Baye's theorem Questions in English

(A) By Bayes' Theorem,the probability $P(B_2|A)$ is given by:
$P(B_2|A) = \frac{P(B_2) \times P(A|B_2)}{P(B_1) \times P(A|B_1) + P(B_2) \times P(A|B_2) + P(B_3) \times P(A|B_3)}$
Substituting the given values:
$P(B_2|A) = \frac{0.30 \times 0.04}{(0.25 \times 0.05) + (0.30 \times 0.04) + (0.45 \times 0.03)}$
$P(B_2|A) = \frac{0.012}{0.0125 + 0.012 + 0.0135}$
$P(B_2|A) = \frac{0.012}{0.038}$
$P(B_2|A) = \frac{12}{38} = \frac{6}{19}$
Thus,the correct option is $A$.

(A) Let $E_1, E_2, E_3$ be the events that the tyre is supplied by companies $C_1, C_2, C_3$ respectively. Let $D$ be the event that the chosen tyre is defective.
Given probabilities are:
$P(E_1) = 0.40, P(E_2) = 0.35, P(E_3) = 0.25$
$P(D|E_1) = 0.02, P(D|E_2) = 0.03, P(D|E_3) = 0.04$
Using Bayes' Theorem,the probability that the defective tyre was supplied by $C_2$ is $P(E_2|D) = \frac{P(E_2)P(D|E_2)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)}$
$P(E_2|D) = \frac{0.35 \times 0.03}{(0.40 \times 0.02) + (0.35 \times 0.03) + (0.25 \times 0.04)}$
$P(E_2|D) = \frac{0.0105}{0.008 + 0.0105 + 0.0100} = \frac{0.0105}{0.0285}$
$P(E_2|D) = \frac{105}{285} = \frac{21}{57} = \frac{7}{19}$

(C) Let $B_1$ and $B_2$ be the events of selecting the first and second box respectively. Since the box is chosen at random,$P(B_1) = P(B_2) = \frac{1}{2}$.
Let $E$ be the event of drawing a black ball.
Let $n_1$ and $n_2$ be the number of black balls in the first and second box respectively. Since each box has $10$ balls,$P(E|B_1) = \frac{n_1}{10}$ and $P(E|B_2) = \frac{n_2}{10}$.
By Bayes' Theorem,the probability that the ball is from the second box given it is black is:
$P(B_2|E) = \frac{P(B_2)P(E|B_2)}{P(B_1)P(E|B_1) + P(B_2)P(E|B_2)} = \frac{\frac{1}{2} \times \frac{n_2}{10}}{\frac{1}{2} \times \frac{n_1}{10} + \frac{1}{2} \times \frac{n_2}{10}} = \frac{n_2}{n_1 + n_2}$.
Given $P(B_2|E) = \frac{1}{5}$,we have $\frac{n_2}{n_1 + n_2} = \frac{1}{5}$,which implies $5n_2 = n_1 + n_2$,or $n_1 = 4n_2$.
Since $n_1$ and $n_2$ are integers between $0$ and $10$,we test possible values for $n_2$:
If $n_2 = 1$,$n_1 = 4$.
If $n_2 = 2$,$n_1 = 8$.
If $n_2 = 0$,$n_1 = 0$ (not possible as a black ball was drawn).
Thus,$n_1$ can be $4$ or $8$.

(A) Let $E_1, E_2, E_3, E_4$ be the events that the boy wrote the words $ABILITY$,$PROBABILITY$,$FACILITY$,and $MOBILITY$ respectively.
Since he chooses one word out of four,$P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}$.
Let $A$ be the event that '$LI$' is left after erasing.
- For $ABILITY$ ($7$ letters),there are $7-1 = 6$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_1) = \frac{1}{6}$.
- For $PROBABILITY$ ($11$ letters),there are $11-1 = 10$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_2) = \frac{1}{10}$.
- For $FACILITY$ ($8$ letters),there are $8-1 = 7$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_3) = \frac{1}{7}$.
- For $MOBILITY$ ($8$ letters),there are $8-1 = 7$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_4) = \frac{1}{7}$.
Using Bayes' theorem:
$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{\sum_{i=1}^{4} P(E_i)P(A|E_i)} = \frac{\frac{1}{4} \times \frac{1}{10}}{\frac{1}{4}(\frac{1}{6} + \frac{1}{10} + \frac{1}{7} + \frac{1}{7})} = \frac{\frac{1}{10}}{\frac{1}{6} + \frac{1}{10} + \frac{2}{7}} = \frac{\frac{1}{10}}{\frac{35+21+60}{210}} = \frac{\frac{1}{10}}{\frac{116}{210}} = \frac{21}{116}$.

