 + 4}{28} \\[5ex] = \dfrac{12}{28} \\[5ex] P(B\:\:\:OR\:\:\:Y) = \dfrac{3}{7} $

Let $E$ be the event that the selected customer ordered a regular coffee without milk.

$ n(S) = 36 \\[3ex] n(E) = 10 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{10}{36} \\[5ex] = \dfrac{5}{18} $

NOTE:
A lattice point is a point with coordinates that are both integers
We are concerned with the lattice points only inside the rectangle

For the $x-coordinate$, the integers inside the rectangle are $0$ through $6$ (both excluded) = $1, 2, 3, 4, 5$

For the $y-coordinate$, the integers inside the rectangle are $0$ through $4$ (both excluded) = $1, 2, 3$

$y\downarrow\:\:(+)\:\:x\rightarrow$	$1$	$2$	$3$	$4$	$5$
$1$	$2$	$\color{red}{3}$	$4$	$\color{red}{5}$	$6$
$2$	$\color{red}{3}$	$4$	$\color{red}{5}$	$6$	$\color{red}{7}$
$3$	$4$	$\color{red}{5}$	$6$	$\color{red}{7}$	$8$

Numbers in red are odd numbers
Let $E$ be the event of selecting odd numbers

$ n(E) = 7 \\[3ex] n(S) = 3 * 5 = 15 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{7}{15} $

$ Let\:\: Member = M \\[3ex] Let\:\:Officer = F \\[3ex] n(S) = 35 \\[3ex] n(M) = 32 \\[3ex] n(F) = 3 \\[3ex] n(Hiroko) = 1 \\[3ex] Hiroko\:\: \epsilon \:\:M \\[3ex] P(Hiroko) = \dfrac{n(Hiroko)}{n(M)} \\[5ex] P(Hiroko) = \dfrac{1}{32} $

Let the:
Event that Veli will arrive late at school = $V$
This implies that the event that Veli will not arrive late at school = $V'$

Event that Bongi will arrive late at school = $B$
This implies that the event that Bongi will not arrive late at school = $B'$

Event that both Veli and Bongi will arrive late at school = $VB$
This implies that the event that neither Veli nor Bongi will arrive late at school = $B'V'$
This also means that the event that both Veli and Bongi will not arrive at school = $B'V'$

$ P(V'B') = 0.7 \\[3ex] (71.1.1) \\[3ex] P(at\:\:least\:\:one\:\:will) + P(none\:\:will) = 1...Complementary\:\:Rule \\[3ex] P(at\:\:least\:\:one\:\:will) + P(V'B') = 1 \\[3ex] P(at\:\:least\:\:one\:\:will) = 1 - P(V'B') \\[3ex] P(at\:\:least\:\:one\:\:will) = 1 - 0.7 \\[3ex] P(at\:\:least\:\:one\:\:will) = 0.3 \\[3ex] (71.1.2) \\[3ex] $

$ Let: \\[3ex] five-dollar\:\:bill = F \\[3ex] n(F) = 5 \\[3ex] ten-dollar\:\:bill = T \\[3ex] n(T) = 7 \\[3ex] twenty-dollar\:\:bill = W \\[3ex] n(W) = 8 \\[3ex] S = \{5F, 7T, 8W\} \\[3ex] n(S) = 5 + 7 + 8 = 20 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] P(W) = \dfrac{8}{20} \\[5ex] P(W) = \dfrac{2}{5} $

Used: Punnett Square
Sample Space for the Four Flips of a Fair Coin

$2$ flips in the Column and $2$ flips in the Row

$2\:\:Flips\:\:\rightarrow$ $2\:\:Flips\:\:\downarrow$	$HH$	$HT$	$TH$	$TT$
$HH$	$HHHH$	$HHHT$	$HHTH$	$HHTT$
$HT$	$HTHH$	$HTHT$	$HTTH$	$HTTT$
$TH$	$THHH$	$THHT$	$THTH$	$THTT$
$TT$	$TTHH$	$TTHT$	$TTTH$	$TTTT$

$ S = \{ \\[3ex] HHHH, HHHT, HHTH, HHTT, \\[3ex] HTHH, HTHT, HTTH, HTTT, \\[3ex] THHH, THHT, THTH, THTT, \\[3ex] TTHH, TTHT, TTTH, TTTT \\[3ex] \} \\[3ex] n(S) = 16 \\[5ex] (a.) \\[3ex] E = \{HHHH, HHTH, HTHH, HTTH\} \\[3ex] n(E) = 4 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{4}{16} \\[5ex] P(E) = \dfrac{1}{4} \\[5ex] $ For the second question:
At least two means two or more
At least two consecutive means: first and second or second and third or third and fourth or
first, second, and third or second, third, and fourth
At least two consecutive flips that come up heads means that:
first and second is $HH$ or second and third is $HH$ or or third and fourth is $HH$ or
first, second, and third is $HH$ or second, third, and fourth is $HH$

$ (b.) \\[3ex] E = \{HHHH, HHHT, HHTH, HHTT, HTHH, THHH, THHT, TTHH\} \\[3ex] n(E) = 8 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{8}{16} \\[5ex] P(E) = \dfrac{1}{2} \\[5ex] $ For the third question:
The second question must be satisfied in addition to the requirement that the first flip is a $T$
So, let us look at the event space in the answer to the second question and select only the events where the first flip results in a tail.

$ (c.) \\[3ex] E = \{THHH, THHT, TTHH\} \\[3ex] P(E) = \dfrac{3}{16} $

We can solve this question in two ways.
Use any method you prefer.

$ Let: \\[3ex] white = W \\[3ex] red = R \\[3ex] black = B \\[3ex] \underline{First\:\:Method} \\[3ex] n(W) = p \\[3ex] n(R) = 12 \\[3ex] n(B) = 16 \\[3ex] n(S) = p + 12 + 16 = p + 28 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] P(W) = \dfrac{p}{p + 28} \\[5ex] P(W) = \dfrac{3}{7} \\[5ex] \rightarrow \dfrac{p}{p + 28} = \dfrac{3}{7} \\[5ex] Cross\:\:multiply \\[3ex] 7p = 3(p + 28) \\[3ex] 7p = 3p + 84 \\[3ex] 7p - 3p = 84 \\[3ex] 4p = 84 \\[3ex] p = \dfrac{84}{4} \\[5ex] p = 21 \\[3ex] \underline{Second\:\:Method} \\[3ex] P(W) = \dfrac{3}{7} \\[5ex] P(W) + P(W') = 1 ...Complementary\:\:Rule \\[3ex] P(W') = 1 - P(W) \\[3ex] P(W') = 1 - \dfrac{3}{4} \\[5ex] P(W') = \dfrac{7}{7} - \dfrac{3}{7} \\[5ex] P(W') = \dfrac{7 - 3}{7} \\[5ex] P(W') = \dfrac{4}{7} \\[5ex] n(W') = n(R) + n(B) \\[3ex] n(W') = 12 + 16 = 28 \\[3ex] P(W') = \dfrac{n(W')}{n(S)} \\[5ex] n(S) * P(W') = n(W') \\[3ex] n(S) = \dfrac{n(W')}{P(W')} \\[5ex] n(S) = n(W') \div P(W') \\[3ex] n(S) = 28 \div \dfrac{4}{7} \\[5ex] n(S) = 28 * \dfrac{7}{4} \\[5ex] n(S) = 7(7) \\[3ex] n(S) = 49 \\[3ex] Additional\:\:white\:\:balls = 49 - 28 \\[3ex] Additional\:\:white\:\:balls = 21 \\[3ex] $ $21$ white balls were introduced in the basket.

(i)
There are two red $10$ in a standard deck of cards: the Diamond $10$ and the Heart $10$

$ n(S) = 52 \\[3ex] E = \{Diamond-10, Heart-10\} \\[3ex] n(E) = 2 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{2}{52} \\[5ex] P(E) = \dfrac{1}{26} \\[5ex] (ii) \\[3ex] Let: \\[3ex] Boy = B \\[3ex] Girl = G \\[3ex] P(B) = \dfrac{1}{2} \\[5ex] P(G) = \dfrac{1}{2}...will\;\;not\;\;be\;\;used \\[5ex] Because\;\; \\[3ex] 1st\;\;child\;\;is\;\;a\;\;boy\;\;AND\;\;2nd\;\;child\;\;is\;\;a\;\;boy \\[3ex] P(BB) = \dfrac{1}{2} * \dfrac{1}{2} ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[5ex] P(BB) = \dfrac{1 * 1}{2 * 2} \\[5ex] P(BB) = \dfrac{1}{4} \\[5ex] (iii) \\[3ex] $ Used: Punnett Square
Sample Space for a Toss of Two Fair Dice

A Die in the Column and A Die in the Row

$1\:\:Die\:\:\rightarrow$ ($+$) $1\:\:Die\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

$ n(S) = 36 \\[5ex] E = \{6, 6, 6, 6, 6\} \\[3ex] n(E) = 5 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{5}{36} $

$ Let: \\[3ex] red = R \\[3ex] green = G \\[3ex] blue = B \\[3ex] n(R) = 3 \\[3ex] n(G) = 5 \\[3ex] n(B) = x \\[3ex] n(S) = 3 + 5 + x \\[3ex] n(S) = 8 + x \\[3ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{3}{8 + x} \\[5ex] Also:\;\;P(R) = \dfrac{1}{6} \\[5ex] (a) \\[3ex] P(R) = P(R) \\[3ex] \implies \dfrac{3}{8 + x} = \dfrac{1}{6} \\[5ex] Cross\;\;Multiply \\[3ex] (8 + x) * 1 = 3 * 6 \\[3ex] 8 + x = 18 \\[3ex] x = 18 - 8 \\[3ex] x = 10 \\[3ex] $ There are ten blue balls in the sack.

$ (b) \\[3ex] n(S) = 8 + x \\[3ex] n(S) = 8 + 10 \\[3ex] n(S) = 18 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} \\[5ex] P(G) = \dfrac{5}{18} \\[5ex] $ The probability of picking a green ball is $\dfrac{5}{18}$

There are at least two options of solving this question.

One option that comes to mind:
Two fair coins tossed two times? (Not two fair coins tossed one time)...
Drawing the Punnett Square or Tree Diagram for this case would be cumbersome.
Yes, it is cumbersome.
So, why not do it this way?
The events are independent.
Look at the keyword in the question: AND
So, let us use the Multiplication Rule for Independent Events
We shall consider these events:
Event One: Two fair dice tossed one time (first throw): selecting a sum of seven
Event Two: Two fair dice tossed one time (second throw): selecting a sum of four
This option is recommended.

