Baye's theorem Questions in English

(A) Given that $A, B, C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$.
Given $P(A) = 0.30$ and $P(B) = 0.50$,we find $P(C) = 1 - (0.30 + 0.50) = 0.20$.
The conditional probabilities are $P(E \mid A) = 0.6$,$P(E \mid B) = 0.3$,and $P(E \mid C) = 0.1$.
Using Bayes' Theorem,the probability $P(C \mid E)$ is given by:
$P(C \mid E) = \frac{P(C) P(E \mid C)}{P(A) P(E \mid A) + P(B) P(E \mid B) + P(C) P(E \mid C)}$
Substituting the values:
$P(C \mid E) = \frac{0.20 \times 0.1}{(0.30 \times 0.6) + (0.50 \times 0.3) + (0.20 \times 0.1)}$
$P(C \mid E) = \frac{0.02}{0.18 + 0.15 + 0.02} = \frac{0.02}{0.35} = \frac{2}{35}$.

(A) Let $E_1$ be the event that a spade is drawn,and $E_2$ be the event that a spade is not drawn. Then $P(E_1) = \frac{13}{52} = \frac{1}{4}$ and $P(E_2) = 1 - \frac{1}{4} = \frac{3}{4}$.
Let $S$ be the event that two balls of the same colour are drawn.
For bag $A$ ($6$ Green,$8$ Red,total $14$): $P(S|E_1) = \frac{{}^6C_2 + {}^8C_2}{{}^{14}C_2} = \frac{15 + 28}{91} = \frac{43}{91}$.
For bag $B$ ($9$ Green,$5$ Red,total $14$): $P(S|E_2) = \frac{{}^9C_2 + {}^5C_2}{{}^{14}C_2} = \frac{36 + 10}{91} = \frac{46}{91}$.
Using Bayes' Theorem,the probability that the balls were drawn from bag $A$ given they are the same colour is:
$P(E_1|S) = \frac{P(E_1)P(S|E_1)}{P(E_1)P(S|E_1) + P(E_2)P(S|E_2)}$
$P(E_1|S) = \frac{\frac{1}{4} \times \frac{43}{91}}{\frac{1}{4} \times \frac{43}{91} + \frac{3}{4} \times \frac{46}{91}} = \frac{43}{43 + 3 \times 46} = \frac{43}{43 + 138} = \frac{43}{181}$.

(A) Let $G$ be the event that the selected ball is green. Let $A, B, C$ be the events of selecting boxes $A, B, C$ respectively.
The probabilities of selecting the boxes are:
$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{6}$
The conditional probabilities of selecting a green ball from each box are:
$P(G|A) = \frac{2}{1+2+3} = \frac{2}{6} = \frac{1}{3}$
$P(G|B) = \frac{3}{2+3+1} = \frac{3}{6} = \frac{1}{2}$
$P(G|C) = \frac{1}{3+1+2} = \frac{1}{6}$
Using the law of total probability,the probability of selecting a green ball $P(G)$ is:
$P(G) = P(A)P(G|A) + P(B)P(G|B) + P(C)P(G|C)$
$P(G) = (\frac{1}{2} \times \frac{1}{3}) + (\frac{1}{3} \times \frac{1}{2}) + (\frac{1}{6} \times \frac{1}{6})$
$P(G) = \frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{6+6+1}{36} = \frac{13}{36}$
Using Bayes' theorem,the probability that the green ball was selected from box $C$ is $P(C|G)$:
$P(C|G) = \frac{P(C)P(G|C)}{P(G)}$
$P(C|G) = \frac{\frac{1}{6} \times \frac{1}{6}}{\frac{13}{36}} = \frac{\frac{1}{36}}{\frac{13}{36}} = \frac{1}{13}$

(D) Let $E$ be the event that the person reports that the die shows $6$.
Let $S$ be the event that the die actually shows $6$,and $S^c$ be the event that the die does not show $6$.
$P(S) = \frac{1}{6}$ and $P(S^c) = \frac{5}{6}$.
Let $T$ be the event that the person speaks the truth. $P(T) = \frac{3}{5}$ and $P(T^c) = \frac{2}{5}$.
$P(E|S)$ is the probability that he reports $6$ given that it is $6$,which is $P(T) = \frac{3}{5}$.
$P(E|S^c)$ is the probability that he reports $6$ given that it is not $6$,which is $P(T^c) = \frac{2}{5}$.
By Bayes' Theorem,$P(S|E) = \frac{P(S)P(E|S)}{P(S)P(E|S) + P(S^c)P(E|S^c)}$.
$P(S|E) = \frac{(\frac{1}{6} \times \frac{3}{5})}{(\frac{1}{6} \times \frac{3}{5}) + (\frac{5}{6} \times \frac{2}{5})} = \frac{\frac{3}{30}}{\frac{3}{30} + \frac{10}{30}} = \frac{3}{13}$.

(A) Let $W$ denote white marbles and $B$ denote black marbles. Bag $P$ contains $5W$ and $3B$. Four marbles are transferred to bag $Q$.
Let $E_1$ be the event that $1W$ and $3B$ are transferred.
Let $E_2$ be the event that $2W$ and $2B$ are transferred.
Let $E_3$ be the event that $3W$ and $1B$ are transferred.
Let $E_4$ be the event that $4W$ and $0B$ are transferred.
Let $A$ be the event that a black marble is drawn from bag $Q$.
The total number of ways to choose $4$ marbles from $8$ is $^8C_4 = 70$.
$P(E_1) = \frac{^5C_1 \times ^3C_3}{70} = \frac{5 \times 1}{70} = \frac{5}{70}$
$P(E_2) = \frac{^5C_2 \times ^3C_2}{70} = \frac{10 \times 3}{70} = \frac{30}{70}$
$P(E_3) = \frac{^5C_3 \times ^3C_1}{70} = \frac{10 \times 3}{70} = \frac{30}{70}$
$P(E_4) = \frac{^5C_4 \times ^3C_0}{70} = \frac{5 \times 1}{70} = \frac{5}{70}$
The conditional probabilities of drawing a black marble from $Q$ are:
$P(A|E_1) = \frac{3}{4}, P(A|E_2) = \frac{2}{4}, P(A|E_3) = \frac{1}{4}, P(A|E_4) = 0$.
Using Bayes' theorem,we find $P(E_1|A)$:
$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{\sum_{i=1}^4 P(E_i)P(A|E_i)}$
$P(E_1|A) = \frac{\frac{5}{70} \times \frac{3}{4}}{\frac{5}{70} \times \frac{3}{4} + \frac{30}{70} \times \frac{2}{4} + \frac{30}{70} \times \frac{1}{4} + \frac{5}{70} \times 0}$
$P(E_1|A) = \frac{15}{15 + 60 + 30 + 0} = \frac{15}{105} = \frac{1}{7}$.

