Baye's theorem Questions in English

(A) If $A, B,$ and $C$ are mutually independent,then $P(A \cap B) = P(A)P(B), P(A \cap C) = P(A)P(C),$ and $P(A \cap B \cap C) = P(A)P(B)P(C)$.
For $S_1$: $P(A \cap (B \cup C)) = P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C) = P(A)P(B) + P(A)P(C) - P(A)P(B)P(C) = P(A)[P(B) + P(C) - P(B)P(C)] = P(A)P(B \cup C)$. Thus,$S_1$ is true.
For $S_2$: $P(A \cap (B \cap C)) = P(A \cap B \cap C) = P(A)P(B)P(C) = P(A)P(B \cap C)$. Thus,$S_2$ is true.

(B) Let the total number of students be $100$.
Since girls constitute $60\%$ of the total students,the number of girls is $60$ and the number of boys is $100 - 60 = 40$.
Number of boys studying Mathematics $= 25\% \text{ of } 40 = \frac{25}{100} \times 40 = 10$.
Number of girls studying Mathematics $= 10\% \text{ of } 60 = \frac{10}{100} \times 60 = 6$.
Total number of students studying Mathematics $= 10 + 6 = 16$.
The probability that the student is a girl,given that the student is studying Mathematics,is calculated using Bayes' Theorem:
$P(\text{Girl} | \text{Maths}) = \frac{\text{Number of girls studying Maths}}{\text{Total number of students studying Maths}} = \frac{6}{16} = \frac{3}{8}$.

(B) Let $A_1$ be the event that the letter came from $LONDON$ and $A_2$ be the event that the letter came from $CLIFTON$. Since the source is not specified,we assume $P(A_1) = P(A_2) = \frac{1}{2}$.
Let $E$ be the event that the two consecutive letters $ON$ are legible on the postmark.
In $LONDON$ (length $6$),there are $5$ pairs of consecutive letters: $(LO, ON, ND, DO, ON)$. Out of these,$2$ are $ON$. Thus,$P(E|A_1) = \frac{2}{5}$.
In $CLIFTON$ (length $7$),there are $6$ pairs of consecutive letters: $(CL, LI, IF, FT, TO, ON)$. Out of these,$1$ is $ON$. Thus,$P(E|A_2) = \frac{1}{6}$.
Using Bayes' Theorem,the probability that it came from $LONDON$ given that $ON$ is legible is:
$P(A_1|E) = \frac{P(A_1)P(E|A_1)}{P(A_1)P(E|A_1) + P(A_2)P(E|A_2)}$
$P(A_1|E) = \frac{\frac{1}{2} \times \frac{2}{5}}{\frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{1}{6}}$
$P(A_1|E) = \frac{\frac{2}{5}}{\frac{2}{5} + \frac{1}{6}} = \frac{\frac{2}{5}}{\frac{12+5}{30}} = \frac{2}{5} \times \frac{30}{17} = \frac{12}{17}$.

(A) Let $E_1, E_2, E_3$ be the events of choosing Bag $I$,Bag $II$,and Bag $III$ respectively. Since the bag is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $A$ be the event that the drawn ball is black.
The probabilities of drawing a black ball from each bag are:
$P(A|E_1) = \frac{3}{2+3} = \frac{3}{5}$
$P(A|E_2) = \frac{1}{4+1} = \frac{1}{5}$
$P(A|E_3) = \frac{7}{3+7} = \frac{7}{10}$
We need to find the probability that the ball was drawn from the bag with the most black balls,which is Bag $III$. Using Bayes' Theorem:
$P(E_3|A) = \frac{P(E_3)P(A|E_3)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$
$P(E_3|A) = \frac{\frac{1}{3} \times \frac{7}{10}}{\frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{7}{10}}$
$P(E_3|A) = \frac{\frac{7}{30}}{\frac{6}{30} + \frac{2}{30} + \frac{7}{30}} = \frac{7}{15}$.

(A) Let $E$ be the event that a six occurs and $E'$ be the event that a six does not occur.
Let $A$ be the event that the man reports that it is a $6$.
We have $P(E) = \frac{1}{6}$ and $P(E') = 1 - \frac{1}{6} = \frac{5}{6}$.
The probability that the man speaks the truth is $P(T) = \frac{3}{4}$,so the probability that he lies is $P(L) = 1 - \frac{3}{4} = \frac{1}{4}$.
If a six occurs $(E)$,he reports it as $6$ if he speaks the truth. Thus,$P(A|E) = \frac{3}{4}$.
If a six does not occur $(E')$,he reports it as $6$ if he lies. Thus,$P(A|E') = \frac{1}{4}$.
Using Bayes' theorem,the probability that it is actually a six given that he reported a six is:
$P(E|A) = \frac{P(E) \times P(A|E)}{P(E) \times P(A|E) + P(E') \times P(A|E')}$
$P(E|A) = \frac{\frac{1}{6} \times \frac{3}{4}}{(\frac{1}{6} \times \frac{3}{4}) + (\frac{5}{6} \times \frac{1}{4})}$
$P(E|A) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8}$.

(D) Let $E_1$ be the event that the ball is drawn from bag $A$,$E_2$ be the event that it is drawn from bag $B$,and $E$ be the event that the drawn ball is red.
We need to find $P(E_2|E)$.
Since both bags are equally likely to be selected,we have $P(E_1) = P(E_2) = \frac{1}{2}$.
Also,the probability of drawing a red ball from bag $A$ is $P(E|E_1) = \frac{3}{5}$,and from bag $B$ is $P(E|E_2) = \frac{5}{9}$.
Using Bayes' theorem:
$P(E_2|E) = \frac{P(E_2) \cdot P(E|E_2)}{P(E_1) \cdot P(E|E_1) + P(E_2) \cdot P(E|E_2)}$
$P(E_2|E) = \frac{\frac{1}{2} \cdot \frac{5}{9}}{\frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{5}{9}}$
$P(E_2|E) = \frac{\frac{5}{18}}{\frac{3}{10} + \frac{5}{18}} = \frac{\frac{5}{18}}{\frac{27 + 25}{90}} = \frac{5}{18} \cdot \frac{90}{52} = \frac{5 \cdot 5}{52} = \frac{25}{52}$.

(C) This problem is solved using Bayes' Theorem.
Let $A$ be the event of selecting bag $A$,and $B$ be the event of selecting bag $B$.
$P(A) = \frac{1}{2}$,$P(B) = \frac{1}{2}$.
Let $G$ be the event of drawing a green ball.
Probability of drawing a green ball from bag $A$ is $P(G|A) = \frac{4}{7}$.
Probability of drawing a green ball from bag $B$ is $P(G|B) = \frac{3}{7}$.
By Bayes' Theorem,the probability that the ball was drawn from bag $B$ given that it is green is:
$P(B|G) = \frac{P(B) \times P(G|B)}{P(A) \times P(G|A) + P(B) \times P(G|B)}$
Substituting the values:
$P(B|G) = \frac{\frac{1}{2} \times \frac{3}{7}}{(\frac{1}{2} \times \frac{4}{7}) + (\frac{1}{2} \times \frac{3}{7})}$
$P(B|G) = \frac{\frac{3}{14}}{\frac{4}{14} + \frac{3}{14}} = \frac{\frac{3}{14}}{\frac{7}{14}} = \frac{3}{7}$.

