Question

A man is known to speak the truth 3 out of 4 times . He throws a die and reports that is a six . find the probability that it is actually a six .​

Answer 1

Answer:

p (it is actually six) = 3

4

Answer 2

Solution ; -

Given that , A man speak 3 out of four time .

Let E be the Event that the man reports that 6 occurs S1 be the event that 6 occurs and S2 be the event that 6 doesn't occurs .

since, P(S1) = 1/6 and : P(S2) = 5/6

P(E/S1) = 3/4 , P(E/S2 ) = 1-3/4 = 1/4

By using Bayes' theorem.

P(S1/E) = P(S1)* P(E/S1) /P(S1) P(E)S1) + P(S2) P(ES2)

P(S1/E) = (1/6* 3/4) /1/6*3/4+5/6*1/4 = 3/8 Answer

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