(A) Let $E$ be the event that the die shows a six,and $E^c$ be the event that the die does not show a six.
Let $A$ be the event that the person reports that it is a six.
We are given:
$P(E) = \frac{1}{6}$
$P(E^c) = 1 - \frac{1}{6} = \frac{5}{6}$
Probability of speaking the truth $P(T) = \frac{3}{4}$,so probability of lying $P(L) = 1 - \frac{3}{4} = \frac{1}{4}$.
If the die shows a six,the person reports a six if he speaks the truth: $P(A|E) = \frac{3}{4}$.
If the die does not show a six,the person reports a six if he lies: $P(A|E^c) = \frac{1}{4}$.
Using Bayes' Theorem,the probability that it is actually a six given that he reported a six is:
$P(E|A) = \frac{P(E) \times P(A|E)}{P(E) \times P(A|E) + P(E^c) \times P(A|E^c)}$
$P(E|A) = \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}}$
$P(E|A) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8}$.

(C) Let $M$ be the event that the employee is a man and $W$ be the event that the employee is a woman. Let $T$ be the event that the employee is a technical assistant.
Given:
$P(M) = 0.70$
$P(W) = 1 - 0.70 = 0.30$
$P(T|M) = 0.30$
$P(T|W) = 0.15$
We need to find $P(M|T)$.
Using Bayes' Theorem:
$P(M|T) = \frac{P(M) \times P(T|M)}{P(M) \times P(T|M) + P(W) \times P(T|W)}$
$P(M|T) = \frac{0.70 \times 0.30}{(0.70 \times 0.30) + (0.30 \times 0.15)}$
$P(M|T) = \frac{0.21}{0.21 + 0.045}$
$P(M|T) = \frac{0.21}{0.255}$
$P(M|T) = \frac{210}{255} = \frac{14}{17}$

(A) Let $E_1, E_2, E_3$ be the events that the bulb is produced by units $A, B, C$ respectively. Let $D$ be the event that the bulb is defective.
Given probabilities are:
$P(E_1) = 0.25, P(E_2) = 0.35, P(E_3) = 0.40$
$P(D|E_1) = 0.05, P(D|E_2) = 0.04, P(D|E_3) = 0.02$
Using the Law of Total Probability,the probability that a bulb is defective is:
$P(D) = P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)$
$P(D) = (0.25 \times 0.05) + (0.35 \times 0.04) + (0.40 \times 0.02)$
$P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345$
Using Bayes' Theorem,the probability that the defective bulb was produced by unit $B$ is:
$P(E_2|D) = \frac{P(E_2)P(D|E_2)}{P(D)}$
$P(E_2|D) = \frac{0.35 \times 0.04}{0.0345} = \frac{0.0140}{0.0345} = \frac{140}{345}$
Dividing numerator and denominator by $5$,we get:
$P(E_2|D) = \frac{28}{69}$

(A) Let $E_1, E_2, E_3$ be the events of choosing families $F_1, F_2, F_3$ respectively. Since the family is chosen randomly,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $G$ be the event that the selected child is a girl.
The probabilities of selecting a girl from each family are:
$P(G|E_1) = \frac{1}{3}$
$P(G|E_2) = \frac{2}{3}$
$P(G|E_3) = \frac{1}{2}$
Using Bayes' Theorem,the probability that the girl is from $F_2$ is $P(E_2|G) = \frac{P(E_2)P(G|E_2)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)}$.
Substituting the values: $P(E_2|G) = \frac{\frac{1}{3} \times \frac{2}{3}}{\frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} + \frac{1}{3} \times \frac{1}{2}} = \frac{\frac{2}{9}}{\frac{1}{9} + \frac{2}{9} + \frac{1}{6}} = \frac{\frac{2}{9}}{\frac{2+4+3}{18}} = \frac{\frac{2}{9}}{\frac{9}{18}} = \frac{2}{9} \times 2 = \frac{4}{9}$.

(C) Let $E_i$ be the event that there are $i$ red balls in the bag,where $i \in \{0, 1, 2, 3, 4, 5\}$.
Since there are equal chances for each case,$P(E_i) = \frac{1}{6}$ for all $i$.
Let $R$ be the event that the ball drawn is red.
If there are $i$ red balls,the probability of drawing a red ball is $P(R|E_i) = \frac{i}{5}$.
Note that $P(R|E_0) = 0$.
By Bayes' Theorem,the probability that there is only $1$ red ball given that the drawn ball is red is:
$P(E_1|R) = \frac{P(R|E_1)P(E_1)}{\sum_{i=0}^{5} P(R|E_i)P(E_i)}$
$P(E_1|R) = \frac{(\frac{1}{5})(\frac{1}{6})}{(\frac{0}{5})(\frac{1}{6}) + (\frac{1}{5})(\frac{1}{6}) + (\frac{2}{5})(\frac{1}{6}) + (\frac{3}{5})(\frac{1}{6}) + (\frac{4}{5})(\frac{1}{6}) + (\frac{5}{5})(\frac{1}{6})}$
$P(E_1|R) = \frac{1}{0+1+2+3+4+5} = \frac{1}{15}$.

(C) Let $D$ be the event that the item is defective and $ND$ be the event that the item is not defective.
Given $P(D) = 0.3$,so $P(ND) = 1 - 0.3 = 0.7$.
Let $R_D$ be the event that the device reports the item as defective and $R_{ND}$ be the event that the device reports the item as not defective.
The device is accurate $80\%$ of the time,so $P(R_D|D) = 0.8$ and $P(R_{ND}|ND) = 0.8$.
Consequently,$P(R_{ND}|D) = 1 - 0.8 = 0.2$ and $P(R_D|ND) = 1 - 0.8 = 0.2$.
We need to find the probability that the item is defective given that the device reports it as not defective,i.e.,$P(D|R_{ND})$.
Using Bayes' Theorem:
$P(D|R_{ND}) = \frac{P(R_{ND}|D) \times P(D)}{P(R_{ND}|D) \times P(D) + P(R_{ND}|ND) \times P(ND)}$
$P(D|R_{ND}) = \frac{0.2 \times 0.3}{(0.2 \times 0.3) + (0.8 \times 0.7)}$
$P(D|R_{ND}) = \frac{0.06}{0.06 + 0.56} = \frac{0.06}{0.62} = \frac{6}{62} = \frac{3}{31}$.

(D) Let $E_1, E_2, E_3$ be the events of selecting sections $A, B$ and $C$ respectively. Let $G$ be the event of selecting a girl.
Given probabilities of selecting sections are $P(E_1) = 0.2, P(E_2) = 0.3, P(E_3) = 0.5$.
The conditional probabilities of selecting a girl from each section are:
$P(G|E_1) = \frac{20}{20+30} = \frac{20}{50} = 0.4$
$P(G|E_2) = \frac{40}{40+20} = \frac{40}{60} = \frac{2}{3}$
$P(G|E_3) = \frac{10}{10+30} = \frac{10}{40} = 0.25$
Using Bayes' Theorem,the probability that the girl belongs to section $A$ is $P(E_1|G) = \frac{P(E_1)P(G|E_1)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)}$.
$P(E_1|G) = \frac{0.2 \times 0.4}{(0.2 \times 0.4) + (0.3 \times \frac{2}{3}) + (0.5 \times 0.25)}$
$P(E_1|G) = \frac{0.08}{0.08 + 0.2 + 0.125} = \frac{0.08}{0.405} = \frac{80}{405} = \frac{16}{81}$.

