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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.

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Let E1, E2 and A be the events defined as follows: 

E1 = Six occurs, 

E2 = Six does not occur 

A = man reports it is a six 

Then, P(E1) = \(\frac{1}{6}\) , P(E2) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\)

 P(A/E1) = Probability of man reporting it a six when six occurs = Probability of speaking truth = \(\frac{3}{4}\)

P(A/E2) = Probability of man reporting a six when six does not occur 

= Probability of not speaking truth = 1-\(\frac{3}{4}\) = \(\frac{1}{4}\)

∴ P(Throw is actually a six) =\(\frac{P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\)                                 ... (Baye’s Theorem

\(\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}\) = \(\frac{\frac{3}{24}}{\frac{8}{24}}\) = \(\frac{3}{8}\)

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