Used: Punnett Square
Sample Space for a Toss of Two Fair Dice (First Throw)

A Die in the Column and A Die in the Row

$1\:\:Die\:\:\rightarrow$ ($+$) $1\:\:Die\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

$ n(S) = 36 \\[5ex] E_1 = \{7, 7, 7, 7, 7, 7\} \\[3ex] n(E_1) = 6 \\[3ex] P(E_1) = \dfrac{n(E_1)}{n(S)} \\[5ex] P(E_1) = \dfrac{6}{36} \\[5ex] P(E_1) = \dfrac{1}{6} \\[5ex] \implies P(sum\;\;of\;\;7\;\;first\;\;throw) = \dfrac{1}{6} \\[5ex] $ Sample Space for a Toss of Two Fair Dice (Second Throw)

A Die in the Column and A Die in the Row

$1\:\:Die\:\:\rightarrow$ ($+$) $1\:\:Die\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$10$
$5$	$6$	$7$	$8$	$9$	$10$	$11$
$6$	$7$	$8$	$9$	$10$	$11$	$12$

$ n(S) = 36 \\[5ex] E_2 = \{4, 4, 4\} \\[3ex] n(E_2) = 3 \\[3ex] P(E_2) = \dfrac{n(E)}{n(S)} \\[5ex] P(E_2) = \dfrac{3}{36} \\[5ex] P(E_2) = \dfrac{1}{12} \\[5ex] \implies P(sum\;\;of\;\;4\;\;second\;\;throw) = \dfrac{1}{12} \\[5ex] therefore \;\; P(sum\;\;of\;\;7\;\;first\;\;throw) \;\;AND\;\; P(sum\;\;of\;\;4\;\;second\;\;throw) \\[5ex] = \dfrac{1}{6} * \dfrac{1}{12} ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[5ex] = \dfrac{1 * 1}{6 * 12} \\[5ex] = \dfrac{1}{72} $

If only one of them hits the target, this means that:
1st Case: $X$ hits it AND $Y$ misses it AND $Z$ misses it
OR
2nd Case: $X$ misses it AND $Y$ hits it AND $Z$ misses it
OR
3rd Case: $X$ misses it AND $Y$ misses it AND $Z$ hits it
So, we have these three cases.

$ P(X) = \dfrac{1}{3}...hits\;\;it \\[5ex] P(X') = 1 - \dfrac{1}{3} = \dfrac{2}{3}...Complementary\;\;Rule...misses\;\;it \\[5ex] P(Y) = \dfrac{1}{5}...hits\;\;it \\[5ex] P(Y') = 1 - \dfrac{1}{5} = \dfrac{4}{5}...Complementary\;\;Rule...misses\;\;it \\[5ex] P(Z) = \dfrac{1}{4}...hits\;\;it \\[5ex] P(Z') = 1 - \dfrac{1}{4} = \dfrac{3}{4}...Complementary\;\;Rule...misses\;\;it \\[5ex] 1st\;\;Case \\[3ex] P(X) * P(Y') * P(Z') \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{1}{3} * \dfrac{4}{5} * \dfrac{3}{4} \\[5ex] = \dfrac{1}{5} \\[5ex] 2nd\;\;Case \\[3ex] P(X') * P(Y) * P(Z') \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{2}{3} * \dfrac{1}{5} * \dfrac{3}{4} \\[5ex] = \dfrac{1}{5} * \dfrac{1}{2} \\[5ex] = \dfrac{1}{10} \\[5ex] 3rd\;\;Case \\[3ex] P(X') * P(Y') * P(Z) \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{2}{3} * \dfrac{4}{5} * \dfrac{1}{4} \\[5ex] = \dfrac{2}{3} * \dfrac{1}{5} \\[5ex] = \dfrac{2}{15} \\[5ex] P(only\;\;one\;\;hits\;\;the\;\;target) \\[3ex] = 1st\;\;Case + 2nd\;\;Case + 3rd\;\;Case \\[3ex] ... Addition\;\;Rule\;\;for\;\;Mutually\;\;Exclusive\;\;Events \\[3ex] Mutually\;\;exclusive\;\;because\;\;only\;\;one\;\;hits\;\;it \\[3ex] Only\;\;one\;\;event\;\;can\;\;occur\;\;at\;\;a\;\;time \\[3ex] Two\;\;or\;\;more\;\;cases\;\;cannot\;\;occur\;\;at\;\;the\;\;same\;\;time \\[3ex] = \dfrac{1}{5} + \dfrac{1}{10} + \dfrac{2}{15} \\[5ex] = \dfrac{6 + 3 + 4}{30} \\[5ex] = \dfrac{13}{30} \\[5ex] = 0.4333333333 \\[3ex] \approx 0.43...to\;\;two\;\;decimal\;\;places \\[3ex] $ The probability that only one of them hits the target is approximately $0.43$

(a.)
Used: Punnett Square
Sample Space for the First Four Prime Numbers and the First Four Square Numbers

$Square\;\;numbers\:\:\rightarrow$ ($+$) $Prime\;\;numbers\:\:\downarrow$	$1$	$4$	$9$	$16$
$2$	$3$	$6$	$11$	$18$
$3$	$4$	$7$	$12$	$19$
$5$	$6$	$9$	$14$	$21$
$7$	$8$	$11$	$16$	$23$

$ n(S) = 4 * 4 = 16 \\[5ex] (b.) \\[3ex] Let\;\;E\;\;be\;\;the\;\;event\;\;the\;\;score\;\;a\;\;prime\;\;number \\[3ex] E = \{3, 7, 11, 11, 19, 23\} \\[3ex] n(E) = 6 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{6}{16} \\[5ex] P(E) = \dfrac{3}{8} \\[5ex] $ The probability that the score is a prime number is $\dfrac{3}{8}$

Some will argue that this is an incomplete question
However, we need to solve this because it is a question that was asked.
Besides, none of the options is: Insufficient information to solve the question.

In analyzing this question:
The entire possibilities are:
1st Case: Team $P$ wins AND Team $Q$ loses OR
2nd Case: Team $P$ loses AND Team $Q$ wins OR
3rd Case: Team $P$ and Team $Q$ game ends in a draw

Team $P$ and Team $Q$ has an equal probability of winning the game.
Let $P$ be the event that Team $P$ wins the game.
$P'$ is the event that Team $P$ loses.
Let $Q$ be the event that Team $Q$ wins the game.
$Q'$ is the event that Team $Q$ loses.

$ P(P) = 50\% = \dfrac{1}{2} \\[5ex] P(Q) = 50\% = \dfrac{1}{2} \\[5ex] P(P') = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\[5ex] P(Q') = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\[5ex] \underline{1st\;\;Case} \\[3ex] P(PQ') = \dfrac{1}{2} * \dfrac{1}{2} \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] P(PQ') = \dfrac{1}{4} \\[5ex] \underline{2nd\;\;Case} \\[3ex] P(P'Q) = \dfrac{1}{2} * \dfrac{1}{2} \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] P(P'Q) = \dfrac{1}{4} \\[5ex] 1st\;\;Case + 2nd\;\;Case + 3rd\;\;Case = 1 \\[3ex] ...Sum\;\;of\;\;all\;\;probabilities \\[3ex] 3rd\;\;Case = P(Draw) \\[3ex] \dfrac{1}{4} + \dfrac{1}{4} + P(Draw) = 1 \\[5ex] \dfrac{1 + 1}{4} + P(Draw) = 1 \\[5ex] \dfrac{2}{4} + P(Draw) = \dfrac{4}{4} \\[5ex] P(Draw) = \dfrac{4}{4} - \dfrac{2}{4} \\[5ex] P(Draw) = \dfrac{4 - 2}{4} \\[5ex] P(Draw) = \dfrac{2}{4} \\[5ex] P(Draw) = \dfrac{1}{2} $

The product, 6 is marked in red.

$ Rectangular\;\;Table\:\:\rightarrow \\[1ex] \;\;\;\;\;* \\[1ex] Standard\;\;Die\:\:\downarrow $	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$1$	$1$	$2$	$3$	$4$	$5$	$\color{red}6$	$7$	$8$
$2$	$2$	$4$	$\color{red}6$	$8$	$10$	$12$	$14$	$16$
$3$	$3$	$\color{red}6$	$9$	$12$	$15$	$18$	$21$	$24$
$4$	$4$	$8$	$12$	$16$	$20$	$24$	$28$	$32$
$5$	$5$	$10$	$15$	$20$	$25$	$30$	$35$	$40$
$6$	$\color{red}6$	$12$	$18$	$24$	$30$	$36$	$42$	$48$

$ n(S) = 6 * 8 = 48 \\[3ex] n(6) = 4 \\[3ex] P(6) = \dfrac{n(6)}{n(S)} \\[5ex] = \dfrac{4}{48} \\[5ex] = \dfrac{1}{12} $

$ A = \{1, 2, 3, 4, 5, 6\} \\[3ex] n(A) = 6 \\[3ex] B \cap C = \{2\} \\[3ex] n(B \cap C) = 1 \\[3ex] P(B \cap C) \\[3ex] = \dfrac{n(B \cap C)}{n(A)} \\[5ex] = \dfrac{1}{6} $

Sample Space for Two Tosses of A Fair Die

A Toss in the Column and a Toss in the Row

$1\:\:Toss\:\:\rightarrow$ ($+$) $1\:\:Toss\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$\color{red}{3}$	$\color{violet}{4}$	$\color{violet}{5}$	$\color{red}{6}$	$7$
$2$	$\color{red}{3}$	$\color{violet}{4}$	$\color{violet}{5}$	$\color{red}{6}$	$7$	$8$
$3$	$\color{violet}{4}$	$\color{violet}{5}$	$\color{red}{6}$	$7$	$8$	$\color{red}{9}$
$4$	$\color{violet}{5}$	$\color{red}{6}$	$7$	$8$	$\color{red}{9}$	$10$
$5$	$\color{red}{6}$	$7$	$8$	$\color{red}{9}$	$10$	$11$
$6$	$7$	$8$	$\color{red}{9}$	$10$	$11$	$\color{red}{12}$

(a.)
Multiples of 3 are marked in red

$ n(S) = 6 * 6 = 36 \\[3ex] Let\;\;E = event\;\;of\;\;multiple\;\;of\;\;3 \\[3ex] n(E) = 12 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{12}{36} = \dfrac{1}{3} \\[5ex] $ (b.)
Sum of the numbers between 3 and 6 are marked in violet

$ Let\;\;E = event\;\;of\;\;numbers\;\;between\;\;3\;\;and\;\;6 \\[3ex] n(E) = 7 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{7}{36} $

(a.) Eating two apples of the same colour implies two red apples OR two green apples
This means that:
The first apple is red AND the second apple is red
OR
The first apple is green AND the second apple is green