(B) Let $T$ be the event that the student watches $TV$ and $B$ be the event that the student reads a book. Let $S$ be the event that the student falls asleep.
Given probabilities are:
$P(T) = \frac{4}{5}$
$P(B) = 1 - P(T) = 1 - \frac{4}{5} = \frac{1}{5}$
$P(S|T) = \frac{3}{4}$
$P(S|B) = \frac{1}{4}$
We need to find $P(T|S)$,the probability that he watched $TV$ given that he is asleep.
Using Bayes' Theorem:
$P(T|S) = \frac{P(T) \times P(S|T)}{P(T) \times P(S|T) + P(B) \times P(S|B)}$
$P(T|S) = \frac{(\frac{4}{5}) \times (\frac{3}{4})}{(\frac{4}{5}) \times (\frac{3}{4}) + (\frac{1}{5}) \times (\frac{1}{4})}$
$P(T|S) = \frac{\frac{12}{20}}{\frac{12}{20} + \frac{1}{20}}$
$P(T|S) = \frac{\frac{12}{20}}{\frac{13}{20}} = \frac{12}{13}$

(C) Let $C$,$B$,and $T$ be the events that the person travels by car,bus,and train respectively. Let $L$ be the event that the person reaches college late,and $L'$ be the event that the person reaches college in time.
Given probabilities: $P(C) = \frac{1}{5}$,$P(B) = \frac{2}{5}$,$P(T) = \frac{3}{5}$.
Probabilities of being late: $P(L|C) = \frac{2}{7}$,$P(L|B) = \frac{4}{7}$,$P(L|T) = \frac{1}{7}$.
Probabilities of being on time: $P(L'|C) = 1 - \frac{2}{7} = \frac{5}{7}$,$P(L'|B) = 1 - \frac{4}{7} = \frac{3}{7}$,$P(L'|T) = 1 - \frac{1}{7} = \frac{6}{7}$.
Using Bayes' Theorem,the probability that he traveled by car given he reached in time is:
$P(C|L') = \frac{P(L'|C)P(C)}{P(L'|C)P(C) + P(L'|B)P(B) + P(L'|T)P(T)}$
$P(C|L') = \frac{(\frac{5}{7} \times \frac{1}{5})}{(\frac{5}{7} \times \frac{1}{5}) + (\frac{3}{7} \times \frac{2}{5}) + (\frac{6}{7} \times \frac{3}{5})}$
$P(C|L') = \frac{\frac{5}{35}}{\frac{5}{35} + \frac{6}{35} + \frac{18}{35}} = \frac{5}{5 + 6 + 18} = \frac{5}{29}$.

(A) Let $B_1$ be the event of choosing a bag from the first group and $B_2$ be the event of choosing a bag from the second group.
Total bags = $2 + 4 = 6$.
$P(B_1) = \frac{2}{6} = \frac{1}{3}$ and $P(B_2) = \frac{4}{6} = \frac{2}{3}$.
Let $B$ be the event that the drawn ball is black.
For the first group,the probability of drawing a black ball is $P(B|B_1) = \frac{5}{3+5} = \frac{5}{8}$.
For the second group,the probability of drawing a black ball is $P(B|B_2) = \frac{4}{6+4} = \frac{4}{10} = \frac{2}{5}$.
Using Bayes' Theorem,the probability that the ball is from the first set of bags given that it is black is:
$P(B_1|B) = \frac{P(B_1)P(B|B_1)}{P(B_1)P(B|B_1) + P(B_2)P(B|B_2)}$
$P(B_1|B) = \frac{\frac{1}{3} \times \frac{5}{8}}{\frac{1}{3} \times \frac{5}{8} + \frac{2}{3} \times \frac{2}{5}} = \frac{\frac{5}{24}}{\frac{5}{24} + \frac{4}{15}} = \frac{\frac{5}{24}}{\frac{25 + 32}{120}} = \frac{5}{24} \times \frac{120}{57} = \frac{5 \times 5}{57} = \frac{25}{57}$.

(B) Let $K$ be the event that the card drawn is a king,and $K^c$ be the event that the card drawn is not a king. Let $R$ be the event that the person reports that the card is a king.
Given:
$P(K) = \frac{4}{52} = \frac{1}{13}$
$P(K^c) = 1 - \frac{1}{13} = \frac{12}{13}$
Let $T$ be the event that the person speaks the truth. $P(T) = \frac{3}{4}$ and $P(T^c) = \frac{1}{4}$.
The probability that the person reports a king given it is a king is $P(R|K) = P(T) = \frac{3}{4}$.
The probability that the person reports a king given it is not a king is $P(R|K^c) = P(T^c) = \frac{1}{4}$.
Using Bayes' Theorem,the probability that it is actually a king given the report is:
$P(K|R) = \frac{P(R|K)P(K)}{P(R|K)P(K) + P(R|K^c)P(K^c)}$
$P(K|R) = \frac{(\frac{3}{4} \times \frac{1}{13})}{(\frac{3}{4} \times \frac{1}{13}) + (\frac{1}{4} \times \frac{12}{13})}$
$P(K|R) = \frac{\frac{3}{52}}{\frac{3}{52} + \frac{12}{52}} = \frac{3}{15} = \frac{1}{5}$

(C) Let $A$ be the event that we draw three green balls. Let $E_k$ be the event that there are $k$ green balls in the bag,where $k \in \{3, 4, 5, 6\}$. Assuming each number of green balls is equally likely,$P(E_k) = \frac{1}{4}$.
The probability of drawing $3$ green balls given $k$ green balls are present is $P(A|E_k) = \frac{{}^k C_3}{{}^6 C_3}$.
Using Bayes' Theorem,the probability that there are $5$ green balls given that $3$ green balls were drawn is:
$P(E_5|A) = \frac{P(A|E_5)P(E_5)}{\sum_{k=3}^{6} P(A|E_k)P(E_k)}$
Since $P(E_k) = \frac{1}{4}$ for all $k$,this simplifies to:
$P(E_5|A) = \frac{{}^5 C_3}{\sum_{k=3}^{6} {}^k C_3} = \frac{10}{1 + 4 + 10 + 20} = \frac{10}{35} = \frac{2}{7}$.

(C) $E_1$: $A$ ball is drawn from bag $A$.
$E_2$: $A$ ball is drawn from bag $B$.
$F$: $A$ ball is found to be red.
Given:
$P(E_1) = \frac{1}{2}$,$P(E_2) = \frac{1}{2}$.
$P(F|E_1) = \frac{3}{5}$ (probability of drawing a red ball from bag $A$).
$P(F|E_2) = \frac{5}{9}$ (probability of drawing a red ball from bag $B$).
Using Bayes' Theorem:
$P(E_2|F) = \frac{P(F|E_2) \cdot P(E_2)}{P(F|E_1) \cdot P(E_1) + P(F|E_2) \cdot P(E_2)}$
$P(E_2|F) = \frac{\frac{5}{9} \times \frac{1}{2}}{\frac{3}{5} \times \frac{1}{2} + \frac{5}{9} \times \frac{1}{2}}$
$P(E_2|F) = \frac{\frac{5}{9}}{\frac{3}{5} + \frac{5}{9}} = \frac{\frac{5}{9}}{\frac{27 + 25}{45}} = \frac{5}{9} \times \frac{45}{52} = \frac{25}{52}$.