(B) Let $E$ be the event that the selected student is a girl. Let $S_i$ be the event that school $B_i$ is selected for $i = 1, 2, 3, 4$.
Since the school is selected at random,$P(S_1) = P(S_2) = P(S_3) = P(S_4) = \frac{1}{4}$.
The conditional probabilities of selecting a girl from each school are $P(E|S_1) = \frac{12}{100}$,$P(E|S_2) = \frac{20}{100}$,$P(E|S_3) = \frac{13}{100}$,and $P(E|S_4) = \frac{17}{100}$.
By Bayes' Theorem,the probability that the school selected is $B_2$ given that the student is a girl is:
$P(S_2|E) = \frac{P(S_2)P(E|S_2)}{\sum_{i=1}^{4} P(S_i)P(E|S_i)}$
$P(S_2|E) = \frac{\frac{1}{4} \times \frac{20}{100}}{\frac{1}{4} \times (\frac{12}{100} + \frac{20}{100} + \frac{13}{100} + \frac{17}{100})}$
$P(S_2|E) = \frac{20}{12 + 20 + 13 + 17} = \frac{20}{62} = \frac{10}{31}$.

(A) Let $R_1$ be the event that a red ball is drawn from $A$ and placed in $B$,and $B_1$ be the event that a black ball is drawn from $A$ and placed in $B$.
$P(R_1) = \frac{6}{10} = \frac{3}{5}$,$P(B_1) = \frac{4}{10} = \frac{2}{5}$.
After transferring a ball,urn $B$ has $11$ balls.
If $R_1$ occurred,$B$ has $5$ red and $6$ black balls. $P(R_2|R_1) = \frac{5}{11}$,$P(B_2|R_1) = \frac{6}{11}$.
If $B_1$ occurred,$B$ has $4$ red and $7$ black balls. $P(R_2|B_1) = \frac{4}{11}$,$P(B_2|B_1) = \frac{7}{11}$.
After transferring back to $A$,urn $A$ has $10$ balls.
If $R_1$ and $R_2$ occurred,$A$ has $6$ red balls. $P(R|R_1, R_2) = \frac{6}{10}$.
If $R_1$ and $B_2$ occurred,$A$ has $5$ red balls. $P(R|R_1, B_2) = \frac{5}{10}$.
If $B_1$ and $R_2$ occurred,$A$ has $7$ red balls. $P(R|B_1, R_2) = \frac{7}{10}$.
If $B_1$ and $B_2$ occurred,$A$ has $6$ red balls. $P(R|B_1, B_2) = \frac{6}{10}$.
The total probability is $P(R) = P(R_1)P(R_2|R_1)P(R|R_1, R_2) + P(R_1)P(B_2|R_1)P(R|R_1, B_2) + P(B_1)P(R_2|B_1)P(R|B_1, R_2) + P(B_1)P(B_2|B_1)P(R|B_1, B_2)$.
$P(R) = (\frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}) + (\frac{6}{10} \times \frac{6}{11} \times \frac{5}{10}) + (\frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}) + (\frac{4}{10} \times \frac{7}{11} \times \frac{6}{10})$.
$P(R) = \frac{180 + 180 + 112 + 168}{1100} = \frac{640}{1100} = \frac{64}{110} = \frac{32}{55}$.

(A) Let $E_1$ be the event of choosing bag $A$ and $E_2$ be the event of choosing bag $B$.
Let $R$ be the event of drawing a red ball.
Since the bags are chosen at random,$P(E_1) = P(E_2) = 1/2$.
The probability of drawing a red ball from bag $A$ is $P(R|E_1) = 3/5$.
The probability of drawing a red ball from bag $B$ is $P(R|E_2) = 5/9$.
Using Bayes' Theorem,the probability that the ball was drawn from bag $B$ given that it is red is:
$P(E_2|R) = \frac{P(E_2)P(R|E_2)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)}$
$P(E_2|R) = \frac{(1/2) \times (5/9)}{(1/2) \times (3/5) + (1/2) \times (5/9)}$
$P(E_2|R) = \frac{5/18}{3/10 + 5/18} = \frac{5/18}{(27+25)/90} = \frac{5/18}{52/90}$
$P(E_2|R) = \frac{5}{18} \times \frac{90}{52} = \frac{5 \times 5}{52} = \frac{25}{52}$.

(D) Let $H$ be the event that the coin shows heads and $T$ be the event that the coin shows tails. $P(H) = P(T) = 1/2$.
Let $W$ be the event that the ball drawn from $U_2$ is white.
If $H$ occurs,$1$ ball is transferred from $U_1$ to $U_2$. $U_1$ has $3W, 2R$. Probability of transferring $W$ is $3/5$,and $R$ is $2/5$.
If $W$ is transferred,$U_2$ has $2W$. $P(W|H_W) = 1$. If $R$ is transferred,$U_2$ has $1W, 1R$. $P(W|H_R) = 1/2$.
$P(W|H) = (3/5 \times 1) + (2/5 \times 1/2) = 3/5 + 1/5 = 4/5$.
If $T$ occurs,$2$ balls are transferred from $U_1$ to $U_2$. Possible transfers: $2W$ (prob $^3C_2/^5C_2 = 3/10$),$1W, 1R$ (prob $(^3C_1 \times ^2C_1)/^5C_2 = 6/10$),$2R$ (prob $^2C_2/^5C_2 = 1/10$).
If $2W$ transferred,$U_2$ has $3W$. $P(W|T_{2W}) = 1$.
If $1W, 1R$ transferred,$U_2$ has $2W, 1R$. $P(W|T_{1W1R}) = 2/3$.
If $2R$ transferred,$U_2$ has $1W, 2R$. $P(W|T_{2R}) = 1/3$.
$P(W|T) = (3/10 \times 1) + (6/10 \times 2/3) + (1/10 \times 1/3) = 3/10 + 4/10 + 1/30 = 9/30 + 12/30 + 1/30 = 22/30 = 11/15$.
Using Bayes' Theorem: $P(H|W) = \frac{P(W|H)P(H)}{P(W|H)P(H) + P(W|T)P(T)} = \frac{(4/5 \times 1/2)}{(4/5 \times 1/2) + (11/15 \times 1/2)} = \frac{4/10}{4/10 + 11/30} = \frac{12/30}{12/30 + 11/30} = 12/23$.

(B) Bag $A$ contains $2$ white,$3$ black,$4$ red balls (Total = $9$).
Bag $B$ contains $3$ white,$4$ black,$5$ red balls (Total = $12$).
Let $W_A, B_A, R_A$ be the events of drawing a white,black,or red ball from bag $A$ respectively.
$P(W_A) = 2/9, P(B_A) = 3/9, P(R_A) = 4/9$.
After transferring a ball to bag $B$,bag $B$ now contains $13$ balls.
If $W_A$ occurs,bag $B$ has $4$ white balls. $P(W_B|W_A) = 4/13$.
If $B_A$ occurs,bag $B$ has $3$ white balls. $P(W_B|B_A) = 3/13$.
If $R_A$ occurs,bag $B$ has $3$ white balls. $P(W_B|R_A) = 3/13$.
Using the Law of Total Probability:
$P(W_B) = P(W_A)P(W_B|W_A) + P(B_A)P(W_B|B_A) + P(R_A)P(W_B|R_A)$
$P(W_B) = (2/9 \times 4/13) + (3/9 \times 3/13) + (4/9 \times 3/13)$
$P(W_B) = (8 + 9 + 12) / 117 = 29/117$.

(D) Let $E_1$ be the event of selecting bag $A$ and $E_2$ be the event of selecting bag $B$.
Let $R$ be the event of selecting a red ball.
We have $P(E_1) = P(E_2) = \frac{1}{2}$.
The probability of drawing a red ball from bag $A$ is $P(R|E_1) = \frac{3}{2+3} = \frac{3}{5}$.
The probability of drawing a red ball from bag $B$ is $P(R|E_2) = \frac{5}{4+5} = \frac{5}{9}$.
Using Bayes' theorem,the probability that the red ball was drawn from bag $B$ is given by:
$P(E_2|R) = \frac{P(E_2) \times P(R|E_2)}{P(E_1) \times P(R|E_1) + P(E_2) \times P(R|E_2)}$
$P(E_2|R) = \frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{9}}$
$P(E_2|R) = \frac{\frac{5}{18}}{\frac{3}{10} + \frac{5}{18}} = \frac{\frac{5}{18}}{\frac{27+25}{90}} = \frac{5}{18} \times \frac{90}{52} = \frac{5 \times 5}{52} = \frac{25}{52}$.

