Question

A man in known to speak the truyth 3 out of 4 times he throws a die and reports that it is six find the probability that it is actually a six

Answer 1

Let us define that,

E1=A speaks truth; E2=A tells a lie; E=A reports a six

Given,

P(E1)=34, P(E2)=14, P(E|E1)=1/6, P(E|E2)=5/6.

The required probability that actually there were six (by Bayes’) is

P(E1|E)=[P(E1)⋅P(E|E1)]/[P(E1)⋅P(E|E1)+P(E2)⋅P(E|E2)]=3/8

Your answer is 3/8

Answer 2

Answer:

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