(C) Let us define the following events:
$E_1$: Student guesses the answer.
$E_2$: Student copies the answer.
$E_3$: Student knows the answer.
$E$: The answer is correct.
Given probabilities:
$P(E_1) = \frac{1}{3}$,$P(E_2) = \frac{1}{6}$.
Since the events are exhaustive,$P(E_3) = 1 - (P(E_1) + P(E_2)) = 1 - (\frac{1}{3} + \frac{1}{6}) = 1 - \frac{1}{2} = \frac{1}{2}$.
Conditional probabilities:
$P(E|E_1) = \frac{1}{4}$ (since there are $4$ choices).
$P(E|E_2) = \frac{1}{8}$ (given).
$P(E|E_3) = 1$ (since he knows the answer).
Using Bayes' Theorem,the probability that he knew the answer given that he answered correctly is:
$P(E_3|E) = \frac{P(E|E_3)P(E_3)}{P(E|E_1)P(E_1) + P(E|E_2)P(E_2) + P(E|E_3)P(E_3)}$
$P(E_3|E) = \frac{1 \times \frac{1}{2}}{(\frac{1}{4} \times \frac{1}{3}) + (\frac{1}{8} \times \frac{1}{6}) + (1 \times \frac{1}{2})}$
$P(E_3|E) = \frac{\frac{1}{2}}{\frac{1}{12} + \frac{1}{48} + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{4+1+24}{48}} = \frac{\frac{1}{2}}{\frac{29}{48}} = \frac{1}{2} \times \frac{48}{29} = \frac{24}{29}$.

(B) Let $S$ be the event that the two cards drawn are spades. Let $M_1$ be the event that the missing card is a spade,and $M_2$ be the event that the missing card is not a spade.
We are given that there are $13$ spades and $39$ non-spades in a pack of $52$ cards.
$P(M_1) = \frac{13}{52} = \frac{1}{4}$ and $P(M_2) = \frac{39}{52} = \frac{3}{4}$.
If $M_1$ occurs,there are $12$ spades left in $51$ cards. The probability of drawing $2$ spades is $P(S|M_1) = \frac{^{12}C_2}{^{51}C_2} = \frac{66}{1275}$.
If $M_2$ occurs,there are $13$ spades left in $51$ cards. The probability of drawing $2$ spades is $P(S|M_2) = \frac{^{13}C_2}{^{51}C_2} = \frac{78}{1275}$.
Using Bayes' Theorem,the probability that the missing card is not a spade given that two spades were drawn is:
$P(M_2|S) = \frac{P(M_2)P(S|M_2)}{P(M_1)P(S|M_1) + P(M_2)P(S|M_2)}$
$P(M_2|S) = \frac{\frac{3}{4} \times \frac{78}{1275}}{\frac{1}{4} \times \frac{66}{1275} + \frac{3}{4} \times \frac{78}{1275}} = \frac{3 \times 78}{66 + 3 \times 78} = \frac{234}{66 + 234} = \frac{234}{300} = \frac{39}{50}$.

(B) Let $E$ be the event that the chosen bulb is defective. Let $M_1, M_2, M_3$ be the events that the bulb is purchased from manufacturers $M_1, M_2, M_3$ respectively.
Given probabilities:
$P(M_1) = 0.25, P(M_2) = 0.45, P(M_3) = 0.30$
$P(E|M_1) = 0.01, P(E|M_2) = 0.01, P(E|M_3) = 0.02$
Using Bayes' Theorem,the probability that the defective bulb is from $M_3$ is:
$P(M_3|E) = \frac{P(M_3) \cdot P(E|M_3)}{P(M_1) \cdot P(E|M_1) + P(M_2) \cdot P(E|M_2) + P(M_3) \cdot P(E|M_3)}$
$P(M_3|E) = \frac{0.30 \times 0.02}{(0.25 \times 0.01) + (0.45 \times 0.01) + (0.30 \times 0.02)}$
$P(M_3|E) = \frac{0.006}{0.0025 + 0.0045 + 0.0060} = \frac{0.006}{0.0130} = \frac{6}{13}$

(B) Let $F$ be the event that a fair coin is drawn and $B$ be the event that a biased coin is drawn. Let $E$ be the event that the first toss is a head and the second toss is a tail.
Given $P(F) = \frac{m}{n}$ and $P(B) = \frac{n-m}{n}$.
For a fair coin,$P(H) = \frac{1}{2}$ and $P(T) = \frac{1}{2}$. Thus,$P(E|F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
For a biased coin,$P(H) = 2P(T)$. Since $P(H) + P(T) = 1$,we have $3P(T) = 1$,so $P(T) = \frac{1}{3}$ and $P(H) = \frac{2}{3}$. Thus,$P(E|B) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$.
Using Bayes' theorem,the probability that the coin is fair given the event $E$ is:
$P(F|E) = \frac{P(F)P(E|F)}{P(F)P(E|F) + P(B)P(E|B)}$
$P(F|E) = \frac{(\frac{m}{n})(\frac{1}{4})}{(\frac{m}{n})(\frac{1}{4}) + (\frac{n-m}{n})(\frac{2}{9})}$
$P(F|E) = \frac{\frac{m}{4}}{\frac{m}{4} + \frac{2(n-m)}{9}} = \frac{9m}{9m + 8(n-m)} = \frac{9m}{9m + 8n - 8m} = \frac{9m}{8n + m}$.

(B) Let $A$ be the event that a prime number occurs on the die.
Let $B$ be the event that a prime number does not occur on the die.
Let $E$ be the event that the man reports that a prime number occurred.
There are $25$ prime numbers between $1$ and $100$.
Therefore,$P(A) = \frac{25}{100} = \frac{1}{4}$ and $P(B) = 1 - P(A) = \frac{75}{100} = \frac{3}{4}$.
The probability that the man speaks the truth is $P(E|A) = \frac{7}{10}$.
The probability that the man lies (reports a prime when it is not) is $P(E|B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Using Bayes' Theorem,the probability that it is actually a prime given that he reported a prime is:
$P(A|E) = \frac{P(A) \times P(E|A)}{P(A) \times P(E|A) + P(B) \times P(E|B)}$
$P(A|E) = \frac{\frac{1}{4} \times \frac{7}{10}}{(\frac{1}{4} \times \frac{7}{10}) + (\frac{3}{4} \times \frac{3}{10})}$
$P(A|E) = \frac{\frac{7}{40}}{\frac{7}{40} + \frac{9}{40}} = \frac{7}{16}$.

(A) Let $E_1$ be the event that die $A$ is chosen (coin shows head) and $E_2$ be the event that die $B$ is chosen (coin shows tail).
Let $R$ be the event that a red face appears in all $n$ throws.
Given $P(E_1) = \frac{1}{2}$ and $P(E_2) = \frac{1}{2}$.
For die $A$,the probability of getting a red face in one throw is $\frac{4}{6} = \frac{2}{3}$. Thus,$P(R \mid E_1) = \left(\frac{2}{3}\right)^n$.
For die $B$,the probability of getting a red face in one throw is $\frac{2}{6} = \frac{1}{3}$. Thus,$P(R \mid E_2) = \left(\frac{1}{3}\right)^n$.
By Bayes' theorem,the probability that die $A$ was used given that red appeared $n$ times is:
$P(E_1 \mid R) = \frac{P(E_1)P(R \mid E_1)}{P(E_1)P(R \mid E_1) + P(E_2)P(R \mid E_2)}$
Substituting the values:
$\frac{32}{33} = \frac{\frac{1}{2} \times (\frac{2}{3})^n}{\frac{1}{2} \times (\frac{2}{3})^n + \frac{1}{2} \times (\frac{1}{3})^n} = \frac{2^n}{2^n + 1^n} = \frac{2^n}{2^n + 1}$.
Solving for $n$:
$32(2^n + 1) = 33(2^n) \implies 32 \cdot 2^n + 32 = 33 \cdot 2^n \implies 2^n = 32$.
Since $32 = 2^5$,we get $n = 5$.