(NONE) Let $H$ be the event that the coin shows head and $T$ be the event that the coin shows tail. $P(H) = P(T) = 1/2$.
Let $W$ be the event that the ball drawn from $U_2$ is white.
Case $1$: If $H$ occurs,$1$ ball is transferred from $U_1$ to $U_2$. $U_1$ has $3W, 2R$. The probability of transferring a white ball is $3/5$ and a red ball is $2/5$.
If $W$ is transferred,$U_2$ now has $2W$. $P(W|H, \text{trans } W) = 1$.
If $R$ is transferred,$U_2$ now has $1W, 1R$. $P(W|H, \text{trans } R) = 1/2$.
$P(W|H) = (3/5 \times 1) + (2/5 \times 1/2) = 3/5 + 1/5 = 4/5$.
Case $2$: If $T$ occurs,$2$ balls are transferred from $U_1$ to $U_2$. The possible pairs are $(2W), (1W, 1R), (0W, 2R)$.
$P(2W) = \binom{3}{2} / \binom{5}{2} = 3/10$.
$P(1W, 1R) = (\binom{3}{1} \times \binom{2}{1}) / \binom{5}{2} = 6/10$.
$P(0W, 2R) = \binom{2}{2} / \binom{5}{2} = 1/10$.
If $2W$ transferred,$U_2$ has $3W$. $P(W|T, 2W) = 1$.
If $1W, 1R$ transferred,$U_2$ has $2W, 1R$. $P(W|T, 1W, 1R) = 2/3$.
If $0W, 2R$ transferred,$U_2$ has $1W, 2R$. $P(W|T, 0W, 2R) = 1/3$.
$P(W|T) = (3/10 \times 1) + (6/10 \times 2/3) + (1/10 \times 1/3) = 3/10 + 4/10 + 1/30 = 7/10 + 1/30 = 22/30 = 11/15$.
Using Bayes' Theorem: $P(H|W) = \frac{P(H)P(W|H)}{P(H)P(W|H) + P(T)P(W|T)} = \frac{1/2 \times 4/5}{1/2 \times 4/5 + 1/2 \times 11/15} = \frac{2/5}{2/5 + 11/30} = \frac{12/30}{12/30 + 11/30} = 12/23$.

(B) Let $A_1$ be the event that the missing card is black,and $A_2$ be the event that the missing card is red. Let $E$ be the event that the first $13$ cards examined are all red.
We need to find $P(A_1|E)$.
Initially,there are $26$ red and $26$ black cards in a pack of $52$. Since one card is missing,$P(A_1) = P(A_2) = \frac{1}{2}$.
If $A_1$ occurs (missing card is black),there are $26$ red and $25$ black cards left. The probability of picking $13$ red cards is $P(E|A_1) = \frac{^{26}C_{13}}{^{51}C_{13}}$.
If $A_2$ occurs (missing card is red),there are $25$ red and $26$ black cards left. The probability of picking $13$ red cards is $P(E|A_2) = \frac{^{25}C_{13}}{^{51}C_{13}}$.
Using Bayes' Theorem:
$P(A_1|E) = \frac{P(E|A_1)P(A_1)}{P(E|A_1)P(A_1) + P(E|A_2)P(A_2)}$
$P(A_1|E) = \frac{\frac{1}{2} \cdot \frac{^{26}C_{13}}{^{51}C_{13}}}{\frac{1}{2} \cdot \frac{^{26}C_{13}}{^{51}C_{13}} + \frac{1}{2} \cdot \frac{^{25}C_{13}}{^{51}C_{13}}}$
$P(A_1|E) = \frac{^{26}C_{13}}{^{26}C_{13} + ^{25}C_{13}} = \frac{\frac{26}{13} \cdot ^{25}C_{12}}{^{26}C_{13} + ^{25}C_{13}} = \frac{2 \cdot ^{25}C_{12}}{2 \cdot ^{25}C_{12} + ^{25}C_{13}} = \frac{2 \cdot ^{25}C_{12}}{2 \cdot ^{25}C_{12} + \frac{13}{13} \cdot ^{25}C_{13}} = \frac{2}{2 + \frac{13}{13}} = \frac{2}{3}$.

(C) Let $E_{1}$ be the event of selecting a bag from the first group and $E_{2}$ be the event of selecting a bag from the second group.
Let $W$ be the event of drawing a white ball.
There are $3$ bags in the first group and $2$ bags in the second group,total $5$ bags.
$P(E_{1}) = \frac{3}{5}$ and $P(E_{2}) = \frac{2}{5}$.
In the first group,each bag has $5$ white and $3$ black balls,so $P(W|E_{1}) = \frac{5}{5+3} = \frac{5}{8}$.
In the second group,each bag has $2$ white and $4$ black balls,so $P(W|E_{2}) = \frac{2}{2+4} = \frac{2}{6} = \frac{1}{3}$.
Using Bayes' Theorem,the probability that the white ball is from the first group is:
$P(E_{1}|W) = \frac{P(E_{1})P(W|E_{1})}{P(E_{1})P(W|E_{1}) + P(E_{2})P(W|E_{2})}$
$P(E_{1}|W) = \frac{\frac{3}{5} \times \frac{5}{8}}{\frac{3}{5} \times \frac{5}{8} + \frac{2}{5} \times \frac{1}{3}}$
$P(E_{1}|W) = \frac{\frac{3}{8}}{\frac{3}{8} + \frac{2}{15}} = \frac{\frac{3}{8}}{\frac{45+16}{120}} = \frac{3}{8} \times \frac{120}{61} = \frac{3 \times 15}{61} = \frac{45}{61}$.

(D) Let $E_1$ be the event of choosing coin $A$ and $E_2$ be the event of choosing coin $B$. Since one coin is chosen randomly,$P(E_1) = 1/2$ and $P(E_2) = 1/2$.
Let $H$ be the event that the coin shows heads on both tosses.
Given $P(H|E_1) = (1/4)^2 = 1/16$ and $P(H|E_2) = (3/4)^2 = 9/16$.
Using Bayes' theorem,the probability that the coin is $A$ given that it showed heads twice is:
$P(E_1|H) = \frac{P(E_1)P(H|E_1)}{P(E_1)P(H|E_1) + P(E_2)P(H|E_2)}$
$P(E_1|H) = \frac{(1/2)(1/16)}{(1/2)(1/16) + (1/2)(9/16)}$
$P(E_1|H) = \frac{1/32}{1/32 + 9/32} = \frac{1/32}{10/32} = 1/10$.

(A) Let $E$ be the event that one white and one red ball are drawn. Let $H_A$ and $H_B$ be the events of selecting box $A$ and box $B$ respectively. $P(H_A) = P(H_B) = \frac{1}{2}$.
The probability of drawing one white and one red ball from box $A$ is $P(E|H_A) = \frac{^2C_1 \times ^3C_1}{^7C_2} = \frac{2 \times 3}{21} = \frac{6}{21} = \frac{2}{7}$.
The probability of drawing one white and one red ball from box $B$ is $P(E|H_B) = \frac{^4C_1 \times ^2C_1}{^9C_2} = \frac{4 \times 2}{36} = \frac{8}{36} = \frac{2}{9}$.
Using Bayes' Theorem,the probability that the balls were drawn from box $B$ given event $E$ is:
$P(H_B|E) = \frac{P(H_B)P(E|H_B)}{P(H_A)P(E|H_A) + P(H_B)P(E|H_B)}$
$P(H_B|E) = \frac{\frac{1}{2} \times \frac{2}{9}}{\frac{1}{2} \times \frac{2}{7} + \frac{1}{2} \times \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{2}{7} + \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{18+14}{63}} = \frac{2}{9} \times \frac{63}{32} = \frac{1}{1} \times \frac{7}{16} = \frac{7}{16}$.

