# 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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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