(D) Let $H_1, H_2, H_3$ be the events of selecting cartons $A, B, C$ respectively. Since a carton is selected at random,$P(H_1) = P(H_2) = P(H_3) = \frac{1}{3}$.
Let $D$ be the event of drawing a defective toy.
The conditional probabilities are $P(D|H_1) = \frac{1}{3}, P(D|H_2) = \frac{1}{4}, P(D|H_3) = \frac{2}{5}$.
By Bayes' theorem,the probability that the toy is from carton $B$ given it is defective is:
$P(H_2|D) = \frac{P(H_2)P(D|H_2)}{P(H_1)P(D|H_1) + P(H_2)P(D|H_2) + P(H_3)P(D|H_3)}$
$P(H_2|D) = \frac{\frac{1}{3} \times \frac{1}{4}}{\frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{2}{5}}$
$P(H_2|D) = \frac{\frac{1}{12}}{\frac{1}{9} + \frac{1}{12} + \frac{2}{15}}$
To simplify,find the common denominator for the denominator: $LCM(9, 12, 15) = 180$.
$P(H_2|D) = \frac{1/12}{(20+15+24)/180} = \frac{1/12}{59/180} = \frac{1}{12} \times \frac{180}{59} = \frac{15}{59}$.

(B) Let $E_1$ be the event that bag $A$ is selected and $E_2$ be the event that bag $B$ is selected. Let $R$ be the event that the two balls drawn are red.
Given $P(E_1) = P(E_2) = \frac{1}{2}$.
The probability of drawing $2$ red balls from bag $A$ is $P(R|E_1) = \frac{^nC_2}{^{n+2}C_2} = \frac{n(n-1)}{(n+2)(n+1)}$.
The probability of drawing $2$ red balls from bag $B$ is $P(R|E_2) = \frac{^2C_2}{^{n+2}C_2} = \frac{1 \times 2}{(n+2)(n+1)} = \frac{2}{(n+2)(n+1)}$.
By Bayes' theorem,$P(E_1|R) = \frac{P(E_1)P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)} = \frac{6}{7}$.
Substituting the values:
$\frac{\frac{1}{2} \cdot \frac{n(n-1)}{(n+2)(n+1)}}{\frac{1}{2} \cdot \frac{n(n-1)}{(n+2)(n+1)} + \frac{1}{2} \cdot \frac{2}{(n+2)(n+1)}} = \frac{6}{7}$.
$\frac{n(n-1)}{n(n-1) + 2} = \frac{6}{7}$.
$7(n^2 - n) = 6(n^2 - n + 2)$.
$7n^2 - 7n = 6n^2 - 6n + 12$.
$n^2 - n - 12 = 0$.
$(n-4)(n+3) = 0$.
Since $n > 0$,we have $n = 4$.

(A) Let $E_1$ be the event of selecting box $B_1$ and $E_2$ be the event of selecting box $B_2$. Since the box is selected at random,$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $R$ be the event that the selected ball is red.
The probability of drawing a red ball from $B_1$ is $P(R|E_1) = \frac{6}{3+6} = \frac{6}{9} = \frac{2}{3}$.
The probability of drawing a red ball from $B_2$ is $P(R|E_2) = \frac{n}{n+8}$.
Using Bayes' theorem,the probability $p$ that the red ball is from $B_2$ is:
$p = P(E_2|R) = \frac{P(E_2)P(R|E_2)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)}$
$p = \frac{\frac{1}{2} \times \frac{n}{n+8}}{\frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{n}{n+8}} = \frac{\frac{n}{n+8}}{\frac{2}{3} + \frac{n}{n+8}} = \frac{3n}{2(n+8) + 3n} = \frac{3n}{5n+16}$.
Since $n \in N$,the minimum value of $n$ is $1$. For $n=1$,$p = \frac{3(1)}{5(1)+16} = \frac{3}{21} = \frac{1}{7}$.
As $n \to \infty$,$p = \lim_{n \to \infty} \frac{3n}{5n+16} = \frac{3}{5}$.
Since $p$ is an increasing function of $n$,the range of $p$ is $\frac{1}{7} \leq p < \frac{3}{5}$.

(B) Let $E$ be the event that the $4$ drawn balls are red. Let $A_k$ be the event that there are $k$ red balls in the bag,where $k \in \{4, 5, 6\}$. Assuming each case is equally likely,$P(A_4) = P(A_5) = P(A_6) = \frac{1}{3}$.
The conditional probabilities are:
$P(E|A_4) = \frac{{}^4C_4}{{}^6C_4} = \frac{1}{15}$
$P(E|A_5) = \frac{{}^5C_4}{{}^6C_4} = \frac{5}{15}$
$P(E|A_6) = \frac{{}^6C_4}{{}^6C_4} = \frac{15}{15} = 1$
By Bayes' theorem,the probability that there are exactly $5$ red balls given that the $4$ drawn balls are red is:
$P(A_5|E) = \frac{P(A_5)P(E|A_5)}{P(A_4)P(E|A_4) + P(A_5)P(E|A_5) + P(A_6)P(E|A_6)}$
$P(A_5|E) = \frac{\frac{1}{3} \times \frac{5}{15}}{\frac{1}{3} \times \frac{1}{15} + \frac{1}{3} \times \frac{5}{15} + \frac{1}{3} \times \frac{15}{15}}$
$P(A_5|E) = \frac{5}{1 + 5 + 15} = \frac{5}{21}$

(A) Let $E_1$ be the event that the student guesses the answer and $E_2$ be the event that the student knows the answer. Let $A$ be the event that the student answers correctly.
Given that the student knows $90 \%$ of the answers,$P(E_2) = \frac{9}{10}$.
Consequently,the probability that the student guesses is $P(E_1) = 1 - P(E_2) = 1 - \frac{9}{10} = \frac{1}{10}$.
If the student knows the answer,the probability of answering correctly is $P(A \mid E_2) = 1$.
If the student guesses,since there are $4$ options and only one is correct,the probability of answering correctly is $P(A \mid E_1) = \frac{1}{4}$.
We need to find the probability that the student was guessing given that they answered correctly,which is $P(E_1 \mid A)$.
Using Bayes' Theorem:
$P(E_1 \mid A) = \frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1) + P(E_2) \cdot P(A \mid E_2)}$
$P(E_1 \mid A) = \frac{\frac{1}{10} \cdot \frac{1}{4}}{(\frac{1}{10} \cdot \frac{1}{4}) + (\frac{9}{10} \cdot 1)}$
$P(E_1 \mid A) = \frac{\frac{1}{40}}{\frac{1}{40} + \frac{9}{10}} = \frac{\frac{1}{40}}{\frac{1+36}{40}} = \frac{1}{37}$.