(b.) Eating two apples, one at a time implies without replacement condition

$ Let: \\[3ex] red = R \\[3ex] green = G \\[3ex] n(R) = 9 \\[3ex] n(G) = 8 \\[3ex] n(S) = 17 \\[3ex] P(RR)\;\;OR\;\;P(GG) \\[3ex] = \left(\dfrac{9}{17} * \dfrac{8}{16}\right) + \left(\dfrac{8}{17} * \dfrac{7}{16}\right) \\[5ex] = \dfrac{9}{34} + \dfrac{7}{34} \\[5ex] = \dfrac{16}{54} \\[5ex] = \dfrac{8}{17} $

$ (a.) \\[3ex] sample\;\;proportion = \dfrac{number\;\;of\;\;prizes}{number\;\;of\;\;tickets} = \dfrac{124}{400} \\[5ex] = \dfrac{31}{100} \\[5ex] $ (b.)
Winning a prize implies having a sum of at least 7
The sum of at least 7 (7 or higher) is marked in red

Sample Space for the two Panels

Panel 1 in the Column and Panel 2 in the Row

$Panel\;2\:\:\rightarrow$ ($+$) $Panel\:1\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$
$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$	$\color{red}{8}$
$3$	$4$	$5$	$6$	$\color{red}{7}$	$\color{red}{8}$	$\color{red}{9}$
$1$	$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$
$1$	$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$
$1$	$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$
$3$	$4$	$5$	$6$	$\color{red}{7}$	$\color{red}{8}$	$\color{red}{9}$
$2$	$3$	$4$	$5$	$6$	$\color{red}{7}$	$\color{red}{8}$

$ n(S) = 8 * 6 = 48 \\[3ex] n(\ge 7) = 14 \\[3ex] P(winning\;\;a\;\;prize) = P(\ge 7) \\[3ex] = \dfrac{n(\ge 7)}{n(S)} \\[5ex] = \dfrac{14}{48} \\[5ex] = \dfrac{7}{24} $

$ S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\} \\[3ex] n(S) = 15 \\[3ex] E = \{1, 2, 3, 4, 5, 6, 7, 8\} \\[3ex] n(E) = 8 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{8}{15} $

8-sided game: First Roll in the Column and Second Roll in the Row

$2nd\:\:Roll\:\:\rightarrow$ ($+$) $1st\:\:Roll\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$1$	$1, 1$	$1, 2$	$1, 3$	$1, 4$	$1, 5$	$1, 6$	$1, 7$	$1, 8$
$2$	$2, 1$	$2, 2$	$2, 3$	$2, 4$	$2, 5$	$2, 6$	$2, 7$	$2, 8$
$3$	$3, 1$	$3, 2$	$3, 3$	$3, 4$	$3, 5$	$3, 6$	$3, 7$	$3, 8$
$4$	$4, 1$	$4, 2$	$4, 3$	$4, 4$	$4, 5$	$4, 6$	$4, 7$	$4, 8$
$5$	$5, 1$	$5, 2$	$5, 3$	$5, 4$	$5, 5$	$5, 6$	$5, 7$	$5, 8$
$6$	$6, 1$	$6, 2$	$6, 3$	$6, 4$	$6, 5$	$6, 6$	$6, 7$	$6, 8$
$7$	$7, 1$	$7, 2$	$7, 3$	$7, 4$	$7, 5$	$7, 6$	$7, 7$	$7, 8$
$8$	$8, 1$	$8, 2$	$8, 3$	$8, 4$	$8, 5$	$8, 6$	$8, 7$	$8, 8$

$ n(S) = 8 * 8 = 64 \\[3ex] E = \{8, 8\} \\[3ex] n(E) = 1 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[3ex] = \dfrac{1}{64} $

$ A \cap B = \{6, 9\} \\[3ex] A \cup B = \{1, 2, 5, 6, 8, 9\} \\[3ex] (A \cup B)' = \{3, 4, 7\} \\[3ex] $ (a) The Venn diagram is:


$ (b) \\[3ex] n(A \cap B) = 2 \\[3ex] n(\mathcal{E}) = 9 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(\mathcal{E})} = \dfrac{2}{9} $

$ n(\mu) = 40 \\[3ex] Let: \\[3ex] Mathematics = M \\[3ex] Accounts = A \\[3ex] Economics = E \\[3ex] n(M) = 18 \\[3ex] n(A) = 19 \\[3ex] n(E) = 16 \\[3ex] n(M\;\;and\;\;A\;\;only) = 5 \\[3ex] n(M\;\;only) = 6 \\[3ex] n(A\;\;only) = 9 \\[3ex] n(A\;\;and\;\;E\;\;only) = 2 \\[3ex] n(M\;\;and\;\;A\;\;and\;\;E) = n(passed\;\;all\;\;three) \\[3ex] $ Let us use a Venn diagram


$ \underline{Accounts} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;A\;\;only) + n(A\;\;and\;\;E\;\;only) + n(A\;\;only) = n(A) \\[3ex] n(passed\;\;all\;\;three) + 5 + 2 + 9 = 19 \\[3ex] n(passed\;\;all\;\;three) + 16 = 19 \\[3ex] n(passed\;\;all\;\;three) = 19 - 16 \\[3ex] n(passed\;\;all\;\;three) = 3 \\[3ex] \underline{Mathematics} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;A\;\;only) + n(M\;\;and\;\;E\;\;only) + n(M\;\;only) = n(M) \\[3ex] 3 + 5 + n(M\;\;and\;\;E\;\;only) + 6 = 18 \\[3ex] 14 + n(M\;\;and\;\;E\;\;only) = 18 n(M\;\;and\;\;E\;\;only) = 18 - 14 \\[3ex] n(M\;\;and\;\;E\;\;only) = 4 \\[3ex] \underline{Economics} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;E\;\;only) + n(A\;\;and\;\;E\;\;only) + n(E\;\;only) = n(E) \\[3ex] 3 + 4 + 2 + n(E\;\;only) = 16 \\[3ex] 9 + n(E\;\;only) = 16 \\[3ex] n(E\;\;only) = 16 - 9 \\[3ex] n(E\;\;only) = 7 \\[3ex] \implies \\[3ex] (a) \\[3ex] n(M\;\;only) + n(A\;\;only) + n(E\;\;only) \\[3ex] + n(M\;\;and\;\;A\;\;only) + n(N\;\;and\;\;E\;\;only) + n(A\;\;and\;\;E\;\;only) \\[3ex] + n(passed\;\;all\;\;three) + n(failed\;\;all\;\;three) = n(\mu) \\[3ex] 6 + 9 + 7 + 5 + 4 + 2 + 3 + n(failed\;\;all\;\;three) = 40 \\[3ex] 36 + n(failed\;\;all\;\;three) = 40 \\[3ex] n(failed\;\;all\;\;three) = 40 - 36 \\[3ex] n(failed\;\;all\;\;three) = 4 \\[3ex] $ 4 students failed all subjects.

The completed Venn diagram is:


(b) Failed at least one of Economics and Mathematics implies:
failed in Economics only
failed in Mathematics only
failed in Economics and Mathematics only

failed in all three subjects

In other words, it implies: not passing anything that has to do with Economics or Mathematics
Because of the Venn diagram (that shows passing for the subjects):
This means:
passing only Accounts
failing all three subjects

$ n(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics) \\[3ex] = n(A\;\;only) + n(failed\;\;all\;\;three) \\[3ex] = 9 + 4 \\[3ex] = 13 \\[3ex] n(U) = 40 \\[3ex] \%(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics) \\[3ex] = \dfrac{n(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics)}{n(U)} * 100 \\[5ex] = \dfrac{13}{40} * 100 \\[5ex] = 32.5\% \\[3ex] (c) \\[3ex] P(failed\;\;in\;\;Accounts) + P(passed\;\;in\;\;Accounts) = 1 ...Complementary\;\;Rule \\[3ex] P(failed\;\;in\;\;Accounts) = 1 - P(passed\;\;in\;\;Accounts) \\[3ex] P(passed\;\;in\;\;Accounts) = \dfrac{n(A)}{n(U)} = \dfrac{19}{40} \\[5ex] \implies \\[3ex] P(passed\;\;in\;\;Accounts) = 1 - \dfrac{19}{40} = \dfrac{40 - 19}{40} = \dfrac{21}{40} \\[5ex] $ Alternatively:
Failed in Accounts means that the student did not pass anything related to Accounts
It is $n(A')$
Because of the Venn diagram (that shows passing for the subjects):
This includes:
passing only Mathematics
passing only Economics
passed Mathematics and Economics only
failing all three subjects

$ n(failed\;\;in\;\;Accounts) \\[3ex] = n(M\;\;only) + n(E\;\;only) + n(M\;\;and\;\;E\;\;only) + n(failed\;\;all\;\;three) \\[3ex] = 6 + 7 + 4 + 4 \\[3ex] = 21 \\[3ex] n(U) = 40 \\[3ex] P(failed\;\;in\;\;Accounts) \\[3ex] = \dfrac{n(failed\;\;in\;\;Accounts)}{n(U)} \\[5ex] = \dfrac{21}{40} $

$ Let: \\[3ex] Land = L \\[3ex] Water = W \\[3ex] n(L) = 1.48 * 10^8 \\[3ex] n(W) = 3.63 * 10^8 \\[3ex] n(S) = n(L) + n(W) \\[3ex] = 1.48 * 10^8 + 3.63 * 10^8 \\[3ex] = 10^8(1.48 + 3.63) \\[3ex] = 10^8 * 5.11 \\[3ex] = 5.11 * 10^8 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] = \dfrac{3.63 * 10^8}{5.11 * 10^8} \\[5ex] = 0.71037182 \\[3ex] = 0.71037182 * 100 \\[3ex] = 71.037182\% \\[3ex] \approx 71\% $

$ Let\:\: Blue = B \\[3ex] Red = R \\[3ex] Green = G \\[3ex] Numbers\:\: as\:\: is \\[3ex] S = \{1B, 2B, 1R, 2R, 3R, 1G, 2G, 3G, 4G, 5G\} \\[3ex] n(S) = 10 \\[3ex] n(R) = 3 \:\: \implies (1R, 2R, 3R) \\[3ex] n(3) = 2 \:\: \implies (3R, 3G) \\[3ex] n(3 \:\:AND\:\: R) = n(3R) = 1 \\[3ex] P(3 \:\:OR\:\: R) = P(R) + P(3) - P(3 \:\:AND\:\: R)...Addition\:\: Rule \\[3ex] P(3 \:\:OR\:\: R) = \dfrac{3}{10} + \dfrac{2}{10} - \dfrac{1}{10} \\[5ex] = \dfrac{3 + 2 - 1}{10} \\[5ex] = \dfrac{4}{10}...Option\;J \\[5ex] = \dfrac{2}{5} $

$ n(S) = n(Outcomes) \\[3ex] = 32 + m + 25 + 40 + 28 + 45 \\[3ex] = m + 170 \\[3ex] (a.) \\[3ex] n(2) = m \\[3ex] P(2) = \dfrac{n(2)}{n(S)} \\[5ex] 0.15 = \dfrac{m}{m + 170} \\[5ex] 0.15(m + 170) = m \\[3ex] 0.15m + 25.5 = m \\[3ex] m = 0.15m + 25.5 \\[3ex] m - 0.15m = 25.5 \\[3ex] 0.85m = 25.5 \\[3ex] m = \dfrac{25.5}{0.85} \\[5ex] m = 30 \\[3ex] (b.) \\[3ex] Number\;\;of\;\;times\;\;the\;\;die\;\;was\;\;rolled \\[3ex] = m + 170 \\[3ex] = 30 + 170 \\[3ex] = 200\;times \\[3ex] (c.) \\[3ex] P(even\;\;number) \\[3ex] = P(2)\;\;OR\;\;P(4)\;\;OR\;\;P(6) \\[3ex] = 0.15 + \dfrac{n(4)}{n(S)} + \dfrac{n(6)}{n(S)} \\[5ex] = 0.15 + \dfrac{40}{200} + \dfrac{45}{200} \\[5ex] = 0.15 + 0.2 + 0.225 \\[3ex] = 0.575 $

When two events are mutually exclusive, both events cannot occur at the same time.
Exactly one event must occur

This means that the first event must occur or the second event must occur.
Both events cannot occur simultaneously.

