    Question

# A man speaks truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

A
38
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B
58
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C
78
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D
112
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Solution

## The correct option is B 38Let E1,E2 and A be the events defined as follows: E1 = six occurs, E2 = six does not occur and A = the man reports that it is a six. We have, P(E1)=16,P(E2)=56 Now P(AE1)=probability that the man reports that there is a six on the die given that six has occured on the die. = probability that the man speak truth = 34 and P(AE2) = probability that the man reports that there is a six on the die given that six has not occured on the die. = probability that the man does not speak truth =1−34=14 We have to find P(E1A) By Baye's rule, we have P(E1A)=P(E1)P(AE1)P(E1)P(AE1)+P(E2)P(AE2)=16×3416×34+56×14=38  Suggest Corrections  0      Similar questions  Related Videos   QUANTITATIVE APTITUDE
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