(B) Let the initial composition of the two balls in the urn be represented by the number of white balls $W$. Since each ball can be white or black,the possible initial states are $0$ white balls $(BB)$,$1$ white ball $(BW)$,or $2$ white balls $(WW)$. Assuming each state is equally likely,the prior probability for each is $P(S_i) = \frac{1}{3}$.
After adding one white ball,the new compositions are:
$1) BB \to BBW$ (Number of white balls = $1$,Total = $3$)
$2) BW \to BWW$ (Number of white balls = $2$,Total = $3$)
$3) WW \to WWW$ (Number of white balls = $3$,Total = $3$)
The probability of drawing a white ball given the state $S_i$ is $P(W|S_i) = \frac{\text{white balls}}{3}$.
$P(W|S_1) = \frac{1}{3}$,$P(W|S_2) = \frac{2}{3}$,$P(W|S_3) = \frac{3}{3} = 1$.
Using the law of total probability: $P(W) = \sum P(W|S_i)P(S_i) = \frac{1}{3} \times (\frac{1}{3} + \frac{2}{3} + 1) = \frac{1}{3} \times (\frac{6}{3}) = \frac{2}{3}$.

(C) Let $H$ be the event of getting a head and $T$ be the event of getting a tail. $P(H) = \frac{1}{2}$ and $P(T) = \frac{1}{2}$.
Case $1$: If head occurs,two dice are rolled. The sum can be $7$ or $8$. The pairs for sum $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ ($6$ outcomes) and for sum $8$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes). Total outcomes = $36$.
$P(7 \text{ or } 8 | H) = \frac{6+5}{36} = \frac{11}{36}$.
Case $2$: If tail occurs,a card is picked from $9$ cards. The favorable numbers are $7$ and $8$.
$P(7 \text{ or } 8 | T) = \frac{2}{9}$.
Total probability $P(7 \text{ or } 8) = P(H) \times P(7 \text{ or } 8 | H) + P(T) \times P(7 \text{ or } 8 | T)$
$= \frac{1}{2} \times \frac{11}{36} + \frac{1}{2} \times \frac{2}{9} = \frac{11}{72} + \frac{1}{9} = \frac{11+8}{72} = \frac{19}{72}$.

(A) The given data can be tabulated as follows:

Type	True / False | Multiple Choice | Total
Easy	$300$ | $500$ | $800$
Difficult	$200$ | $400$ | $600$
Total	$500$ | $900$ | $1400$

Let $E$ be the event that the question is easy and $M$ be the event that the question is a multiple choice question.
Total number of multiple choice questions $= 900$.
Number of easy multiple choice questions $= 500$.
We need to find the conditional probability $P(E | M)$,which is the probability that the question is easy given that it is a multiple choice question.
Using the formula for conditional probability:
$P(E | M) = \frac{n(E \cap M)}{n(M)}$
Where $n(E \cap M)$ is the number of easy multiple choice questions and $n(M)$ is the total number of multiple choice questions.
$P(E | M) = \frac{500}{900} = \frac{5}{9}$.
Therefore,the required probability is $\frac{5}{9}$.

(A) Let $E_1$ be the event of choosing Bag $I$,$E_2$ be the event of choosing Bag $II$,and $A$ be the event of drawing a red ball.
Then $P(E_1) = P(E_2) = \frac{1}{2}$.
The probability of drawing a red ball from Bag $I$ is $P(A|E_1) = \frac{3}{7}$.
The probability of drawing a red ball from Bag $II$ is $P(A|E_2) = \frac{5}{11}$.
We need to find $P(E_2|A)$. By Bayes' theorem:
$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
Substituting the values:
$P(E_2|A) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}}$
$P(E_2|A) = \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}} = \frac{\frac{5}{22}}{\frac{33 + 35}{154}} = \frac{5}{22} \times \frac{154}{68} = \frac{5 \times 7}{68} = \frac{35}{68}$.

(A) Let $E_1, E_2$,and $E_3$ be the events that boxes $I, II$,and $III$ are chosen,respectively.
Since the boxes are chosen at random,$P(E_1) = P(E_2) = P(E_3) = 1/3$.
Let $A$ be the event that the coin drawn is gold.
Then,the conditional probabilities are:
$P(A|E_1) = 2/2 = 1$ (both coins are gold in box $I$).
$P(A|E_2) = 0/2 = 0$ (no gold coins in box $II$).
$P(A|E_3) = 1/2$ (one gold coin in box $III$).
We want to find the probability that the other coin is also gold,which means we must have chosen box $I$.
By Bayes' theorem,$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$.
Substituting the values:
$P(E_1|A) = \frac{(1/3 \times 1)}{(1/3 \times 1) + (1/3 \times 0) + (1/3 \times 1/2)} = \frac{1/3}{1/3 + 1/6} = \frac{1/3}{3/6} = \frac{1/3}{1/2} = 2/3$.

(A) Let $E$ denote the event that the person selected actually has $HIV$ and $A$ be the event that the person's $HIV$ test is diagnosed as positive. We need to find $P(E|A)$. Also,$E^{\prime}$ denotes the event that the person selected does not have $HIV$.
Clearly,$\{E, E^{\prime}\}$ is a partition of the sample space.
We are given:
$P(E) = 0.1\% = \frac{0.1}{100} = 0.001$
$P(E^{\prime}) = 1 - P(E) = 0.999$
$P(A|E) = P(\text{Person tested as } HIV \text{ positive given that he/she has } HIV) = 90\% = 0.9$
$P(A|E^{\prime}) = P(\text{Person tested as } HIV \text{ positive given that he/she does not have } HIV) = 1\% = 0.01$
By Bayes' theorem:
$P(E|A) = \frac{P(E) P(A|E)}{P(E) P(A|E) + P(E^{\prime}) P(A|E^{\prime})}$
$P(E|A) = \frac{0.001 \times 0.9}{(0.001 \times 0.9) + (0.999 \times 0.01)}$
$P(E|A) = \frac{0.0009}{0.0009 + 0.00999} = \frac{0.0009}{0.01089} = \frac{90}{1089} \approx 0.083$
Thus,the probability that the person actually has $HIV$ is approximately $0.083$.

(A) Let events $B_1, B_2, B_3$ be defined as follows:
$B_1$: The bolt is manufactured by machine $A$.
$B_2$: The bolt is manufactured by machine $B$.
$B_3$: The bolt is manufactured by machine $C$.
Clearly,$B_1, B_2, B_3$ are mutually exclusive and exhaustive events.
Let $E$ be the event that the bolt is defective.
Given probabilities:
$P(B_1) = 0.25, P(B_2) = 0.35, P(B_3) = 0.40$.
Conditional probabilities of defective bolts:
$P(E|B_1) = 0.05, P(E|B_2) = 0.04, P(E|B_3) = 0.02$.
By Bayes' Theorem,the probability that the bolt is manufactured by machine $B$ given it is defective is:
$P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(B_1)P(E|B_1) + P(B_2)P(E|B_2) + P(B_3)P(E|B_3)}$
$P(B_2|E) = \frac{0.35 \times 0.04}{(0.25 \times 0.05) + (0.35 \times 0.04) + (0.40 \times 0.02)}$
$P(B_2|E) = \frac{0.0140}{0.0125 + 0.0140 + 0.0080} = \frac{0.0140}{0.0345}$
$P(B_2|E) = \frac{140}{345} = \frac{28}{69}$.