(B) Let $E_1$ be the event that the student knows the answer and $E_2$ be the event that the student guesses the answer.
Given $P(E_1) = 9/10$,then $P(E_2) = 1 - 9/10 = 1/10$.
Let $E$ be the event that the answer is correct.
If the student knows the answer,the probability of answering correctly is $P(E|E_1) = 1$.
If the student guesses,since there are $4$ options and $1$ is correct,the probability of answering correctly is $P(E|E_2) = 1/4$.
Using Bayes' theorem,the probability that the student was guessing given that the answer is correct is $P(E_2|E) = \frac{P(E|E_2)P(E_2)}{P(E|E_1)P(E_1) + P(E|E_2)P(E_2)}$.
Substituting the values:
$P(E_2|E) = \frac{(1/4) \times (1/10)}{(1) \times (9/10) + (1/4) \times (1/10)} = \frac{1/40}{9/10 + 1/40} = \frac{1/40}{36/40 + 1/40} = \frac{1/40}{37/40} = \frac{1}{37}$.

(B) Let the possible compositions of white balls in the bag be $E_1$ ($4$ white),$E_2$ ($3$ white,$1$ other),and $E_3$ ($2$ white,$2$ others). Assuming each composition is equally likely,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $E$ be the event that two drawn balls are white.
$P(E|E_1) = \frac{^4C_2}{^4C_2} = 1$
$P(E|E_2) = \frac{^3C_2}{^4C_2} = \frac{3}{6} = \frac{1}{2}$
$P(E|E_3) = \frac{^2C_2}{^4C_2} = \frac{1}{6}$
Using Bayes' Theorem:
$P(E_1|E) = \frac{P(E_1)P(E|E_1)}{P(E_1)P(E|E_1) + P(E_2)P(E|E_2) + P(E_3)P(E|E_3)}$
$P(E_1|E) = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{6}} = \frac{1}{1 + \frac{1}{2} + \frac{1}{6}} = \frac{1}{\frac{6+3+1}{6}} = \frac{6}{10} = \frac{3}{5}$

(D) Let $E_1$ be the event of getting a $5$ and $E_2$ be the event of not getting a $5$. Let $A$ be the event that the boy reports that it is a $5$.
Given: $P(E_1) = \frac{1}{6}$,$P(E_2) = \frac{5}{6}$.
The boy speaks the truth with probability $\frac{3}{5}$,so he lies with probability $1 - \frac{3}{5} = \frac{2}{5}$.
$P(A|E_1) = \frac{3}{5}$ (He reports $5$ when it is $5$)
$P(A|E_2) = \frac{2}{5}$ (He reports $5$ when it is not $5$)
Using Bayes' Theorem:
$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$
$P(E_1|A) = \frac{\frac{1}{6} \times \frac{3}{5}}{(\frac{1}{6} \times \frac{3}{5}) + (\frac{5}{6} \times \frac{2}{5})} = \frac{3/30}{3/30 + 10/30} = \frac{3}{13}$.
Thus,the correct option is $D$.

(D) Total number of balls in the urn = $7 + 5 + 3 = 15$.
Given that the first ball drawn is red and the second ball drawn is white,these two balls are removed from the urn.
Number of balls remaining in the urn = $15 - 2 = 13$.
Remaining balls consist of:
Red balls = $7 - 1 = 6$.
White balls = $5 - 1 = 4$.
Black balls = $3$.
Total remaining balls = $6 + 4 + 3 = 13$.
We need to find the probability that the third ball drawn is not red.
The number of balls that are not red = $4 \text{ (white)} + 3 \text{ (black)} = 7$.
Therefore,the probability that the third ball is not red = $\frac{\text{Number of non-red balls}}{\text{Total remaining balls}} = \frac{7}{13}$.

(C) Given $P(B) = 0.4$ and $P(A \cap \bar{B}) = 0.5$.
We know that $P(A) = P(A \cap B) + P(A \cap \bar{B})$.
Let $P(A \cap B) = x$. Then $P(A) = x + 0.5$.
Also,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = (x + 0.5) + 0.4 - x = 0.9$.
Now,consider the term $P\left(\frac{B}{A \cup \bar{B}}\right)$.
By definition,$P\left(\frac{B}{A \cup \bar{B}}\right) = \frac{P(B \cap (A \cup \bar{B}))}{P(A \cup \bar{B})}$.
Since $B \cap (A \cup \bar{B}) = (B \cap A) \cup (B \cap \bar{B}) = (B \cap A) \cup \emptyset = A \cap B$,we have $P(B \cap (A \cup \bar{B})) = P(A \cap B) = x$.
Also,$P(A \cup \bar{B}) = P(A) + P(\bar{B}) - P(A \cap \bar{B}) = (x + 0.5) + (1 - 0.4) - 0.5 = x + 0.6$.
Substituting these into the given equation: $0.9 + \frac{x}{x + 0.6} = 1.15$.
$\frac{x}{x + 0.6} = 1.15 - 0.9 = 0.25 = \frac{1}{4}$.
$4x = x + 0.6 \implies 3x = 0.6 \implies x = 0.2$.
Therefore,$P(A) = x + 0.5 = 0.2 + 0.5 = 0.7$.

(A) Let $E_1, E_2, E_3$ be the events of selecting urns $A, B, C$ respectively. Since the urns are similar,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
Let $F$ be the event of drawing a red ball.
The probabilities of drawing a red ball from each urn are:
$P(F|E_1) = \frac{2}{5}$
$P(F|E_2) = \frac{3}{5}$
$P(F|E_3) = \frac{1}{5}$
Using Bayes' Theorem,the probability that the ball is drawn from urn $C$ given that it is red is:
$P(E_3|F) = \frac{P(F|E_3) \cdot P(E_3)}{P(F|E_1) \cdot P(E_1) + P(F|E_2) \cdot P(E_2) + P(F|E_3) \cdot P(E_3)}$
$P(E_3|F) = \frac{\frac{1}{5} \cdot \frac{1}{3}}{\frac{2}{5} \cdot \frac{1}{3} + \frac{3}{5} \cdot \frac{1}{3} + \frac{1}{5} \cdot \frac{1}{3}}$
$P(E_3|F) = \frac{\frac{1}{15}}{\frac{2}{15} + \frac{3}{15} + \frac{1}{15}} = \frac{\frac{1}{15}}{\frac{6}{15}} = \frac{1}{6}$.

