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# A man is known to speak 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.

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Solution

## Let E be the event that the man reports that six occurs in throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur. Then, P(S1)= Probability that six occurs =16 P(S2)= Probability that six doesn't occur =56 P(E|S1)= Probability that the man reports six when six has actually occurred = ​​Probability that the man speaks the truth =34 P(E|S2)= Probability that the man reports six when six has not actually occurred = ​​Probability that the man does not speaks the truth =1−34=14 Therefore, by Bayes' Theorem, Probability that the report of the man that six has occurred is actually a six is given by, P(S1|E)=P(S1)P(E|S1)P(S1)P(E|S1)+P(S2)P(E|S2) =16×3416×34+56×14 =38  Suggest Corrections  1      Similar questions
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