(A) Let $E$ be the event that the doctor arrives late. Let $T_1, T_2, T_3,$ and $T_4$ be the events that the doctor comes by train,bus,scooter,and other means of transport,respectively.
Given probabilities:
$P(T_1) = \frac{3}{10}, P(T_2) = \frac{1}{5}, P(T_3) = \frac{1}{10}, P(T_4) = \frac{2}{5}$.
Conditional probabilities of being late:
$P(E|T_1) = \frac{1}{4}, P(E|T_2) = \frac{1}{3}, P(E|T_3) = \frac{1}{12}, P(E|T_4) = 0$.
Using Bayes' Theorem,the probability that he comes by train given he is late is:
$P(T_1|E) = \frac{P(T_1)P(E|T_1)}{\sum_{i=1}^{4} P(T_i)P(E|T_i)}$
Denominator calculation:
$P(E) = (\frac{3}{10} \times \frac{1}{4}) + (\frac{1}{5} \times \frac{1}{3}) + (\frac{1}{10} \times \frac{1}{12}) + (\frac{2}{5} \times 0)$
$P(E) = \frac{3}{40} + \frac{1}{15} + \frac{1}{120} + 0 = \frac{9 + 8 + 1}{120} = \frac{18}{120} = \frac{3}{20}$.
Numerator calculation:
$P(T_1)P(E|T_1) = \frac{3}{10} \times \frac{1}{4} = \frac{3}{40}$.
Result:
$P(T_1|E) = \frac{3/40}{3/20} = \frac{3}{40} \times \frac{20}{3} = \frac{1}{2}$.

(A) Let $E$ be the event that the man reports that a six occurs.
Let $S_{1}$ be the event that a six actually occurs.
Let $S_{2}$ be the event that a six does not occur.
We have:
$P(S_{1}) = 1/6$
$P(S_{2}) = 5/6$
$P(E|S_{1})$ is the probability that the man reports a six when a six has occurred (he speaks the truth) $= 3/4$.
$P(E|S_{2})$ is the probability that the man reports a six when a six has not occurred (he lies) $= 1 - 3/4 = 1/4$.
By Bayes' theorem,the probability that it is actually a six given that he reports a six is:
$P(S_{1}|E) = \frac{P(S_{1})P(E|S_{1})}{P(S_{1})P(E|S_{1}) + P(S_{2})P(E|S_{2})}$
$P(S_{1}|E) = \frac{(1/6) \times (3/4)}{(1/6) \times (3/4) + (5/6) \times (1/4)}$
$P(S_{1}|E) = \frac{3/24}{3/24 + 5/24} = \frac{3/24}{8/24} = 3/8$.
Thus,the required probability is $3/8$.

(A) Let $E_{1}$ and $E_{2}$ be the events of selecting the first bag and the second bag respectively.
$P(E_{1}) = P(E_{2}) = 1/2$.
Let $A$ be the event of drawing a red ball.
$P(A | E_{1}) = \text{Probability of drawing a red ball from the first bag} = 4/8 = 1/2$.
$P(A | E_{2}) = \text{Probability of drawing a red ball from the second bag} = 2/8 = 1/4$.
We need to find the probability that the ball was drawn from the first bag,given that it is red,which is $P(E_{1} | A)$.
By Bayes' theorem:
$P(E_{1} | A) = \frac{P(E_{1}) \cdot P(A | E_{1})}{P(E_{1}) \cdot P(A | E_{1}) + P(E_{2}) \cdot P(A | E_{2})}$
$P(E_{1} | A) = \frac{(1/2) \cdot (1/2)}{(1/2) \cdot (1/2) + (1/2) \cdot (1/4)}$
$P(E_{1} | A) = \frac{1/4}{1/4 + 1/8} = \frac{1/4}{3/8} = \frac{1}{4} \cdot \frac{8}{3} = 2/3$.

(A) Let $E_{1}$ be the event that the student is a hostler and $E_{2}$ be the event that the student is a day scholar. Let $A$ be the event that the chosen student gets an $A$ grade.
Given probabilities:
$P(E_{1}) = 60 \% = 0.6$
$P(E_{2}) = 40 \% = 0.4$
$P(A | E_{1}) = 30 \% = 0.3$
$P(A | E_{2}) = 20 \% = 0.2$
We need to find $P(E_{1} | A)$. By Bayes' theorem:
$P(E_{1} | A) = \frac{P(E_{1}) \cdot P(A | E_{1})}{P(E_{1}) \cdot P(A | E_{1}) + P(E_{2}) \cdot P(A | E_{2})}$
Substituting the values:
$P(E_{1} | A) = \frac{0.6 \times 0.3}{(0.6 \times 0.3) + (0.4 \times 0.2)}$
$P(E_{1} | A) = \frac{0.18}{0.18 + 0.08}$
$P(E_{1} | A) = \frac{0.18}{0.26}$
$P(E_{1} | A) = \frac{18}{26} = \frac{9}{13}$

(A) Let $E_{1}$ and $E_{2}$ be the respective events that the student knows the answer and he guesses the answer.
Let $A$ be the event that the answer is correct.
Given probabilities are $P(E_{1}) = \frac{3}{4}$ and $P(E_{2}) = \frac{1}{4}$.
The probability that the student answers correctly given that he knows the answer is $P(A|E_{1}) = 1$.
The probability that the student answers correctly given that he guesses is $P(A|E_{2}) = \frac{1}{4}$.
We need to find $P(E_{1}|A)$. By Bayes' theorem:
$P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$
$P(E_{1}|A) = \frac{\frac{3}{4} \times 1}{(\frac{3}{4} \times 1) + (\frac{1}{4} \times \frac{1}{4})}$
$P(E_{1}|A) = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} = \frac{\frac{3}{4}}{\frac{12+1}{16}} = \frac{\frac{3}{4}}{\frac{13}{16}}$
$P(E_{1}|A) = \frac{3}{4} \times \frac{16}{13} = \frac{12}{13}$.

(A) Let $E_{1}$ be the event that a person has the disease and $E_{2}$ be the event that a person does not have the disease.
Given that $P(E_{1}) = 0.1 \% = 0.001$.
Since $E_{1}$ and $E_{2}$ are complementary events,$P(E_{2}) = 1 - P(E_{1}) = 1 - 0.001 = 0.999$.
Let $A$ be the event that the blood test result is positive.
$P(A | E_{1}) = 99 \% = 0.99$ (Probability of positive result given the person has the disease).
$P(A | E_{2}) = 0.5 \% = 0.005$ (Probability of positive result given the person is healthy).
We need to find $P(E_{1} | A)$. By Bayes' theorem:
$P(E_{1} | A) = \frac{P(E_{1}) \cdot P(A | E_{1})}{P(E_{1}) \cdot P(A | E_{1}) + P(E_{2}) \cdot P(A | E_{2})}$
Substituting the values:
$P(E_{1} | A) = \frac{0.001 \times 0.99}{(0.001 \times 0.99) + (0.999 \times 0.005)}$
$= \frac{0.00099}{0.00099 + 0.004995}$
$= \frac{0.00099}{0.005985}$
$= \frac{990}{5985}$
$= \frac{22}{133}$

(A) Let $E_{1}$,$E_{2}$,and $E_{3}$ be the events of choosing a two-headed coin,a biased coin,and an unbiased coin,respectively.
Since one coin is chosen at random,$P(E_{1}) = P(E_{2}) = P(E_{3}) = \frac{1}{3}$.
Let $A$ be the event that the coin shows heads.
For a two-headed coin,$P(A|E_{1}) = 1$.
For a biased coin,$P(A|E_{2}) = 75\% = \frac{75}{100} = \frac{3}{4}$.
For an unbiased coin,$P(A|E_{3}) = \frac{1}{2}$.
Using Bayes' theorem,the probability that the coin is two-headed given that it shows heads is $P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2}) + P(E_{3})P(A|E_{3})}$.
Substituting the values: $P(E_{1}|A) = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}$.
$P(E_{1}|A) = \frac{\frac{1}{3}}{\frac{1}{3} (1 + \frac{3}{4} + \frac{1}{2})} = \frac{1}{\frac{4+3+2}{4}} = \frac{1}{\frac{9}{4}} = \frac{4}{9}$.