(C) Let $E_1$ be the event that the person suffers from the disease and $E_2$ be the event that the person does not suffer from the disease. Let $A$ be the event that the diagnostic test is positive.
Given:
$P(E_1) = 0.5 \% = 0.005$
$P(E_2) = 99.5 \% = 0.995$
$P(A|E_1) = 0.95$
$P(A|E_2) = 0.10$
We need to find $P(E_1|A)$. Using Bayes' Theorem:
$P(E_1|A) = \frac{P(E_1) \times P(A|E_1)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)}$
$P(E_1|A) = \frac{0.005 \times 0.95}{(0.005 \times 0.95) + (0.995 \times 0.10)}$
$P(E_1|A) = \frac{0.00475}{0.00475 + 0.0995}$
$P(E_1|A) = \frac{0.00475}{0.10425} = \frac{475}{10425} \approx 0.0455$

(B) Let $U_1$ and $U_2$ be the events of selecting Bag $I$ and Bag $II$ respectively.
Since the bags are chosen at random,$P(U_1) = P(U_2) = \frac{1}{2}$.
Let $R$ be the event of drawing a red ball.
The probability of drawing a red ball from Bag $I$ is $P(R|U_1) = \frac{3}{3+4} = \frac{3}{7}$.
The probability of drawing a red ball from Bag $II$ is $P(R|U_2) = \frac{5}{5+6} = \frac{5}{11}$.
Using Bayes' Theorem,the probability that the ball was drawn from Bag $II$ given that it is red is:
$P(U_2|R) = \frac{P(U_2) \times P(R|U_2)}{P(U_1) \times P(R|U_1) + P(U_2) \times P(R|U_2)}$.
Substituting the values:
$P(U_2|R) = \frac{\frac{1}{2} \times \frac{5}{11}}{\frac{1}{2} \times \frac{3}{7} + \frac{1}{2} \times \frac{5}{11}} = \frac{\frac{5}{22}}{\frac{3}{14} + \frac{5}{22}} = \frac{\frac{5}{22}}{\frac{33+35}{154}} = \frac{5}{22} \times \frac{154}{68} = \frac{5 \times 7}{68} = \frac{35}{68}$.

(C) Let $M$ be the event that the person is a man and $W$ be the event that the person is a woman. Since the population is divided into men and women,we assume $P(M) = \frac{1}{2}$ and $P(W) = \frac{1}{2}$.
Let $C$ be the event that the person is colour blind.
Given: $P(C|M) = \frac{1}{20}$ and $P(C|W) = \frac{1}{10}$.
We need to find the probability that the person is a man given that they are colour blind,i.e.,$P(M|C)$.
Using Bayes' Theorem:
$P(M|C) = \frac{P(M) \cdot P(C|M)}{P(M) \cdot P(C|M) + P(W) \cdot P(C|W)}$
$P(M|C) = \frac{\frac{1}{2} \cdot \frac{1}{20}}{\frac{1}{2} \cdot \frac{1}{20} + \frac{1}{2} \cdot \frac{1}{10}}$
$P(M|C) = \frac{\frac{1}{40}}{\frac{1}{40} + \frac{1}{20}} = \frac{\frac{1}{40}}{\frac{1+2}{40}} = \frac{1}{3}$.

(A) Let $E_1$ be the event that the student guesses the answer,$E_2$ be the event that the student knows the answer,and $E_3$ be the event that he copies the answer. Let $A$ be the event that the answer is correct.
Given that,
$P(E_1) = \frac{1}{3}, P(E_3) = \frac{1}{12}$.
Since the events are exhaustive,$P(E_2) = 1 - P(E_1) - P(E_3) = 1 - \frac{1}{3} - \frac{1}{12} = \frac{12-4-1}{12} = \frac{7}{12}$.
The probability that the answer is correct if he guesses is $P(A|E_1) = \frac{1}{4}$ (since there are $4$ choices).
The probability that the answer is correct if he knows the answer is $P(A|E_2) = 1$.
The probability that the answer is correct if he copies is $P(A|E_3) = \frac{1}{6}$.
Using Bayes' theorem,the probability that he knew the answer given that he answered correctly is:
$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$
$P(E_2|A) = \frac{\frac{7}{12} \times 1}{(\frac{1}{3} \times \frac{1}{4}) + (\frac{7}{12} \times 1) + (\frac{1}{12} \times \frac{1}{6})}$
$P(E_2|A) = \frac{\frac{7}{12}}{\frac{1}{12} + \frac{7}{12} + \frac{1}{72}} = \frac{\frac{7}{12}}{\frac{6 + 42 + 1}{72}} = \frac{7}{12} \times \frac{72}{49} = \frac{6}{7}$.

(D) Let $E_1$ be the event that the weak student knows the answer,and $E_2$ be the event that the weak student guesses the answer. Let $A$ be the event that the weak student gets the correct answer.
We are given that the student knows $20 \%$ of the answers,so $P(E_1) = 0.20$. The probability of guessing is $P(E_2) = 1 - 0.20 = 0.80$.
If the student knows the answer,the probability of getting it correct is $P(A|E_1) = 1$.
Since there are $4$ options and only one is correct,the probability of guessing the correct answer is $P(A|E_2) = \frac{1}{4} = 0.25$.
We need to find the probability that the student was guessing given that they got the answer correct,which is $P(E_2|A)$.
Using Bayes' Theorem:
$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$
$P(E_2|A) = \frac{0.80 \times 0.25}{(0.20 \times 1) + (0.80 \times 0.25)}$
$P(E_2|A) = \frac{0.20}{0.20 + 0.20} = \frac{0.20}{0.40} = 0.5$.

(A) Let $E_1$ be the event of selecting bag $P$ and $E_2$ be the event of selecting bag $Q$.
Since the bags are selected at random,$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $A$ be the event of drawing a red ball.
The probability of drawing a red ball from bag $P$ is $P(A|E_1) = \frac{5}{3+5} = \frac{5}{8}$.
The probability of drawing a red ball from bag $Q$ is $P(A|E_2) = \frac{6}{4+6} = \frac{6}{10} = \frac{3}{5}$.
Using Bayes' Theorem,the probability that the ball is from bag $Q$ given that it is red is $P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$.
Substituting the values:
$P(E_2|A) = \frac{\frac{1}{2} \times \frac{6}{10}}{\frac{1}{2} \times \frac{5}{8} + \frac{1}{2} \times \frac{6}{10}} = \frac{\frac{6}{10}}{\frac{5}{8} + \frac{6}{10}} = \frac{\frac{3}{5}}{\frac{25+24}{40}} = \frac{3}{5} \times \frac{40}{49} = \frac{24}{49}$.

(A) Let $E_1$ and $E_2$ be the events that the selected student is a woman and a man,respectively. Let $A$ be the event that the selected student is taller than $1.8 \ m$.
Given:
$P(E_1) = 60/100 = 0.6$
$P(E_2) = 40/100 = 0.4$
$P(A|E_1) = 1/100 = 0.01$
$P(A|E_2) = 4/100 = 0.04$
By Bayes' Theorem,the probability that the student is a woman given that they are taller than $1.8 \ m$ is:
$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$
$P(E_1|A) = \frac{0.6 \times 0.01}{(0.6 \times 0.01) + (0.4 \times 0.04)}$
$P(E_1|A) = \frac{0.006}{0.006 + 0.016} = \frac{0.006}{0.022} = \frac{6}{22} = \frac{3}{11}$

(D) Let the probabilities of $A$,$B$,and $C$ passing the test be $P(T_A) = k$,$P(T_B) = 2k$,and $P(T_C) = 3k$ respectively,based on the ratio $1:2:3$.
Let $I$ be the event that a person faces the interview successfully. The conditional probabilities are $P(I|T_A) = 0.8$,$P(I|T_B) = 0.7$,and $P(I|T_C) = 0.6$.
We need to find the probability that $A$ is selected,given that one of them is selected. This is calculated using Bayes' Theorem:
$P(A|I) = \frac{P(T_A) \times P(I|T_A)}{P(T_A) \times P(I|T_A) + P(T_B) \times P(I|T_B) + P(T_C) \times P(I|T_C)}$
$P(A|I) = \frac{k \times 0.8}{k \times 0.8 + 2k \times 0.7 + 3k \times 0.6}$
$P(A|I) = \frac{0.8k}{0.8k + 1.4k + 1.8k} = \frac{0.8k}{4.0k} = \frac{0.8}{4} = \frac{1}{5}$.

