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Worked Examples · Example 21

Q.A man is known to speak 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 Bayes' theorem problem. The probability the die actually shows a six, given the report, is 38\dfrac{3}{8}.

Let SS be the event "the die shows a six" and RR the event "the man reports a six".

P(S)=16,P(S′)=56.P(S)=\frac16,\qquad P(S')=\frac56.

He speaks truth 33 of 44 times, so if it is a six he correctly reports six with probability 34\tfrac34, and if it is not a six he (lying) reports six with probability 14\tfrac14:

P(R∣S)=34,P(R∣S′)=14.P(R\mid S)=\frac34,\qquad P(R\mid S')=\frac14.

By Bayes' theorem, …

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