(A) Let $E_{1}, E_{2},$ and $E_{3}$ be the events that the driver is a scooter driver,a car driver,and a truck driver,respectively.
Let $A$ be the event that the person meets with an accident.
Total number of drivers $= 2000 + 4000 + 6000 = 12000$.
$P(E_{1}) = \frac{2000}{12000} = \frac{1}{6}$.
$P(E_{2}) = \frac{4000}{12000} = \frac{1}{3}$.
$P(E_{3}) = \frac{6000}{12000} = \frac{1}{2}$.
Given probabilities of accidents are $P(A|E_{1}) = 0.01 = \frac{1}{100}$,$P(A|E_{2}) = 0.03 = \frac{3}{100}$,and $P(A|E_{3}) = 0.15 = \frac{15}{100}$.
Using Bayes' theorem,the probability that the driver is a scooter driver given he met with an accident is $P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2}) + P(E_{3})P(A|E_{3})}$.
$P(E_{1}|A) = \frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{1}{2} \times \frac{15}{100}}$.
$P(E_{1}|A) = \frac{\frac{1}{600}}{\frac{1}{600} + \frac{3}{300} + \frac{15}{200}} = \frac{\frac{1}{600}}{\frac{1 + 6 + 45}{600}} = \frac{1}{52}$.

(B) Let $E_{1}$ and $E_{2}$ be the events that the item is produced by machines $A$ and $B$ respectively. Let $X$ be the event that the item is defective.
Given:
$P(E_{1}) = 60 \% = 60/100 = 3/5$
$P(E_{2}) = 40 \% = 40/100 = 2/5$
$P(X | E_{1}) = 2 \% = 2/100$
$P(X | E_{2}) = 1 \% = 1/100$
We need to find $P(E_{2} | X)$. By Bayes' theorem:
$P(E_{2} | X) = \frac{P(E_{2}) \cdot P(X | E_{2})}{P(E_{1}) \cdot P(X | E_{1}) + P(E_{2}) \cdot P(X | E_{2})}$
Substituting the values:
$P(E_{2} | X) = \frac{(2/5) \cdot (1/100)}{(3/5) \cdot (2/100) + (2/5) \cdot (1/100)}$
$P(E_{2} | X) = \frac{2/500}{6/500 + 2/500}$
$P(E_{2} | X) = \frac{2/500}{8/500}$
$P(E_{2} | X) = 2/8 = 1/4$

(A) Let $E_{1}$ and $E_{2}$ be the events that the first group and the second group win the competition,respectively.
Let $A$ be the event of introducing a new product.
Given:
$P(E_{1}) = 0.6$
$P(E_{2}) = 0.4$
$P(A | E_{1}) = 0.7$
$P(A | E_{2}) = 0.3$
We need to find the probability that the new product was introduced by the second group,which is $P(E_{2} | A)$.
By Bayes' theorem:
$P(E_{2} | A) = \frac{P(E_{2}) P(A | E_{2})}{P(E_{1}) P(A | E_{1}) + P(E_{2}) P(A | E_{2})}$
Substituting the values:
$P(E_{2} | A) = \frac{0.4 \times 0.3}{(0.6 \times 0.7) + (0.4 \times 0.3)}$
$P(E_{2} | A) = \frac{0.12}{0.42 + 0.12}$
$P(E_{2} | A) = \frac{0.12}{0.54}$
$P(E_{2} | A) = \frac{12}{54} = \frac{2}{9}$

(A) Let $E_{1}$ be the event that the outcome on the die is $5$ or $6$ and $E_{2}$ be the event that the outcome on the die is $1, 2, 3,$ or $4$.
$P(E_{1}) = \frac{2}{6} = \frac{1}{3}$ and $P(E_{2}) = \frac{4}{6} = \frac{2}{3}$.
Let $A$ be the event of getting exactly one head.
$P(A | E_{1})$ is the probability of getting exactly one head by tossing a coin three times: $\binom{3}{1} (\frac{1}{2})^1 (\frac{1}{2})^2 = \frac{3}{8}$.
$P(A | E_{2})$ is the probability of getting exactly one head in a single throw of a coin: $\frac{1}{2}$.
We need to find $P(E_{2} | A)$. By Bayes' theorem:
$P(E_{2} | A) = \frac{P(E_{2}) P(A | E_{2})}{P(E_{1}) P(A | E_{1}) + P(E_{2}) P(A | E_{2})}$
$P(E_{2} | A) = \frac{(\frac{2}{3}) (\frac{1}{2})}{(\frac{1}{3}) (\frac{3}{8}) + (\frac{2}{3}) (\frac{1}{2})}$
$P(E_{2} | A) = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{3+8}{24}} = \frac{\frac{1}{3}}{\frac{11}{24}} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}$.

(A) Let $E_1, E_2,$ and $E_3$ be the events that the items are produced by operators $A, B,$ and $C$ respectively.
$P(E_1) = 50\% = 0.5, P(E_2) = 30\% = 0.3, P(E_3) = 20\% = 0.2$.
Let $X$ be the event that the item produced is defective.
$P(X|E_1) = 1\% = 0.01, P(X|E_2) = 5\% = 0.05, P(X|E_3) = 7\% = 0.07$.
We need to find $P(E_1|X)$. By Bayes' theorem:
$P(E_1|X) = \frac{P(E_1)P(X|E_1)}{P(E_1)P(X|E_1) + P(E_2)P(X|E_2) + P(E_3)P(X|E_3)}$
$P(E_1|X) = \frac{0.5 \times 0.01}{(0.5 \times 0.01) + (0.3 \times 0.05) + (0.2 \times 0.07)}$
$P(E_1|X) = \frac{0.005}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034} = \frac{5}{34}$.

(A) Let $E_{1}$ be the event that the lost card is a diamond and $E_{2}$ be the event that the lost card is not a diamond.
Let $A$ be the event that two cards drawn from the remaining $51$ cards are both diamonds.
Out of $52$ cards,$13$ are diamonds and $39$ are not diamonds.
$P(E_{1}) = \frac{13}{52} = \frac{1}{4}$ and $P(E_{2}) = \frac{39}{52} = \frac{3}{4}$.
If $E_{1}$ occurs,there are $12$ diamonds left in $51$ cards. The probability of drawing $2$ diamonds is $P(A|E_{1}) = \frac{^{12}C_{2}}{^{51}C_{2}} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550}$.
If $E_{2}$ occurs,there are $13$ diamonds left in $51$ cards. The probability of drawing $2$ diamonds is $P(A|E_{2}) = \frac{^{13}C_{2}}{^{51}C_{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550}$.
Using Bayes' theorem,$P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$.
$P(E_{1}|A) = \frac{\frac{1}{4} \times \frac{132}{2550}}{\frac{1}{4} \times \frac{132}{2550} + \frac{3}{4} \times \frac{156}{2550}} = \frac{132}{132 + 3 \times 156} = \frac{132}{132 + 468} = \frac{132}{600}$.
Dividing by $12$,we get $\frac{11}{50}$.

(B) Let $H$ be the event that a head appears and $T$ be the event that a tail appears.
Let $R_H$ be the event that $A$ reports a head.
We are given $P(\text{Truth}) = \frac{4}{5}$ and $P(\text{False}) = \frac{1}{5}$.
If a head appears,$A$ reports a head if $A$ speaks the truth. Thus,$P(R_H | H) = \frac{4}{5}$.
If a tail appears,$A$ reports a head if $A$ lies. Thus,$P(R_H | T) = \frac{1}{5}$.
Since the coin is fair,$P(H) = \frac{1}{2}$ and $P(T) = \frac{1}{2}$.
We want to find $P(H | R_H)$. By Bayes' Theorem:
$P(H | R_H) = \frac{P(R_H | H) P(H)}{P(R_H | H) P(H) + P(R_H | T) P(T)}$
$P(H | R_H) = \frac{(\frac{4}{5}) \cdot (\frac{1}{2})}{(\frac{4}{5}) \cdot (\frac{1}{2}) + (\frac{1}{5}) \cdot (\frac{1}{2})}$
$P(H | R_H) = \frac{\frac{4}{10}}{\frac{4}{10} + \frac{1}{10}} = \frac{4/10}{5/10} = \frac{4}{5}$.
Thus,the correct option is $B$.