If we say that exactly one event may occur, then there is the possibility that none of the events occur.
This goes against the formula for mutually exclusive events
Assume events A and B are two mutually exclusive events

$ P(A \;\;OR\;\; B) = P(A) + P(B) - P(A \;\;AND\;\; B) \\[3ex] P(A \cup B) = P(A) + P(B) - P(A \cap B) \\[3ex] But\;\; P(A \cap B) = 0...mutually\;\;exclusive\;\;events \\[3ex] \implies \\[3ex] P(A \cup B) = P(A) + P(B) \\[3ex] $ The formula does not take into account that none of the events may occur
Hence, only one event must occur.

$ n(\mu) = 200 \\[3ex] Let: \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] Mathematics = M \\[3ex] n(P) = 70 \\[3ex] n(C) = 90 \\[3ex] n(M) = 100 \\[3ex] n(neither) = 24 \\[3ex] n(P\;\;and\;\;C) = 23 \\[3ex] n(C\;\;and\;\;M) = 41 \\[3ex] n(all\;\;three) = 8 \\[3ex] n(P\;\;and\;\;M\;\;only) = d \\[3ex] $ (a) The Venn diagram is:


$ n(P\;\;and\;\;C\;\;only) \\[3ex] = 23 - 8 \\[3ex] = 15 \\[3ex] n(C\;\;and\;\;M\;\;only) \\[3ex] = 41 - 8 \\[3ex] = 33 \\[3ex] n(C\;\;only) \\[3ex] = 90 - (15 + 8 + 33) \\[3ex] = 90 - 56 \\[3ex] = 34 \\[3ex] n(P\;\;only) \\[3ex] = 70 - (d + 8 + 15) \\[3ex] = 70 - (d + 23) \\[3ex] = 70 - d - 23 \\[3ex] = 47 - d \\[3ex] n(M\;\;only) \\[3ex] = 100 - (d + 8 + 33) \\[3ex] = 100 - (d + 41) \\[3ex] = 100 - d - 41 \\[3ex] = 59 - d \\[3ex] \implies \\[3ex] (47 - d) + (59 - d) + 34 + d + 15 + 33 + 8 + 24 = 200 \\[3ex] 47 - d + 59 - d + d + 34 + 15 + 33 + 8 + 24 = 200 \\[3ex] 220 - d = 200 \\[3ex] 220 - 200 = d \\[3ex] 20 = d \\[3ex] d = 20 \\[3ex] \implies \\[3ex] n(P\;\;and\;\;M\;\;only) = 20 \\[3ex] (b) \\[3ex] (i) \\[3ex] n(P\;\;only) = 47 - d = 47 - 20 = 27 \\[3ex] n(\mu) = 200 \\[3ex] P(P\;\;only) = \dfrac{n(P\;\;only)}{n(\mu)} = \dfrac{27}{200} \\[5ex] (ii) \\[3ex] n(exactly\;\;2\;\;subjects) \\[3ex] = n(P\;\;and\;\;C\;\;only) + n(C\;\;and\;\;M\;\;only) + n(P\;\;and\;\;M\;\;only) \\[3ex] = 15 + 33 + 20 \\[3ex] = 68 \\[3ex] P(exactly\;\;2\;\;subjects) \\[3ex] = \dfrac{n(exactly\;\;2\;\;subjects)}{n(\mu)} \\[5ex] = \dfrac{68}{200} \\[5ex] = \dfrac{17}{50} $

$ \mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \\[3ex] A = \{2, 8, 10, 14\} \\[3ex] B = \{6, 8, 20\} \\[3ex] C = \{8, 18, 20, 22\} \\[3ex] n(all\;\;three) = n(A \cap B \cap C) = \{8\} \\[3ex] A' = \{4, 6, 12, 16, 18, 20, 22, 24\} \\[3ex] B' = \{2, 4, 10, 12, 14, 16, 18, 22, 24\} \\[3ex] C' = \{2, 4, 6, 10, 12, 14, 16, 24\} \\[3ex] \underline{A\;\;and\;\;B\;\;only} \\[3ex] A \cap B = \{8\} \\[3ex] A \cap B \cap C' = \{\} \\[3ex] \underline{A\;\;and\;\;C\;\;only} \\[3ex] A \cap C = \{8\} \\[3ex] A \cap C \cap B' = \phi \\[3ex] \underline{B\;\;and\;\;C\;\;only} \\[3ex] B \cap C = \{8, 20\} \\[3ex] B \cap C \cap A' = \{20\} \\[3ex] \underline{A\;\;only} \\[3ex] B' \cap C' = \{2, 4, 10, 12, 14, 16, 24\} \\[3ex] A \cap B' \cap C' = \{2, 10, 14\} \\[3ex] \underline{B\;\;only} \\[3ex] A' \cap C' = \{4, 6, 12, 16, 24\} \\[3ex] B \cap A' \cap C' = \{6\} \\[3ex] \underline{C\;\;only} \\[3ex] A' \cap B' = \{4, 12, 16, 18, 22, 24\} \\[3ex] C \cap A' \cap B' = \{18, 22\} \\[3ex] $ (a) The Venn diagram is:


$ A \cap B = \{8\} \\[3ex] n(A \cap B) = 1 \\[3ex] \mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \\[3ex] n(\mathcal{E}) = 12 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(\mathcal{E})} = \dfrac{1}{12} $

Two numbers are selected: With Replacement condition
Prime numbers = {5, 7}
At least one prime number means one or two prime numbers for example: {5, 5}, {5, 6}, {5, 7}, etc.
Because there are several options involved, it is better to use a Punnett Square
The set containing at least one prime number is marked in red.

First Number in the Column and Second Number in the Row

$2nd\:\:Number\:\:\rightarrow$ $1st\:\:Number\:\:\downarrow$	$5$	$6$	$7$	$8$	$9$
$5$	$\color{red}{5, 5}$	$\color{red}{5, 6}$	$\color{red}{5, 7}$	$\color{red}{5, 8}$	$\color{red}{5, 9}$
$6$	$\color{red}{6, 5}$	$6, 6$	$\color{red}{6, 7}$	$6, 8$	$6, 9$
$7$	$\color{red}{7, 5}$	$\color{red}{7, 6}$	$\color{red}{7, 7}$	$\color{red}{7, 8}$	$\color{red}{7, 9}$
$8$	$\color{red}{8, 5}$	$8, 6$	$\color{red}{8, 7}$	$8, 8$	$8, 9$
$9$	$\color{red}{9, 5}$	$9, 6$	$\color{red}{9, 7}$	$9, 8$	$9, 9$

$ n(at\;\;least\;\;one\;\;prime\;\;number) = n(E) = 16 \\[3ex] n(S) = 5 * 5 = 25 \\[3ex] P(at\;\;least\;\;one\;\;prime\;\;number) \\[3ex] = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{16}{25} $

(a.)
The missing values are marked in red

Absolute Value of:

$ Die\:\:1\;\rightarrow \\[1ex] \;\;\;- \\[1ex] Die\:\:2\;\downarrow $	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$0$	$1$	$2$	$3$	$4$	$5$
$2$	$\color{red}1$	$0$	$1$	$2$	$3$	$4$
$3$	$2$	$1$	$\color{red}0$	$\color{red}1$	$2$	$\color{red}3$
$4$	$3$	$2$	$1$	$\color{red}0$	$1$	$2$
$5$	$\color{red}4$	$3$	$2$	$1$	$0$	$1$
$6$	$5$	$4$	$3$	$2$	$1$	$0$

$ (b.) \\[3ex] P(NOT\;\;0) + P(0) = 1...Complementary\;\;Rule \\[3ex] P(0) = \dfrac{n(0)}{n(S)} \\[5ex] n(S) = 6 * 6 = 36 \\[3ex] n(0) = 6 \\[3ex] P(0) = \dfrac{6}{36} = \dfrac{1}{6} \\[5ex] P(NOT\;\;0) \\[3ex] = 1 - P(0) \\[5ex] = 1 - \dfrac{1}{6} \\[5ex] = \dfrac{6}{6} - \dfrac{1}{6} \\[5ex] = \dfrac{5}{6} $

We can solve this question in two ways
Use any method you prefer

Let the number of additional red marbles = $r$

$ \underline{First\:\:Approach} \\[3ex] \underline{Initial} \\[3ex] n(S) = 24 \\[3ex] n(Red) = 8 \\[3ex] \underline{Updated} \\[3ex] n(Red) = 8 + r \\[3ex] n(S) = 24 + r \\[3ex] P(Red) = \dfrac{8 + r}{24 + r} \\[5ex] P(Red) = \dfrac{3}{5} \\[5ex] \therefore \dfrac{8 + r}{24 + r} = \dfrac{3}{5} \\[5ex] Cross\:\: Multiply\:\: method \\[3ex] 5(8 + r) = 3(24 + r) \\[3ex] 40 + 5r = 72 + 3r \\[3ex] 5r - 3r = 72 - 40 \\[3ex] 2r = 32 \\[3ex] r = \dfrac{32}{2} \\[5ex] r = 16 \\[5ex] \underline{Second\:\:Approach} \\[3ex] P(Red) = \dfrac{3}{5} \\[5ex] P(Red) + P(Red') = 1 ...Complementary\:\:Rule \\[3ex] P(Red') = 1 - P(Red) \\[3ex] P(Red') = 1 - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5}{5} - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5 - 3}{5} \\[5ex] P(Red') = \dfrac{2}{5} \\[5ex] n(Red') = n(Yellow) + n(Green) \\[3ex] n(Red') = 9 + 7 = 16 \\[3ex] P(Red') = \dfrac{n(Red')}{n(S)} \\[5ex] n(S) * P(Red') = n(Red') \\[3ex] n(S) = \dfrac{n(Red')}{P(Red')} \\[5ex] n(S) = n(Red') \div P(Red') \\[3ex] n(S) = 16 \div \dfrac{2}{5} \\[5ex] n(S) = 16 * \dfrac{5}{2} \\[5ex] n(S) = 8(5) \\[3ex] n(S) = 40 \\[3ex] Additional\:\:red\:\:balls = 40 - (8 + 9 + 7) \\[3ex] Additional\:\:red\:\:balls = 40 - 24 \\[3ex] Additional\:\:red\:\:balls = 16 \\[3ex] $ $16$ red marbles need to be added so that the probability of drawing a red marble is $\dfrac{3}{5}$