(A) Let $E_1$ be the event of choosing a fair coin and $E_2$ be the event of choosing a two-headed coin.
$P(E_1) = \frac{8}{10} = \frac{4}{5}$ and $P(E_2) = \frac{2}{10} = \frac{1}{5}$.
Let $A$ be the event that the coin shows heads $6$ times in $6$ tosses.
$P(A|E_1) = (\frac{1}{2})^6 = \frac{1}{64}$.
$P(A|E_2) = 1^6 = 1$.
Using Bayes' Theorem,the probability that the selected coin is two-headed given that it showed heads $6$ times is:
$P(E_2|A) = \frac{P(E_2) \times P(A|E_2)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)}$.
$P(E_2|A) = \frac{\frac{1}{5} \times 1}{\frac{4}{5} \times \frac{1}{64} + \frac{1}{5} \times 1} = \frac{\frac{1}{5}}{\frac{1}{80} + \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{1+16}{80}} = \frac{1}{5} \times \frac{80}{17} = \frac{16}{17}$.

(A) Let $A_i$ (for $i=1, 2, 3, 4$) be the event that the urn contains $i+1$ white balls. Let $B$ be the event that two white balls are drawn.
We need to find $P(A_4 | B)$.
Since the four events $A_1, A_2, A_3, A_4$ are equally likely,$P(A_1) = P(A_2) = P(A_3) = P(A_4) = \frac{1}{4}$.
$P(B | A_i)$ is the probability that two white balls are drawn given the urn contains $i+1$ white balls.
$P(B | A_1) = \frac{^2C_2}{^5C_2} = \frac{1}{10}$.
$P(B | A_2) = \frac{^3C_2}{^5C_2} = \frac{3}{10}$.
$P(B | A_3) = \frac{^4C_2}{^5C_2} = \frac{6}{10} = \frac{3}{5}$.
$P(B | A_4) = \frac{^5C_2}{^5C_2} = 1$.
By Bayes' Theorem:
$P(A_4 | B) = \frac{P(A_4) P(B | A_4)}{\sum_{i=1}^4 P(A_i) P(B | A_i)} = \frac{\frac{1}{4} \cdot 1}{\frac{1}{4} \left( \frac{1}{10} + \frac{3}{10} + \frac{6}{10} + \frac{10}{10} \right)} = \frac{1}{\frac{20}{10}} = \frac{1}{2}$.

(C) Let $E$ be the event that the selected item is defective. Let $A, B, C$ be the events that the item was produced by machines $A, B, C$ respectively.
Given probabilities:
$P(A) = \frac{2500}{10000} = 0.25$
$P(B) = \frac{3500}{10000} = 0.35$
$P(C) = \frac{4000}{10000} = 0.40$
Conditional probabilities of defect:
$P(E|A) = \frac{2}{100} = 0.02$
$P(E|B) = \frac{3}{100} = 0.03$
$P(E|C) = \frac{5}{100} = 0.05$
Using Bayes' theorem,the probability that the item was produced by machine $C$ given it is defective is:
$P(C|E) = \frac{P(E|C) \cdot P(C)}{P(E|A) \cdot P(A) + P(E|B) \cdot P(B) + P(E|C) \cdot P(C)}$
$P(C|E) = \frac{0.05 \cdot 0.40}{(0.02 \cdot 0.25) + (0.03 \cdot 0.35) + (0.05 \cdot 0.40)}$
$P(C|E) = \frac{0.0200}{0.0050 + 0.0105 + 0.0200} = \frac{0.0200}{0.0355} = \frac{200}{355} = \frac{40}{71}$

(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 $P(E_1) = 9/10$ and $P(E_2) = 1 - 9/10 = 1/10$.
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:
$P(E_2|E) = \frac{(1/4) \times (1/10)}{(1) \times (9/10) + (1/4) \times (1/10)} = \frac{1/40}{9/10 + 1/40} = \frac{1/40}{(36+1)/40} = \frac{1}{37}$.