(A) Let $A$ be the event that a black ball is selected,and $E_1, E_2, E_3, E_4$ be the events that boxes $I, II, III, IV$ are selected respectively.
Since the boxes are chosen at random,$P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}$.
The probabilities of drawing a black ball from each box are:
$P(A|E_1) = \frac{3}{3+4+5+6} = \frac{3}{18} = \frac{1}{6}$
$P(A|E_2) = \frac{2}{2+2+2+2} = \frac{2}{8} = \frac{1}{4}$
$P(A|E_3) = \frac{1}{1+2+3+1} = \frac{1}{7}$
$P(A|E_4) = \frac{4}{4+3+1+5} = \frac{4}{13}$
Using Bayes' theorem,the probability that the ball is from box $III$ given it is black is:
$P(E_3|A) = \frac{P(E_3)P(A|E_3)}{\sum_{i=1}^{4} P(E_i)P(A|E_i)}$
$P(E_3|A) = \frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4}(\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13})} = \frac{\frac{1}{7}}{\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}}$
Calculating the denominator: $\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13} = \frac{364 + 546 + 312 + 672}{2184} = \frac{1894}{2184} \approx 0.8672$
$P(E_3|A) = \frac{1/7}{1894/2184} = \frac{312}{1894} \approx 0.16475 \approx 0.165$.

(A) Let $B_{1}$ be the event that the machine is correctly set up and $B_{2}$ be the event that the machine is incorrectly set up.
Let $A$ be the event that the machine produces $2$ acceptable items.
Given:
$P(B_{1}) = 0.8$
$P(B_{2}) = 1 - 0.8 = 0.2$
$P(A|B_{1}) = 0.9 \times 0.9 = 0.81$
$P(A|B_{2}) = 0.4 \times 0.4 = 0.16$
Using Bayes' Theorem,the probability that the machine is correctly set up given it produced $2$ acceptable items is:
$P(B_{1}|A) = \frac{P(B_{1})P(A|B_{1})}{P(B_{1})P(A|B_{1}) + P(B_{2})P(A|B_{2})}$
$P(B_{1}|A) = \frac{0.8 \times 0.81}{(0.8 \times 0.81) + (0.2 \times 0.16)}$
$P(B_{1}|A) = \frac{0.648}{0.648 + 0.032} = \frac{0.648}{0.680} = \frac{648}{680} = \frac{81}{85} \approx 0.9529$

(A) Let $M$ be the event that the selected person is male and $W$ be the event that the selected person is female. Since there are equal numbers of males and females,$P(M) = P(W) = 0.5$.
Let $G$ be the event that the selected person has grey hair.
Given: $P(G|M) = 0.05$ and $P(G|W) = 0.0025$.
We need to find $P(M|G)$.
Using Bayes' Theorem:
$P(M|G) = \frac{P(G|M)P(M)}{P(G|M)P(M) + P(G|W)P(W)}$
$P(M|G) = \frac{0.05 \times 0.5}{(0.05 \times 0.5) + (0.0025 \times 0.5)}$
$P(M|G) = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525}$
$P(M|G) = \frac{500}{525} = \frac{20}{21}$

(A) Let $E_A, E_B, E_C,$ and $E_D$ be the events of selecting boxes $A, B, C,$ and $D$ respectively. Since one box is selected at random,$P(E_A) = P(E_B) = P(E_C) = P(E_D) = \frac{1}{4}$.
Let $R$ be the event of drawing a red marble.
The conditional probabilities of drawing a red marble from each box are:
$P(R|E_A) = \frac{1}{1+6+3} = \frac{1}{10}$
$P(R|E_B) = \frac{6}{6+2+2} = \frac{6}{10}$
$P(R|E_C) = \frac{8}{8+1+1} = \frac{8}{10}$
$P(R|E_D) = \frac{0}{0+6+4} = 0$
By the Law of Total Probability,$P(R) = P(E_A)P(R|E_A) + P(E_B)P(R|E_B) + P(E_C)P(R|E_C) + P(E_D)P(R|E_D)$
$P(R) = \frac{1}{4} \times \frac{1}{10} + \frac{1}{4} \times \frac{6}{10} + \frac{1}{4} \times \frac{8}{10} + \frac{1}{4} \times 0 = \frac{1+6+8}{40} = \frac{15}{40} = \frac{3}{8}$.
Using Bayes' Theorem,the probability that the red marble was drawn from box $X$ is $P(E_X|R) = \frac{P(E_X)P(R|E_X)}{P(R)}$.
For box $A$: $P(E_A|R) = \frac{(1/4)(1/10)}{3/8} = \frac{1/40}{3/8} = \frac{1}{40} \times \frac{8}{3} = \frac{1}{15}$.
For box $B$: $P(E_B|R) = \frac{(1/4)(6/10)}{3/8} = \frac{6/40}{3/8} = \frac{6}{40} \times \frac{8}{3} = \frac{2}{5}$.
For box $C$: $P(E_C|R) = \frac{(1/4)(8/10)}{3/8} = \frac{8/40}{3/8} = \frac{8}{40} \times \frac{8}{3} = \frac{8}{15}$.

(A) Let $A$ be the event that the patient has a heart attack. Let $E_{1}$ be the event that the patient chooses the meditation and yoga course,and $E_{2}$ be the event that the patient chooses the drug prescription.
Given $P(E_{1}) = P(E_{2}) = \frac{1}{2}$.
The initial probability of a heart attack is $0.40$.
If the patient chooses $E_{1}$,the risk is reduced by $30 \%$,so the new probability is $P(A|E_{1}) = 0.40 \times (1 - 0.30) = 0.40 \times 0.70 = 0.28$.
If the patient chooses $E_{2}$,the risk is reduced by $25 \%$,so the new probability is $P(A|E_{2}) = 0.40 \times (1 - 0.25) = 0.40 \times 0.75 = 0.30$.
Using Bayes' Theorem,the probability that the patient followed the meditation and yoga course given they had a heart attack is:
$P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$
$P(E_{1}|A) = \frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30} = \frac{0.28}{0.28 + 0.30} = \frac{0.28}{0.58} = \frac{28}{58} = \frac{14}{29}$.

(A) Let $E_{1}$ be the event that a red ball is transferred from Bag $I$ to Bag $II$,and $E_{2}$ be the event that a black ball is transferred from Bag $I$ to Bag $II$.
The probabilities are $P(E_{1}) = \frac{3}{7}$ and $P(E_{2}) = \frac{4}{7}$.
Let $A$ be the event that the ball drawn from Bag $II$ is red.
If $E_{1}$ occurs,Bag $II$ now contains $5$ red and $5$ black balls. Thus,$P(A|E_{1}) = \frac{5}{10} = \frac{1}{2}$.
If $E_{2}$ occurs,Bag $II$ now contains $4$ red and $6$ black balls. Thus,$P(A|E_{2}) = \frac{4}{10} = \frac{2}{5}$.
We need to find $P(E_{2}|A)$. By Bayes' Theorem:
$P(E_{2}|A) = \frac{P(E_{2})P(A|E_{2})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$
$P(E_{2}|A) = \frac{(\frac{4}{7}) \times (\frac{2}{5})}{(\frac{3}{7}) \times (\frac{1}{2}) + (\frac{4}{7}) \times (\frac{2}{5})}$
$P(E_{2}|A) = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}} = \frac{\frac{8}{35}}{\frac{15+16}{70}} = \frac{8}{35} \times \frac{70}{31} = \frac{16}{31}$.

