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Q.A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is '1'. Find the probability that it is actually 1.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2020Subjective· 6mImportance★★★★★
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Apply Bayes' theorem with E1E_1 = die actually shows 1, E2E_2 = die shows something else, and AA = man reports '1'.

Let E1E_1: the die actually shows 11, so P(E1)=16P(E_1)=\dfrac16. Let E2E_2: the die shows any other number, so P(E2)=56P(E_2)=\dfrac56.

Let AA: the man reports that it is 11.

Since the man speaks the truth 33 out of 55 times, P(truth)=35P(\text{truth})=\dfrac35 and P(lie)=25P(\text{lie})=\dfrac25.

P(A∣E1)=P(truth)=35P(A|E_1) = P(\text{truth}) = \dfrac35

If the die does NOT show 11 and he lies, he could (with equal likelihood) claim any of the other 55 faces, so the chance he specifically says '1' while lying is 15\dfrac15:

P(A∣E2)=P(lie)×15=25×15=225P(A|E_2) = P(\text{lie})\times\dfrac15 = \dfrac25\times\dfrac15 = \dfrac2{25}

By Bayes' theorem: …

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