(A) Let $A, B, C, D$ be the events that the person goes to the office by car,scooter,bus,and train respectively. Then $P(A) = 1/7, P(B) = 3/7, P(C) = 2/7, P(D) = 1/7$.
Let $L$ be the event that he reaches late and $E$ be the event that he reaches in time. Then $P(E|A) = 1 - P(L|A) = 1 - 2/9 = 7/9$. Similarly,$P(E|B) = 1 - 1/9 = 8/9, P(E|C) = 1 - 4/9 = 5/9, P(E|D) = 1 - 1/9 = 8/9$.
We need to find $P(A|E)$. By Bayes' Theorem:
$P(A|E) = \frac{P(A)P(E|A)}{P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) + P(D)P(E|D)}$
$P(A|E) = \frac{(1/7)(7/9)}{(1/7)(7/9) + (3/7)(8/9) + (2/7)(5/9) + (1/7)(8/9)}$
$P(A|E) = \frac{7/63}{7/63 + 24/63 + 10/63 + 8/63} = \frac{7}{7 + 24 + 10 + 8} = \frac{7}{49} = 1/7$.

(C) Let $S$ be the event that a person is a smoker and $NS$ be the event that a person is a non-smoker.
Let $D$ be the event that death is due to lung cancer.
Given: $P(S) = 0.20$,$P(NS) = 0.80$,and $P(D) = 0.006$.
According to the problem,$P(D|S) = 10 \times P(D|NS)$,which implies $P(D|NS) = \frac{1}{10} P(D|S)$.
Using the Law of Total Probability:
$P(D) = P(S) \cdot P(D|S) + P(NS) \cdot P(D|NS)$
$0.006 = 0.20 \cdot P(D|S) + 0.80 \cdot \left( \frac{1}{10} P(D|S) \right)$
$0.006 = 0.20 \cdot P(D|S) + 0.08 \cdot P(D|S)$
$0.006 = 0.28 \cdot P(D|S)$
$P(D|S) = \frac{0.006}{0.28} = \frac{6}{280} = \frac{3}{140}$.

(C) Let $E_1$ be the event that the student does not know the answer,and $E_2$ be the event that the student knows the answer. Let $E$ be the event that the student answers the question correctly.
We are given $P(E_2) = p$ and $P(E_1) = 1 - p$.
If the student knows the answer,the probability of answering correctly is $P(E|E_2) = 1$.
If the student does not know the answer,he chooses one of the $5$ alternatives randomly,so the probability of answering correctly is $P(E|E_1) = \frac{1}{5}$.
We want to find the probability that the student knew the answer (did not tick randomly) given that he answered correctly,which is $P(E_2|E)$.
By Bayes' Theorem:
$P(E_2|E) = \frac{P(E_2) P(E|E_2)}{P(E_1) P(E|E_1) + P(E_2) P(E|E_2)}$
Substituting the values:
$P(E_2|E) = \frac{p \times 1}{(1 - p) \times \frac{1}{5} + p \times 1}$
$P(E_2|E) = \frac{p}{\frac{1 - p + 5p}{5}}$
$P(E_2|E) = \frac{5p}{1 + 4p}$

(D) Let $E$ be the event of getting heads.
Let $E_{1}$ be the event of choosing the biased coin and $E_{2}$ be the event of choosing the unbiased coin.
Since a coin is selected at random,$P(E_{1}) = \frac{1}{2}$ and $P(E_{2}) = \frac{1}{2}$.
The probability of getting heads given the unbiased coin is $P(E|E_{2}) = \frac{1}{2}$.
The probability of getting heads given the biased coin is $P(E|E_{1}) = \frac{3}{4}$.
Using Bayes' Theorem,the probability that the unbiased coin was selected given that it shows heads is:
$P(E_{2}|E) = \frac{P(E_{2}) \cdot P(E|E_{2})}{P(E_{2}) \cdot P(E|E_{2}) + P(E_{1}) \cdot P(E|E_{1})}$
$P(E_{2}|E) = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{3}{4}}$
$P(E_{2}|E) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{3}{8}} = \frac{\frac{1}{4}}{\frac{2+3}{8}} = \frac{\frac{1}{4}}{\frac{5}{8}} = \frac{1}{4} \times \frac{8}{5} = \frac{2}{5}$.

(B) Let $F$ denote the fair coin,$T$ denote the two-headed coin,and $H$ denote the event that the outcome is a head.
Given probabilities are $P(F) = \frac{3}{4}$ and $P(T) = 1 - \frac{3}{4} = \frac{1}{4}$.
The probability of getting a head given the fair coin is $P(H|F) = \frac{1}{2}$.
The probability of getting a head given the two-headed coin is $P(H|T) = 1$.
We need to find the probability that the two-headed coin was chosen given that the outcome is a head,$P(T|H)$.
By Bayes' theorem:
$P(T|H) = \frac{P(H|T) \cdot P(T)}{P(H|T) \cdot P(T) + P(H|F) \cdot P(F)}$
$P(T|H) = \frac{1 \cdot \frac{1}{4}}{1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}}$
$P(T|H) = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{3}{8}} = \frac{\frac{1}{4}}{\frac{2+3}{8}} = \frac{\frac{1}{4}}{\frac{5}{8}} = \frac{1}{4} \times \frac{8}{5} = \frac{2}{5}$.