(C) Let $E_1$ be the event that the missing card is a spade,and $E_2$ be the event that the missing card is not a spade.
$P(E_1) = \frac{13}{52} = \frac{1}{4}$ and $P(E_2) = \frac{39}{52} = \frac{3}{4}$.
Let $A$ be the event that two cards drawn from the remaining $51$ cards are spades.
If $E_1$ occurs,there are $12$ spades left in $51$ cards. Thus,$P(A|E_1) = \frac{^{12}C_2}{^{51}C_2} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550}$.
If $E_2$ occurs,there are $13$ spades left in $51$ cards. Thus,$P(A|E_2) = \frac{^{13}C_2}{^{51}C_2} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550}$.
We need to find $P(E_2|A)$. By Bayes' Theorem:
$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
$P(E_2|A) = \frac{\frac{3}{4} \times \frac{156}{2550}}{\frac{1}{4} \times \frac{132}{2550} + \frac{3}{4} \times \frac{156}{2550}} = \frac{3 \times 156}{132 + 3 \times 156} = \frac{468}{132 + 468} = \frac{468}{600} = \frac{39}{50}$.

(A) Let $B_{1}$ be the event that Box-$I$ is selected and $B_{2}$ be the event that Box-$II$ is selected.
$P(B_{1}) = P(B_{2}) = \frac{1}{2}$.
Let $E$ be the event that the selected card is a non-prime number.
In Box-$I$ (cards $1$ to $30$),the prime numbers are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$ ($10$ primes). Thus,there are $30 - 10 = 20$ non-prime numbers. So,$P(E|B_{1}) = \frac{20}{30} = \frac{2}{3}$.
In Box-$II$ (cards $31$ to $50$),the prime numbers are $\{31, 37, 41, 43, 47\}$ ($5$ primes). Thus,there are $20 - 5 = 15$ non-prime numbers. So,$P(E|B_{2}) = \frac{15}{20} = \frac{3}{4}$.
Using Bayes' Theorem,the probability that the card was drawn from Box-$I$ given it is non-prime is:
$P(B_{1}|E) = \frac{P(B_{1})P(E|B_{1})}{P(B_{1})P(E|B_{1}) + P(B_{2})P(E|B_{2})}$
$P(B_{1}|E) = \frac{\frac{1}{2} \times \frac{2}{3}}{\frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{3}{4}} = \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} = \frac{\frac{1}{3}}{\frac{8+9}{24}} = \frac{1}{3} \times \frac{24}{17} = \frac{8}{17}$.

(C) Let the events be defined as follows:
$A$: The person chosen is a smoker and non-vegetarian.
$B$: The person chosen is a smoker and vegetarian.
$C$: The person chosen is a non-smoker and vegetarian.
$E$: The person chosen has a chest disorder.
Given probabilities:
$P(A) = \frac{160}{400}$,$P(B) = \frac{100}{400}$,$P(C) = \frac{140}{400}$.
Conditional probabilities of having the disorder:
$P(E|A) = \frac{35}{100}$,$P(E|B) = \frac{20}{100}$,$P(E|C) = \frac{10}{100}$.
Using Bayes' Theorem,we need to find $P(A|E)$:
$P(A|E) = \frac{P(A) \cdot P(E|A)}{P(A) \cdot P(E|A) + P(B) \cdot P(E|B) + P(C) \cdot P(E|C)}$
Substituting the values:
$P(A|E) = \frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100} + \frac{100}{400} \times \frac{20}{100} + \frac{140}{400} \times \frac{10}{100}}$
$P(A|E) = \frac{160 \times 35}{(160 \times 35) + (100 \times 20) + (140 \times 10)}$
$P(A|E) = \frac{5600}{5600 + 2000 + 1400} = \frac{5600}{9000} = \frac{56}{90} = \frac{28}{45}$.

(C) Let $E_1$ be the event of selecting Bag $A$ and $E_2$ be the event of selecting Bag $B$.
$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $A$ be the event that the drawn balls are $1$ red and $1$ black.
For Bag $A$ (total $6$ balls: $2W, 1B, 3R$): $P(A|E_1) = \frac{{}^3C_1 \times {}^1C_1}{{}^6C_2} = \frac{3 \times 1}{15} = \frac{1}{5}$.
For Bag $B$ (total $n+5$ balls: $nW, 3B, 2R$): $P(A|E_2) = \frac{{}^2C_1 \times {}^3C_1}{{}^{n+5}C_2} = \frac{6}{\frac{(n+5)(n+4)}{2}} = \frac{12}{(n+5)(n+4)}$.
Using Bayes' Theorem: $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac{6}{11}$.
$\frac{\frac{1}{2} \times \frac{1}{5}}{\frac{1}{2} \times \frac{1}{5} + \frac{1}{2} \times \frac{12}{(n+5)(n+4)}} = \frac{6}{11}$.
$\frac{\frac{1}{5}}{\frac{1}{5} + \frac{12}{(n+5)(n+4)}} = \frac{6}{11}$.
$11 = 6 + \frac{60}{(n+5)(n+4)} \times 5 \Rightarrow 5 = \frac{60}{(n+5)(n+4)} \Rightarrow (n+5)(n+4) = 12$.
Since $n$ must be a positive integer,we test values. For $n=4$,$(9)(8) = 72 \neq 12$. Re-evaluating: $\frac{1/10}{1/10 + 6/((n+5)(n+4))} = \frac{6}{11} \Rightarrow 1.1 = 0.6 + \frac{60}{(n+5)(n+4)} \Rightarrow 0.5 = \frac{60}{(n+5)(n+4)} \Rightarrow (n+5)(n+4) = 120$.
Solving $n^2 + 9n + 20 = 120 \Rightarrow n^2 + 9n - 100 = 0$. This does not yield an integer. Checking the calculation: $P(A|E_1) = 3/15 = 1/5$. $P(A|E_2) = 6/((n+5)(n+4)/2) = 12/((n+5)(n+4))$. $\frac{1/10}{1/10 + 6/((n+5)(n+4))} = \frac{6}{11} \Rightarrow \frac{1}{1 + 60/((n+5)(n+4))} = \frac{6}{11} \Rightarrow 11 = 6 + \frac{360}{(n+5)(n+4)} \Rightarrow 5 = \frac{360}{(n+5)(n+4)} \Rightarrow (n+5)(n+4) = 72$.
$(n+5)(n+4) = 9 \times 8 \Rightarrow n+4 = 8 \Rightarrow n = 4$.

(B) Let $E_1, E_2, E_3$ be the events that the transferred ball is red,black,or white respectively.
$P(E_1) = \frac{3}{10}, P(E_2) = \frac{4}{10}, P(E_3) = \frac{3}{10}$.
Let $A$ be the event that the ball drawn from Bag $II$ is black.
If $E_1$ occurs,Bag $II$ has $3$ red,$5$ black,$2$ white balls. $P(A|E_1) = \frac{5}{10}$.
If $E_2$ occurs,Bag $II$ has $2$ red,$6$ black,$2$ white balls. $P(A|E_2) = \frac{6}{10}$.
If $E_3$ occurs,Bag $II$ has $2$ red,$5$ black,$3$ white balls. $P(A|E_3) = \frac{5}{10}$.
By Bayes' Theorem,$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$.
$P(E_1|A) = \frac{(\frac{3}{10} \times \frac{5}{10})}{(\frac{3}{10} \times \frac{5}{10}) + (\frac{4}{10} \times \frac{6}{10}) + (\frac{3}{10} \times \frac{5}{10})} = \frac{15}{15 + 24 + 15} = \frac{15}{54} = \frac{5}{18}$.