(C) Let $E_1, E_2, E_3$ be the events that the refrigerator is from companies $C_1, C_2, C_3$ respectively. Let $D$ be the event that the refrigerator is defective.
Given probabilities are:
$P(E_1) = 0.25, P(E_2) = 0.35, P(E_3) = 0.40$
$P(D|E_1) = 0.03, P(D|E_2) = 0.02, P(D|E_3) = 0.01$
Using Bayes' Theorem,the probability that the defective refrigerator is from $C_2$ is:
$P(E_2|D) = \frac{P(E_2) \cdot P(D|E_2)}{P(E_1) \cdot P(D|E_1) + P(E_2) \cdot P(D|E_2) + P(E_3) \cdot P(D|E_3)}$
$P(E_2|D) = \frac{0.35 \times 0.02}{(0.25 \times 0.03) + (0.35 \times 0.02) + (0.40 \times 0.01)}$
$P(E_2|D) = \frac{0.0070}{0.0075 + 0.0070 + 0.0040} = \frac{0.0070}{0.0185}$
$P(E_2|D) = \frac{70}{185} = \frac{14}{37}$

(D) Let $E_1, E_2, E_3, E_4$ be the events of selecting boxes $A, B, C, D$ respectively. Let $F$ be the event that the selected fuse is defective.
$P(E_1) = \frac{5000}{11000} = \frac{5}{11}, P(E_2) = \frac{3000}{11000} = \frac{3}{11}, P(E_3) = \frac{2000}{11000} = \frac{2}{11}, P(E_4) = \frac{1000}{11000} = \frac{1}{11}$.
The conditional probabilities of selecting a defective fuse are:
$P(F|E_1) = \frac{3}{100}, P(F|E_2) = \frac{2}{100}, P(F|E_3) = \frac{1}{100}, P(F|E_4) = \frac{0.5}{100} = \frac{1}{200}$.
Using Bayes' Theorem,the probability that the defective fuse came from box $D$ is $P(E_4|F) = \frac{P(E_4)P(F|E_4)}{\sum_{i=1}^{4} P(E_i)P(F|E_i)}$.
$P(E_4|F) = \frac{\frac{1}{11} \times \frac{1}{200}}{\frac{5}{11} \times \frac{3}{100} + \frac{3}{11} \times \frac{2}{100} + \frac{2}{11} \times \frac{1}{100} + \frac{1}{11} \times \frac{1}{200}}$.
$P(E_4|F) = \frac{\frac{1}{2200}}{\frac{15}{1100} + \frac{6}{1100} + \frac{2}{1100} + \frac{1}{2200}} = \frac{\frac{1}{2200}}{\frac{30+12+4+1}{2200}} = \frac{1}{47}$.

(C) Let $E_1$ be the event that the student answers without guessing,and $E_2$ be the event that the student answers by guessing. Given $P(E_2) = 3/7$,so $P(E_1) = 1 - 3/7 = 4/7$.
Let $A$ be the event that at least $3$ questions are answered correctly.
For $E_1$ (without guessing),the probability of success is $p = 2/3$. Using binomial distribution $B(4, 2/3)$:
$P(A|E_1) = \binom{4}{3} (2/3)^3 (1/3)^1 + \binom{4}{4} (2/3)^4 = 4 \cdot (8/27) \cdot (1/3) + 16/81 = 32/81 + 16/81 = 48/81 = 16/27$.
For $E_2$ (guessing),the probability of success is $p = 1/2$. Using binomial distribution $B(4, 1/2)$:
$P(A|E_2) = \binom{4}{3} (1/2)^4 + \binom{4}{4} (1/2)^4 = 5/16$.
Using Bayes' Theorem,we need $P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$.
$P(E_1|A) = \frac{(4/7) \cdot (16/27)}{(4/7) \cdot (16/27) + (3/7) \cdot (5/16)} = \frac{64/189}{64/189 + 15/112} = \frac{64/189}{7168/21168 + 2835/21168} = \frac{64/189}{10003/21168} = \frac{64 \cdot 112}{10003} = \frac{7168}{10003} = \frac{1024}{1429}$.

(A) Let $E_1$,$E_2$,and $E_3$ be the events that the selected examinee is a graduate,a post-graduate,and a doctorate holder,respectively.
Total examinees = $5000 + 2000 + 1000 = 8000$.
$P(E_1) = \frac{5000}{8000} = \frac{5}{8}$,$P(E_2) = \frac{2000}{8000} = \frac{2}{8} = \frac{1}{4}$,$P(E_3) = \frac{1000}{8000} = \frac{1}{8}$.
Let $A$ be the event that the examinee passed the examination.
$P(A|E_1) = \frac{2}{3}$,$P(A|E_2) = \frac{3}{4}$,$P(A|E_3) = \frac{4}{5}$.
Using Bayes' Theorem,the probability that the examinee is a post-graduate given that they passed is:
$P(E_2|A) = \frac{P(E_2) \times P(A|E_2)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2) + P(E_3) \times P(A|E_3)}$
$P(E_2|A) = \frac{\frac{2}{8} \times \frac{3}{4}}{\frac{5}{8} \times \frac{2}{3} + \frac{2}{8} \times \frac{3}{4} + \frac{1}{8} \times \frac{4}{5}}$
$P(E_2|A) = \frac{\frac{6}{32}}{\frac{10}{24} + \frac{6}{32} + \frac{4}{40}} = \frac{\frac{3}{16}}{\frac{5}{12} + \frac{3}{16} + \frac{1}{10}}$
Finding a common denominator $(240)$:
$P(E_2|A) = \frac{\frac{45}{240}}{\frac{100}{240} + \frac{45}{240} + \frac{24}{240}} = \frac{45}{100 + 45 + 24} = \frac{45}{169}$.

(A) Total number of coins $= 2n+1$.
Number of two-headed coins $= n$.
Number of fair coins $= n+1$.
Let $H$ be the event that the toss results in a head.
The probability of picking a two-headed coin is $\frac{n}{2n+1}$,and for such a coin,$P(H) = 1$.
The probability of picking a fair coin is $\frac{n+1}{2n+1}$,and for such a coin,$P(H) = \frac{1}{2}$.
Using the law of total probability:
$P(H) = \left(\frac{n}{2n+1}\right) \times 1 + \left(\frac{n+1}{2n+1}\right) \times \frac{1}{2} = \frac{31}{42}$
$\frac{2n + n + 1}{2(2n+1)} = \frac{31}{42}$
$\frac{3n+1}{4n+2} = \frac{31}{42}$
$42(3n+1) = 31(4n+2)$
$126n + 42 = 124n + 62$
$2n = 20$
$n = 10$