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Question

# 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:

A
38
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B
25
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C
47
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D
324
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Solution

## The correct option is A 38Let E1 be the event it is actually 6 and E2 be the event of not actually 6 on the die. Let A be the event that the man speaks truth P(E1)=16,P(E2)=56 P(AE1)=34,P(AE2)=14 Using Bayes' Theorem, P(E1/A)=(P(E1).P(A/E1)P(E1).P(A/E1)+P(E2).P(A/E2))=⎡⎢ ⎢ ⎢ ⎢⎣(16×34)(16×34)+(56×14)⎤⎥ ⎥ ⎥ ⎥⎦=38

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