(D) Let $E$ be the event that $3$ black balls are drawn. Let $H_k$ be the hypothesis that the bag contains $k$ red balls and $(10-k)$ black balls.
By Bayes' Theorem,$P(H_1 | E) = \frac{P(E | H_1) P(H_1)}{\sum_{k=0}^{7} P(E | H_k) P(H_k)}$.
Assuming each composition $k$ is equally likely,$P(H_k) = \frac{1}{11}$.
$P(E | H_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$.
$P(H_1 | E) = \frac{\binom{9}{3}}{\sum_{k=0}^{7} \binom{10-k}{3}} = \frac{\binom{9}{3}}{\binom{11}{4}} = \frac{84}{330} = \frac{14}{55}$.

(B) Let $E$ be the event that the die shows a six and $E'$ be the event that the die does not show a six.
$P(E) = 1/6$ and $P(E') = 5/6$.
Let $A$ be the event that the man reports that it is a six.
Given that the man speaks the truth $4/5$ of the time,the probability of reporting a six when it is a six is $P(A|E) = 4/5$.
The probability of reporting a six when it is not a six (i.e.,he lies) is $P(A|E') = 1 - 4/5 = 1/5$.
Using Bayes' Theorem,the probability that it was actually a six given that he reported a six is:
$P(E|A) = \frac{P(A|E)P(E)}{P(A|E)P(E) + P(A|E')P(E')}$
$P(E|A) = \frac{(4/5) \times (1/6)}{(4/5) \times (1/6) + (1/5) \times (5/6)}$
$P(E|A) = \frac{4/30}{4/30 + 5/30} = \frac{4/30}{9/30} = 4/9$.

(B) Let $K$ be the event that the letter is from $KANPUR$ and $A$ be the event that the letter is from $ANANTPUR$.
The word $KANPUR$ has $6$ letters,which gives $5$ pairs of consecutive letters: $(KA, AN, NP, PU, UR)$. Out of these,$1$ pair is $AN$. Therefore,$P(AN|K) = 1/5$.
The word $ANANTPUR$ has $8$ letters,which gives $7$ pairs of consecutive letters: $(AN, NA, AN, NT, TP, PU, UR)$. Out of these,$2$ pairs are $AN$. Therefore,$P(AN|A) = 2/7$.
Assuming the prior probabilities are equal,$P(K) = P(A) = 1/2$.
Using Bayes' Theorem,the probability that the letter came from $ANANTPUR$ given that $AN$ is visible is:
$P(A|AN) = \frac{P(AN|A)P(A)}{P(AN|A)P(A) + P(AN|K)P(K)}$
$P(A|AN) = \frac{(2/7) \times (1/2)}{(2/7) \times (1/2) + (1/5) \times (1/2)}$
$P(A|AN) = \frac{2/7}{2/7 + 1/5} = \frac{2/7}{17/35} = \frac{2}{7} \times \frac{35}{17} = \frac{10}{17}$.

(B) Total number of coins = $N+1$.
Probability of selecting a fair coin = $\frac{N}{N+1}$.
Probability of selecting the two-headed coin = $\frac{1}{N+1}$.
Probability of getting a 'Head' = $P(H|Fair) \cdot P(Fair) + P(H|TwoHead) \cdot P(TwoHead)$.
Since a fair coin has a $1/2$ probability of 'Head' and the two-headed coin has a $1$ probability of 'Head',we have:
$P(H) = \frac{1}{2} \cdot \frac{N}{N+1} + 1 \cdot \frac{1}{N+1} = \frac{N}{2(N+1)} + \frac{2}{2(N+1)} = \frac{N+2}{2(N+1)}$.
Given that $P(H) = \frac{9}{16}$,we set up the equation:
$\frac{N+2}{2(N+1)} = \frac{9}{16}$.
Cross-multiplying gives: $16(N+2) = 18(N+1)$.
$16N + 32 = 18N + 18$.
$32 - 18 = 18N - 16N$.
$14 = 2N$.
$N = 7$.