Sample Space for Two Throws of a Fair Die

First Throw in the Column and Second Throw in the Row

$2nd\:\:Throw\:\:\rightarrow$ ($+$) $1st\:\:Throw\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$
$2$	$3$	$4$	$5$	$6$	$7$	$8$
$3$	$4$	$5$	$6$	$7$	$8$	$9$
$4$	$5$	$6$	$7$	$8$	$9$	$\color{red}{10}$
$5$	$6$	$7$	$8$	$9$	$\color{red}{10}$	$\color{red}{11}$
$6$	$7$	$8$	$9$	$\color{red}{10}$	$\color{red}{11}$	$\color{red}{12}$

$ n(S) = 6 * 6 = 36 \\[3ex] Let\;\;E = event\;\;of\;\;selecting\;\;at\;\;least\;\;10 \\[3ex] E = \{10, 10, 11, 10, 11, 12\} \\[3ex] n(E) = 6 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{6}{36} \\[5ex] = \dfrac{1}{6} $

$ (a) \\[3ex] n(guitar\;\;players) = n(G) = 21 + 8 = 29 \\[3ex] $ 29 people are guitar players

$ (b) \\[3ex] n(singers\;\;but\;\;not\;\;guitar\;\;players) \\[3ex] = n(singers\;\;only) \\[3ex] = 4 \\[3ex] $ 4 people are singers but not guitar players

$ (c) \\[3ex] n(universal\;\;set) = n(\xi) = 50 \\[3ex] n(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player) = 17 \\[3ex] P(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player) \\[3ex] = \dfrac{n(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player)}{n(\xi)} \\[5ex] = \dfrac{17}{50} $

$ Number\;\;of\;\;participants = 1000 \\[3ex] 1000\;\;Yq77\;\;tests\;\;@\;\;\$2500/test = 1000(2500) = 2,500,000 \\[3ex] 1000\;\;Sam77\;\;tests\;\;@\;\;\$50/test = 1000(50) = 50,000 \\[3ex] Total\;\;cost = \$2,500,000 + \$50,000 = \$2,550,000 $

$ Number\;\;of\;\;participants = 1000 \\[3ex] Number\;\;of\;\;positive\;\;Yq77\;\;test = 590 + 25 = 615 \\[3ex] \%\;\;of\;\;positive\;\;Yq77\;\;test \\[3ex] = \dfrac{615}{1000} * 100 \\[5ex] = 61.5\% $

Yq77 test is 100% accurate but Sam77 test is not.
So, an incorrect result by the Sam77 test implies that the:
(a.) volunteer had a positive Sam77 test but a negative Yq77 test (10 volunteers)
(b.) volunteer had a negative Sam77 test but a positive Yq77 test (25 volunteers)

10 + 25 = 35
This means that 35 volunteers received an incorrect result from the Sam77 test.

The question deals with Conditional Probability.
It can be reworded as:
What is the probability that the chosen volunteer does not possess Yq77 given that the Sam77 test result is positive?

$ Let: \\[3ex] positive\;\;Sam77\;\;test = A \\[3ex] positive\;\;Yq77\;\;test = Y \\[3ex] \implies negative\;\;Yq77\;\;test = Y' \\[3ex] P(Y' | A) = \dfrac{n(Y' \cap A)}{n(A)} \\[5ex] = \dfrac{10}{590 + 10} \\[5ex] = \dfrac{10}{600} \\[5ex] = 0.0166666667 \\[3ex] \approx 0.017 $

(a.)
One or both picture cards mean at least one picture card
Let event E = At least one picture card are:
2 and Queen, 2 and King, 4 and Queen, 4 and King, 6 and Queen, 6 and King, 8 and Queen, 8 and King, Queen and King

$ n(S) = 15 \\[3ex] n(E) = 9 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{9}{15} \\[5ex] = \dfrac{3}{5} \\[5ex] $ (b.)
We can solve this question using at least thwo approaches
Use any approach you prefer.

First Approach
Event E = selecting at least one picture card
Event E' = selecting no picture card
Both are complementary events.

$ P(E') + P(E) = 1...Complementary\;\;Rule \\[3ex] P(E') = 1 - P(E) \\[3ex] = 1 - \dfrac{3}{5} \\[5ex] = \dfrac{5}{5} - \dfrac{3}{5} \\[5ex] = \dfrac{5 - 3}{5} \\[5ex] = \dfrac{2}{5} \\[5ex] $ Second Approach
Event E' = selecting no picture card

$ n(E') = 6 \\[3ex ] P(E') = \dfrac{n(E')}{n(S)} \\[5ex] = \dfrac{6}{15} \\[5ex] = \dfrac{2}{5} $

$ Let: \\[3ex] red = R \\[3ex] blue = B \\[3ex] n(R) = 5 \\[3ex] n(B) = 4 \\[3ex] n(S) = 5 + 4 = 9 \\[3ex] \underline{With\;\;Replacement\;\;condition} \\[3ex] P(R\;\;AND\;\;B) \\[3ex] = P(R) * P(B) \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{n(R)}{n(S)} * \dfrac{n(B)}{n(S)} \\[5ex] = \dfrac{5}{9} * \dfrac{4}{9} \\[5ex] = \dfrac{20}{81} $

(a.) This is an example of a theoretical probability because the event is not based on an experiment.

(b.) This is an example of an empirical probability because it is based on an experiment.

(c.) This is an example of an empirical probability because it is based on an experiment.

(d.) This is a theoretical probability. It is not based on an experiment.

$ Let\;\;event\;\;of\;\;head\;\;facing\;\;up = H \\[3ex] P(H) = \dfrac{1}{2} \\[5ex] P(HHH) = \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2} \\[5ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;events \\[3ex] = \dfrac{1}{8} $

(a.) Choosing both students from the same grade and the same high school implies selecting:
Two 10th-grade students from the North high school
OR
Two 10th-grade students from the South high school
OR
Two 11th-grade students from the North high school
OR
Two 11th-grade students from the South high school

(b.) Because both students are chosen simultaneously at random, this is a case of without replacement condition

$ n(S) = 47 + 93 + 73 + 57 = 270 \\[3ex] P(choosing\;\;2\;\;students\;\;from\;\;same\;\;grade\;\;and\;\;same\;\;high\;\;school) \\[3ex] = \dfrac{47(46)}{270(269)} + \dfrac{93(92)}{270(269)} + \dfrac{73(72)}{270(269)} + \dfrac{57(56)}{270(269)} $

$ (114.1.1) \\[3ex] P(A) + P(not\;\;A) = 1 ...Complementary\;\;Rule \\[3ex] P(A) + 0.4 = 1 \\[3ex] P(A) = 1 - 0.4 \\[3ex] P(A) = 0.6 \\[3ex] P(A\;\;and\;\;B) = P(A) * P(B) \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = 0.6 * 0.3 \\[3ex] = 0.18 \\[5ex] (114.2.1) \\[3ex] a = \dfrac{n(H \cap D \cap C)}{n(S)} \\[5ex] = \dfrac{15}{150} = \dfrac{1}{10} = 0.1 \\[3ex] (114.2.2) \\[3ex] P(at\;\;least\;\;one\;\;activity) + P(no\;\;activity) = 1 \\[3ex] ...Complementary\;\;Rule \\[3ex] 0.7 + m = 1 \\[3ex] m = 1 - 0.7 \\[3ex] m = 0.3 \\[3ex] (114.2.3) \\[3ex] P(only\;\;1\;\;activity) = 0.24 + b + b = 2b + 0.24 \\[3ex] P(only\;\;2\;\;activities) = 0.02 + 0.14 + 0.12 = 0.28 \\[3ex] P(all\;\;three\;\;activities) = a = 0.1 \\[3ex] P(no\;\;activity) = m = 0.3 \\[3ex] P(total\;\;probability) = 1 \\[3ex] \implies \\[3ex] 2b + 0.24 + 0.28 + 0.1 + 0.3 = 1 \\[3ex] 2b + 0.92 = 1 \\[3ex] 2b = 1 - 0.92 \\[3ex] 2b = 0.08 \\[3ex] b = \dfrac{0.08}{2} \\[5ex] b = 0.04 \\[3ex] P(only\;\;chess) = \dfrac{n(only\;\;chess)}{n(S)} \\[5ex] b = \dfrac{n(only\;\;chess)}{150} \\[5ex] n(only\;\;chess) \\[3ex] = 150 * b \\[3ex] = 150 * 0.04 \\[3ex] = 6 \\[3ex] $ 6 learners play only chess

$ Let\;\;the\;\;number\;\;of\;\;candy\;\;pieces = n \\[3ex] \implies \\[3ex] n(S) = n \\[3ex] n(brown) = \dfrac{1}{3}n \\[5ex] n(other\;\;colors) = remaining \\[3ex] = 1 - \dfrac{1}{3}n = \dfrac{2}{3}n \\[5ex] n(red) \\[3ex] = \dfrac{2}{3}n \div 5...even\;\;distribution \\[5ex] = \dfrac{2}{3}n * \dfrac{1}{5} \\[5ex] = \dfrac{2n}{15} \\[5ex] P(red) = \dfrac{n(red)}{n(S)} \\[5ex] = n(red) \div n(S) \\[3ex] = \dfrac{2n}{15} \div n \\[5ex] = \dfrac{2n}{15} * \dfrac{1}{n} \\[5ex] = \dfrac{2}{15} $

A marble is randomly removed from the bag and then returned to the bag.
This is a case of "with replacement" - independent events

$ Let\:\: red = R \\[3ex] yellow = Y \\[3ex] green = G \\[3ex] n(S) = 64 \\[3ex] P(R) = \dfrac{5}{8} \\[5ex] P(Y) = \dfrac{1}{4} \\[5ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] \rightarrow \dfrac{5}{8} = \dfrac{n(R)}{64} \\[5ex] 8 * 8 = 64 \\[3ex] \therefore n(R) = 5 * 8 = 40 \\[3ex] P(Y) = \dfrac{n(Y)}{n(S)} \\[5ex] \rightarrow \dfrac{1}{4} = \dfrac{n(Y)}{64} \\[5ex] 4 * 16 = 64 \\[3ex] \therefore n(Y) = 1 * 16 = 16 \\[3ex] n(G) + n(R) + n(Y) = 64 \\[3ex] n(G) + 40 + 16 = 64 \\[3ex] n(G) + 56 = 64 \\[3ex] n(G) = 64 - 56 \\[3ex] n(G) = 8 \\[3ex] $ There are $8$ green marbles in the bag.

