A man is known to speak the truth 3 out of 4 | vedclass

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Question

(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 $6$.

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(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,the man reports it as a six if he speaks the truth,so $P(A|E) = \frac{3}{4}$.If a six does not occur,the man reports it as a six if he lies,so $P(A|E') = \frac{1}{4}$.Using Bayes' theorem,the probability that it is actually a six given that he reported a six is:$P(E|A) = \frac{P(E) \cdot P(A|E)}{P(E) \cdot P(A|E) + P(E') \cdot P(A|E')}$$P(E|A) = \frac{\frac{1}{6} \cdot \frac{3}{4}}{\frac{1}{6} \cdot \frac{3}{4} + \frac{5}{6} \cdot \frac{1}{4}}$$P(E|A) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8}$.

$A$ man is known to speak the truth $3$ out of $4$ times. He throws a die and reports that it is a six. The probability that it is actually a six,is

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