Let us analyze this question.
Based on the last sentence, it is possible to have CPAs who do not have business degrees.
So, we have:
120 employees, say E (given)
made up of:
those with business degrees, say B (85 employees...given); and those with non-business degrees, B' (not given)
Those with business degrees consists of:
CPAs, say C (75 employees...given); and non-CPAs, C' (not given)
Similarly, those with non-business degrees are made up of:
CPAs, C (not given); and non-CPAs, C' (14...given)

$ n(E) = 120...given \\[3ex] n(E) = n(B) + n(B') \\[3ex] n(B) = 85...given \\[3ex] n(C) = 75 ...given \\[3ex] n(B') = n(E) - n(B) = 120 - 85 = 35...found \\[3ex] n(B') = n(C) + n(C') \\[3ex] n(C') = 14...given \\[3ex] n(C) = n(B') - n(C') = 35 - 14 = 21...found \\[3ex] $ Let us represent this information as a tree diagram...visualize it


$ n(C) = 75 + 21 ...given + found \\[3ex] n(C) = 96 \\[3ex] n(S) = n(E) = 120 \\[3ex] P(C) = \dfrac{n(C)}{n(S)} \\[5ex] = \dfrac{96}{120} $

$ Let: \\[3ex] swim\;\;in\;\;sea = W \\[3ex] go\;\;to\;\;the\;\;funfair = G \\[3ex] n(swim) = n(W) = 13 \\[3ex] n(go) = n(G) = 17 \\[3ex] n(both) = n(W \cap G) = p n(swim\;\;only) = 13 - p \\[3ex] n(go\;\;only) = 17 - p \\[3ex] n(neither) = n((W \cup G)') = 2 \\[3ex] $ The Venn diagram is:


$ 13 - p + 17 - p + p + 2 = 20 \\[3ex] 32 - p = 20 \\[3ex] 32 - 20 = p \\[3ex] p = 12 \\[3ex] \implies \\[3ex] n(both) = 12 \\[3ex] n(swim\;\;only) \\[3ex] = 13 - p \\[3ex] = 13 - 12 \\[3ex] = 1 \\[3ex] n(go\;\;only) \\[3ex] = 17 - p \\[3ex] = 17 - 12 \\[3ex] = 5 \\[3ex] $ The completed Venn diagram is:


$ (b) \\[3ex] P(both) = \dfrac{n(both)}{n(\varepsilon)} \\[5ex] = \dfrac{12}{20} \\[5ex] = \dfrac{3}{5} \\[5ex] (c) \\[3ex] $ Swimming in the sea or going to the funfair without doing both are:
those who do only one: swim the sea only OR go to the funfair only

$ n(only\;\;one) \\[3ex] = 1 + 5 = 6 \\[3ex] P(only\;\;one) \\[3ex] = \dfrac{n(only\;\;one)}{n(\varepsilon)} \\[5ex] = \dfrac{6}{20} \\[5ex] = \dfrac{3}{10} \\[5ex] $ Selecting somone who swims but does not go to the funfair means that the person swims only

$ (d) \\[3ex] n(swim) = 13 \\[3ex] n(swim\;\;only) = 1 \\[3ex] P(swim\;\;only) = \dfrac{n(swim\;\;only)}{n(swim)} = \dfrac{1}{13} $

$ (a.) \\[3ex] n(\xi) = 135 \\[3ex] n(B \cap F) = 3 \\[3ex] n(B) = 15 \\[3ex] \implies \\[3ex] n(B\;\;ONLY) = n(B \cap F') \\[3ex] = 15 - 3 = 12 \\[3ex] n(B') = n(\xi) - n(B) \\[3ex] = 135 - 15 \\[3ex] = 120 \\[3ex] n(F \cap B') = \dfrac{1}{4} * 120 = 30 \\[5ex] n(B' \cap F') \\[3ex] = n(neither\;\;B\;\;nor\;\;F) \\[3ex] = 135 - (12 + 30 + 3) \\[3ex] = 135 - 45 \\[3ex] = 90 \\[3ex] $
$ (b.) \\[3ex] n(B) = 15 \\[3ex] n(\xi) = 135 \\[3ex] P(B) = \dfrac{n(B)}{n(\xi)} \\[5ex] = \dfrac{15}{135} \\[5ex] = \dfrac{1}{9} $

(a.) Kim chooses a chocolate and eats it.
Sam then chooses another chocolate and eats it.
This is a case of two selections, one at a time, without replacement

(b.) Choosing chocolates with different centres implies that:
Kim chose peppermint (P) AND Sam chose caramel (C)
OR
Kim chose caramel (C) AND Sam chose peppermint (P)

$ n(P) = 3 \\[3ex] n(C) = 5 \\[3ex] n(S) = 3 + 5 = 8 \\[3ex] \underline{Without\;\;Replacement\;\;condition} \\[3ex] P(P\;\;AND\;\;C) \;\;OR\;\; P(C\;\;AND\;\;P) \\[3ex] = \left(\dfrac{3}{8} * \dfrac{5}{7}\right) + \left(\dfrac{5}{8} * \dfrac{3}{7}\right) \\[5ex] = \dfrac{15}{56} + \dfrac{15}{56} \\[5ex] = \dfrac{15 + 15}{56} \\[5ex] = \dfrac{30}{56} \\[5ex] = \dfrac{15}{28} $

$ (a) \\[3ex] Let: \\[3ex] Mathematics = M \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] n(Physics\;\;ONLY) = k \\[3ex] n(Chemistry\;\;ONLY) = 2k \\[3ex] $
$ (i) \\[3ex] k + 2k + 13 + 2 + 2 + 5 + 3 = 40 \\[3ex] 3k + 25 = 40 \\[3ex] 3k = 40 - 25 \\[3ex] 3k = 15 \\[3ex] k = \dfrac{15}{3} \\[5ex] k = 5 \\[3ex] $ 5 students study only Physics

$ (ii) \\[3ex] n(only\;\;one\;\;subject) \\[3ex] = n(only\;\;Mathematics) + n(only\;\;Physics) + n(only\;\;Chemistry) \\[3ex] = 13 + k + 2k \\[3ex] = 13 + 5 + 2(5) \\[3ex] = 18 + 10 \\[3ex] = 28 \\[3ex] $ 28 students study only one subject

$ n(exactly\;\;two\;\;subjects) \\[3ex] = n(only\;\;Mathematics\;\;and\;\;Physics) + n(only\;\;Mathematics\;\;and\;\;Chemistry) + n(only\;\;Physics\;\;and\;\;Chemistry) \\[3ex] = 2 + 5 + 2 \\[3ex] = 9 \\[3ex] n(\xi) = 40 \\[3ex] P(exactly\;\;two\;\;subjects) \\[3ex] = \dfrac{n(exactly\;\;two\;\;subjects)}{n(\xi)} \\[5ex] = \dfrac{9}{40} $

$ \underline{First\;\;Scenario:\;\; Days} \\[3ex] n(S) = n(Days) = 7 \\[3ex] n(Tuesday) = 1 \\[3ex] P(Tuesday) = \dfrac{n(Tuesday)}{n(Days)} = \dfrac{1}{7} \\[5ex] \underline{Second\;\;Scenario:\;\;Months} \\[3ex] n(S) = n(Months) = 12 \\[3ex] n(January) = 1 \\[3ex] P(January) = \dfrac{n(January)}{n(Months)} = \dfrac{1}{12} \\[5ex] P(1st\;\;Scenario\;\;AND\;\;2nd\;\;Scenario) \\[3ex] = \dfrac{1}{7} * \dfrac{1}{12} ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{1}{84} $

$ (a) \\[3ex] Male\;\;teachers\;\;with\;\;blue\;\;eyes \\[3ex] = n(M \cap B) \\[3ex] = \dfrac{1}{3} * 42 \\[5ex] = 14 \\[3ex] Male\;\;teachers = n(M) = 42 \\[3ex] n(Male\;\;teachers\;\;only) = 42 - 14 = 28 \\[3ex] Female\;\;teachers\;\;with\;\;blue\;\;eyes \\[3ex] = \dfrac{1}{4} * 44 \\[5ex] = 11 \\[3ex] Teachers\;\;with\;\;blue\;\;eyes \\[3ex] = n(blue\;\;eyes) \\[3ex] = 14 + 11 = 25 \\[3ex] n(Blue\;\;eyes\;\;only) = 25 - 14 = 11 \\[3ex] Let\;\;n(neither\;\;male\;\;nor\;\;blue\;\;eyes) = k \\[3ex] 28 + 14 + 11 + k = 86 \\[3ex] 53 + k = 86 \\[3ex] k = 86 - 53 \\[3ex] k = 33 \\[3ex] $ The Venn Diagram is:


$ n(sample\;\;space) = n(B) = 25 \\[3ex] n(M \cap B) = 14 \\[3ex] P(M \cap B) = \dfrac{n(M \cap B)}{n(B)} = \dfrac{14}{25} $

$ n(Grade\;3) = 20 \\[3ex] n(1\;child\;\;in\;\;Grade\;3) = 1 \\[3ex] n(Grade\;4) = 15 \\[3ex] n(1\;child\;\;in\;\;Grade\;4) = 1 \\[3ex] P(1\;\;child\;\;Grade\;3\;\;AND\;\;1\;\;child\;\;Grade\;4) \\[3ex] = \dfrac{1}{20} * \dfrac{1}{15} \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{1}{300} $

$ S = \{2G, 4R, 3W, 1B\} \\[3ex] n(S) = 2 + 4 + 3 + 1 = 10 \\[3ex] n(G) = 2 \\[3ex] \underline{Without\:\: Replacement - Dependent\:\: Events} \\[3ex] P(G \:\:AND\:\: G) = P(GG) \\[3ex] P(GG) = \dfrac{2}{10} * \dfrac{1}{9} ...Multiplication\:\: Rule \\[5ex] P(GG) = \dfrac{1}{5} * \dfrac{1}{9} \\[5ex] P(GG) = \dfrac{1 * 1}{5 * 9} \\[5ex] P(GG) = \dfrac{1}{45} $

As noted, we are concerned with those three shaded areas
But, first; let us find the probability of neither A nor B
The sum of all the probabilities is 1

$ P(A \cup B)' = P(neither) \\[3ex] 0.3 + 0.15 + 0.35 + P(neither) = 1 \\[3ex] 0.8 + P(neither) = 1 \\[3ex] P(neither) = 1 - 0.8 \\[3ex] P(neither) = 0.2 \\[3ex] $ So, we need to add these shaded areas


$ P(A' \cup B) \\[3ex] = 0.15 + 0.35 + 0.2 \\[3ex] = 0.7 $

The set containing (1, 1) is indicated in red

Sample Space for Two Rolls of a Fair Die

First Roll in the Column and Second Roll in the Row

$2nd\:\:Roll\:\:\rightarrow$ $1st\:\:Roll\:\:\downarrow$	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$\color{red}{(1, 1)}$	$(1, 2)$	$(1, 3)$	$(1, 4)$	$(1, 5)$	$(1, 6)$
$2$	$(2, 1)$	$(2, 2)$	$(2, 3)$	$(2, 4)$	$(2, 5)$	$(2, 6)$
$3$	$(3, 1)$	$(3, 2)$	$(3, 3)$	$(3, 4)$	$(3, 5)$	$(3, 6)$
$4$	$(4, 1)$	$(4, 2)$	$(4, 3)$	$(4, 4)$	$(4, 5)$	$(4, 6)$
$5$	$(5, 1)$	$(5, 2)$	$(5, 3)$	$(5, 4)$	$(5, 5)$	$(5, 6)$
$6$	$(6, 1)$	$(6, 2)$	$(6, 3)$	$(6, 4)$	$(6, 5)$	$(6, 6)$

$ n(1, 1) = 1 \\[3ex] n(S) = 6 * 6 = 36 \\[3ex] P(1, 1) = \dfrac{n(1, 1)}{S} = \dfrac{1}{36} $

We can solve this question in at least two ways.
Use any approach you prefer.

First Method: Punnett Square
6 tosses of a fair coin
6 tosses = 4 tosses + 2 tosses
At most 3 heads = 3 heads or less
This implies: 3 heads, 2 heads, 1 head, and 0 head
These will be in dark blue color
Sample Space for the Six Tosses of 1 Fair Coin

4 tosses in the Column and 2 tosses in the Row

$2\:\:Tosses\:\:\rightarrow$ $4\:\:Tosses\:\:\downarrow$	$HH$	$HT$	$TH$	$TT$
$HHHH$	$HHHHHH$	$HHHHHT$	$HHHHTH$	$HHHHTT$
$HHHT$	$HHHTHH$	$HHHTHT$	$HHHTTH$	$HHHTTT$
$HHTH$	$HHTHHH$	$HHTHHT$	$HHTHTH$	$HHTHTT$
$HHTT$	$HHTTHH$	$HHTTHT$	$HHTTTH$	$HHTTTT$
$HTHH$	$HTHHHH$	$HTHHHT$	$HTHHTH$	$HTHHTT$
$HTHT$	$HTHTHH$	$HTHTHT$	$HTHTTH$	$HTHTTT$
$HTTH$	$HTTHHH$	$HTTHHT$	$HTTHTH$	$HTTHTT$
$HTTT$	$HTTTHH$	$HTTTHT$	$HTTTTH$	$HTTTTT$
$THHH$	$THHHHH$	$THHHHT$	$THHHTH$	$THHHTT$
$THHT$	$THHTHH$	$THHTHT$	$THHTTH$	$THHTTT$
$THTH$	$THTHHH$	$THTHHT$	$THTHTH$	$THTHTT$
$THTT$	$THTTHH$	$THTTHT$	$THTTTH$	$THTTTT$
$TTHH$	$TTHHHH$	$TTHHHT$	$TTHHTH$	$TTHHTT$
$TTHT$	$TTHTHH$	$TTHTHT$	$TTHTTH$	$TTHTTT$
$TTTH$	$TTTHHH$	$TTTHHT$	$TTTHTH$	$TTTHTT$
$TTTT$	$TTTTHH$	$TTTTHT$	$TTTTTH$	$TTTTTT$

$ n(S) = 16 * 4 = 64 \\[3ex] Let\;\;E = At\;\;most\;\;3H = dark\;blue\;\;color \\[3ex] n(E) = 42 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{42}{64} \\[5ex] = 0.65625 \\[3ex] $ You can review the Second Method to solve the question:
Question (3.) using Binomial Distribution Formula

We can solve this question in at least two ways.
Use any approach you prefer.

First Approach: Punnett Square
The sets containing exactly 2 tails are indicated in red

Sample Space for Three Tosses of a Fair Coin

Two Tosses in the Column and One Toss in the Row

$One\:\:Toss\:\:\rightarrow$ $Two\:\:Tosses\:\:\downarrow$	$H$	$T$
$HH$	$HHH$	$HHT$
$HT$	$HTH$	$\color{red}{HTT}$
$TH$	$THH$	$\color{red}{THT}$
$TT$	$\color{red}{TTH}$	$TTT$

$ n(exactly\;\;2\;\;tails) = 3 \\[3ex] n(S) = 4 * 2 = 8 \\[3ex] P(exactly\;\;2\;\;tails) \\[3ex] = \dfrac{n(exactly\;\;2\;\;tails)}{n(S)} \\[5ex] = \dfrac{3}{8} \\[5ex] $ You can review the Second Approach to solve the question:
Question (8.) using Binomial Distribution Formula

Let:
minivan = M
sedans = E
hatchbacks = H

Things to note:
(a.) Renting 1 of each of the 3 cars means that the cars could be rented in any order
This means that Thalia could rent it in any of these order:
M AND E AND H
OR
M AND H AND E
OR
E AND M AND H
OR
E AND H AND M
OR
H AND M AND E
OR
H AND E AND M

(b.) The renting of each car is done without replacement because:
When Thalia rents 1 car of a type from the 16 cars, she will need to rent 1 car of another type from the remaining 15 cars, then she will rent 1 car of the other type from the remaining 14 cars.

$ n(S) = 16 \\[3ex] n(M) = 6 \\[3ex] n(E) = 7 \\[3ex] n(H) = 3 \\[3ex] P(1\;\;car\;\;of\;\;each\;\;type) \\[3ex] = P(M-E-H) + P(M-H-E) + P(E-M-H) + P(E-H-M) + P(H-M-E) + P(H-E-M) \\[5ex] P(M-E-H) = \dfrac{6}{16} * \dfrac{7}{15} * \dfrac{3}{14} = \dfrac{3}{80} \\[5ex] P(M-H-E) = \dfrac{6}{16} * \dfrac{3}{15} * \dfrac{7}{14} = \dfrac{3}{80} \\[5ex] The\;\;rest\;\;of\;\;the\;\;probabilities\;\;are\;\;the\;\;same \\[3ex] But\;\;just\;\;to\;\;show\;\;you \\[3ex] P(E-M-H) = \dfrac{7}{16} * \dfrac{6}{15} * \dfrac{3}{14} = \dfrac{3}{80} \\[5ex] P(E-H-M) = \dfrac{7}{16} * \dfrac{3}{15} * \dfrac{6}{14} = \dfrac{3}{80} \\[5ex] P(H-M-E) = \dfrac{3}{16} * \dfrac{6}{15} * \dfrac{7}{14} = \dfrac{3}{80} \\[5ex] P(H-E-M) = \dfrac{3}{16} * \dfrac{7}{15} * \dfrac{6}{14} = \dfrac{3}{80} \\[5ex] P( = \dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} \\[5ex] = \dfrac{3}{80} * 6 \\[5ex] = \dfrac{3}{40} * 3 \\[5ex] = \dfrac{9}{40} $

When the 1st member (a senior) is selected, selecting the 2nd member is now dependent on the selection of the 1st member...Without Replacement condition

$ Let: \\[3ex] junior = J \\[3ex] senior = E \\[3ex] n(J) = 9 \\[3ex] n(E) = 11 \\[3ex] n(S) = 9 + 11 = 20 \\[3ex] P(J | E) = \dfrac{P(J \cap E)}{P(E)}...Conditional\;\;Probability \\[5ex] P(J \cap E) = P(J) * P(E) \\[3ex] = \dfrac{9}{20} * \dfrac{11}{19}...Dependent\;\;Events \\[5ex] = \dfrac{99}{380} \\[5ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{11}{20} \\[5ex] \implies \\[3ex] P(J | E) \\[3ex] = \dfrac{99}{380} \div \dfrac{11}{20} \\[5ex] = \dfrac{99}{380} * \dfrac{20}{11} \\[5ex] = \dfrac{9}{19} $

$ \mathcal{E} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\} \\[3ex] A = \{3, 9, 15, 21, 27\} \\[3ex] A' = \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29\} \\[3ex] B = \{5, 15, 25\} \\[3ex] B' = \{1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29\} \\[3ex] A \cap B = \{15\} \\[3ex] A \cup B = \{3, 5, 9, 15, 21, 25, 27\} \\[3ex] (A \cup B)' = \{1, 7, 11, 13, 17, 19, 23, 29\} \\[3ex] \underline{A\;\;only} \\[3ex] A \cap B' = \{3, 9, 21, 27\} \\[3ex] \underline{B\;\;only} \\[3ex] B \cap A' = \{5, 25\} \\[3ex] $ (a) The Venn diagram is:


$ (b) \\[3ex] A \cup B = \{3, 5, 9, 15, 21, 25, 27\} \\[3ex] n(A \cup B) = 7 \\[3ex] \mathcal{E} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\} \\[3ex] n(\mathcal{E}) = 15 \\[3ex] P(A \cup B) = \dfrac{n(A \cup B)}{n(\mathcal{E})} = \dfrac{7}{15} $

$ A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] n(A) = 9 \\[3ex] B \cap C = \{4, 8\} \\[3ex] n(B \cap C) = 2 \\[3ex] P(B \cap C) \\[3ex] = \dfrac{n(B \cap C)}{n(A)} \\[5ex] = \dfrac{2}{9} $

We can solve this in at least two ways
Choose whatever approach you prefer

$ Let: \\[3ex] red = R \\[3ex] blue = B \\[3ex] white = W \\[3ex] S = \{3R, 5B, 10W\} \\[3ex] n(S) = 3 + 5 + 10 = 18 \\[5ex] \underline{First\:\: Method: Quantitative\:\: Reasoning} \\[3ex] NOT\:\:White \implies R\:\:AND\:\:B \\[3ex] n(R) = 3 \\[3ex] n(B) = 5 \\[3ex] n(R\:\:AND\:\:B) = 3 + 5 = 8 \\[3ex] P(NOT\:\:White) = P(R\:\:AND\:\:B) = \dfrac{n(R\:\:AND\:\:B)}{n(S)} \\[5ex] P(NOT\:\:White) = \dfrac{8}{18} \\[5ex] P(NOT\:\:White) = \dfrac{4}{9} \\[5ex] \underline{Second\:\: Method: Complementary\:\:Rule} \\[3ex] NOT\:\:white = W' \\[3ex] P(White) + P(NOT\:\:White) = 1 ... Complementary\:\: Rule \\[3ex] P(W) + P(W') = 1 \\[3ex] n(W) = 10 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} = \dfrac{10}{18} \\[5ex] P(W') = 1 - P(W) \\[3ex] P(W') = 1 - \dfrac{10}{18} \\[5ex] P(W') = \dfrac{18}{18} - \dfrac{10}{18} \\[5ex] P(W') = \dfrac{18 - 10}{18} \\[5ex] P(W') = \dfrac{8}{18} \\[5ex] P(W') = \dfrac{4}{9} $

Things to NOTE:
(a.) Choosing two owners from the same age bracket implies choosing:
1 owner from ages 16 - 25 AND 1 owner from ages 16 - 25
OR
1 owner from ages 26 - 45 AND 1 owner from ages 26 - 45
OR
1 owner from ages 46 - 60 AND 1 owner from ages 46 - 60

(b.) Because an owner cannot be chosen twice, this is a case of Without Replacement

$ P(2\;\;owners\;\;from\;\;ages\;\;16 - 25) = \dfrac{80}{335} * \dfrac{79}{334} \\[5ex] P(2\;\;owners\;\;from\;\;ages\;\;26 - 45) = \dfrac{155}{335} * \dfrac{154}{334} \\[5ex] P(2\;\;owners\;\;from\;\;ages\;\;46 - 60) = \dfrac{100}{335} * \dfrac{99}{334} \\[5ex] \implies \\[3ex] P(2\;\;owners\;\;from\;\;same\;\;age\;\;bracket) \\[3ex] = \dfrac{80}{335} * \dfrac{79}{334} + \dfrac{155}{335} * \dfrac{154}{334} + \dfrac{100}{335} * \dfrac{99}{334} $

$ n(sample\;\;space) = n(P) = x + 35...number\;\;of\;\;Physics\;\;students \\[3ex] n(P \cap C) = x \\[3ex] P(P \cap C) = \dfrac{n(P \cap C)}{n(P)} = \dfrac{5}{12} \\[5ex] \implies \\[3ex] \dfrac{x}{x + 35} = \dfrac{5}{12} \\[5ex] 12x = 5(x + 35) \\[3ex] 12x = 5x + 175 \\[3ex] 12x - 5x = 175 \\[3ex] 7x = 175 \\[3ex] x = \dfrac{175}{7} \\[5ex] x = 25 \\[3ex] 47 + 35 + x + y = 150 \\[3ex] 82 + 25 + y = 150 \\[3ex] 107 + y = 150 \\[3ex] y = 150 - 107 \\[3ex] y = 43 \\[3ex] n(neither) = n(P \cup C)' = y = 43 \\[3ex] n(sample\;\;space) = n(\xi) = 150 \\[3ex] P(P \cup C)' = \dfrac{n(P \cup C)'}{n(\xi)} \\[5ex] = \dfrac{y}{150} \\[5ex] = \dfrac{43}{150} $

$ P(A\:\:\:AND\:\:\:B) = P(A) * P(B) \\[3ex] ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(A\:\:\:AND\:\:\:B) = 0.20(0.40) \\[3ex] P(A\:\:\:AND\:\:\:B) = 0.08 $

$ 30 - x + 40 - x + x + 20 = 80 \\[3ex] 90 - x = 80 \\[3ex] 90 - 80 = x \\[3ex] 10 = x \\[3ex] x = 10 \\[3ex] n(Music\;\;only) \\[3ex] = 30 - x \\[3ex] = 30 - 10 \\[3ex] = 20 \\[3ex] P(Music\;\;only) = \dfrac{n(Music\;\;only)}{n(\mu)} = \dfrac{20}{80} = \dfrac{1}{4} = 0.25 $

$ \varepsilon = \{2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] P = \{2, 4, 6, 8\} \\[3ex] Q = \{3, 6, 9\} \\[3ex] P \cap Q = \{6\} \\[3ex] P \cup Q = \{2, 3, 4, 6, 8, 9\} \\[3ex] (P \cup Q)' = \{5, 7\} \\[3ex] $ (a)
The Venn diagram is:


(b)
Let the set: R = {odd and not divisible by 3}

$ R = \{5, 7\} \\[3ex] n(R) = 2 \\[3ex] n(\varepsilon) = 8 \\[3ex] P(R) = \dfrac{n(R)}{n(\varepsilon)} = \dfrac{2}{8} = \dfrac{1}{4} $

It rained in Franklin City on Monday

The probability that it will rain there on Tuesday (the following day) is $0.70$

Let $T$ be the event that it will rain on Tuesday

This implies that $T'$ is the event that it will not rain on Tuesday... this is what we are asked to find

$ P(T) = 0.70 \\[3ex] P(T) + P(T') = 1 ...Complementary\:\:Rule \\[3ex] P(T') = 1 - P(T) \\[3ex] P(T') = 1 - 0.70 \\[3ex] P(T') = 0.30 $

$ 6 + 2x + 5 + (x + 2) = 31 \\[3ex] 11 + 2x + x + 2 = 31 \\[3ex] 3x + 13 = 31 \\[3ex] 3x = 31 - 13 \\[3ex] 3x = 18 \\[3ex] x = \dfrac{18}{3} \\[5ex] x = 6 \\[3ex] (a) \\[3ex] n(sample\;\;space) = n(\xi) = 31 \\[3ex] n(event\;\;space) = n(D) \\[3ex] = 2x + 5 \\[3ex] = 2(6) + 5 \\[3ex] = 12 + 5 \\[3ex] = 17 \\[3ex] P(D) = \dfrac{n(D)}{n(\xi)} = \dfrac{17}{31} \\[5ex] (b) \\[3ex] n(sample\;\;space) = n(C) = 6 + 5 = 11 \\[3ex] n(event\;\;space) = n(C \cap D) = 5 \\[3ex] P(C \cap D) = \dfrac{n(C \cap D)}{n(C)} = \dfrac{5}{11} $

$ Area\;\;of\;\;rectangular\;\;box = 6(13) = 78\;square\;\;inches \\[3ex] Area\;\;of\;\;painted\;\;square = 4(4) = 16\;square\;\;inches \\[3ex] P(bead\;\;falls\;\;on\;\;painted\;\;square) \\[3ex] = \dfrac{16}{78} \\[5ex] = 0.2051282051 \\[3ex] \approx 0.21 $

$ \underline{Given:\;\;in\;\;English} \\[3ex] n(\varepsilon) = 30 \\[3ex] n(all\;\;three\;\;fruits) = 2 \\[3ex] n(oranges\;\;but\;\;not\;\;apples\;\;or\;\;bananas) = 5 \\[3ex] n(oranges\;\;and\;\;bananas) = 6 \\[3ex] n(apples) = 17 \\[3ex] n(bananas) = 10 \\[3ex] \underline{Given:\;\;in\;\;Diagram} \\[3ex] n(apples\;\;and\;\;oranges\;\;but\;\;not\;\;bananas) = 8 \\[3ex] n(apples\;\;and\;\;bananas\;\;but\;\;not\;\;oranges) = 1 \\[3ex] \underline{Determine} \\[3ex] n(oranges\;\;and\;\;bananas\;\;but\;\;not\;\;apples) \\[3ex] = 6 - 2 \\[3ex] = 4 \\[3ex] n(apples\;\;but\;\;not\;\;oranges\;\;or\;\;bananas) \\[3ex] = 17 - (1 + 8 + 2) \\[3ex] = 17 - 11 \\[3ex] = 6 \\[3ex] n(bananas\;\;but\;\;not\;\;apples\;\;or\;\;oranges) \\[3ex] = 10 - (1 + 2 + 4) \\[3ex] = 10 - 7 \\[3ex] = 3 \\[3ex] n(neither\;\;apples\;\;nor\;\;oranges\;\;nor\;\;bananas) = p \\[3ex] 6 + 5 + 3 + 1 + 8 + 4 + 2 + p = 30 \\[3ex] 29 + p = 30 \\[3ex] p = 30 - 29 \\[3ex] p = 1 \\[3ex] $ (a) The Venn diagram is:


$ (b) \\[3ex] n(neither\;\;fruit) = p = 1 \\[3ex] n(\varepsilon) = 30 \\[3ex] P(neither\;\;fruit) = \dfrac{n(neither\;\;fruit)}{n(\varepsilon)} = \dfrac{1}{30} \\[5ex] (c) \\[3ex] \underline{From\;\;Apples} \\[3ex] n(apples) = 17 \\[3ex] n(apples\;\;but\;\;not\;\;oranges) = 6 + 1 = 7 \\[3ex] P(apples\;\;but\;\;not\;\;oranges) \\[3ex] = \dfrac{n(apples\;\;but\;\;not\;\;oranges)}{n(apples)} \\[5ex] = \dfrac{7}{17} $

$ n(\xi) = 120 \\[3ex] Let: \\[3ex] bags = B \\[3ex] shoes = H \\[3ex] n(B \cap H) = n(both) = 45 \\[3ex] n(B) = n(bags) = b \\[3ex] \rightarrow n(H) = n(shoes) = 11 + b \\[3ex] n(bags\;\;only) = b - 45 \\[3ex] n(shoes\;\;only) = (11 + b) - 45 = 11 + b - 45 = b - 34 \\[3ex] $ (a)
The Venn diagram is:


$ (b - 45) + (b - 34) + 45 = 120 \\[3ex] b - 45 + b - 34 + 45 = 120 \\[3ex] 2b = 120 + 34 \\[3ex] 2b = 154 \\[3ex] b = \dfrac{154}{2} \\[5ex] b = 77 \\[3ex] (b) \\[3ex] n(shoes) = n(H) \\[3ex] = (b - 34) + 45 \\[3ex] = b - 34 + 45 \\[3ex] = 77 - 34 + 45 \\[3ex] = 88 \\[3ex] $ 88 customers bought shoes

$ (c) \\[3ex] n(bags) = n(B) \\[3ex] = (b - 45) + 45 \\[3ex] = 77 - 45 + 45 \\[3ex] = 77 \\[3ex] n(\xi) = 120 \\[3ex] P(bags) = \dfrac{n(bags)}{n\xi} = \dfrac{77}{120} $

$ P(A\:\:\:OR\:\:\:B) = P(A) + P(B) \\[3ex] ...Addition\:\:Rule\:\:for\:\:Mutually\:\:Exclusive\:\:Events \\[3ex] P(A\:\:\:OR\:\:\:B) = P(A) + P(B) \\[3ex] P(A\:\:\:OR\:\:\:B) = 0.2 + 0.6 \\[3ex] P(A\:\:\:OR\:\:\:B) = 0.8 $