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in Mathematics by (60.8k points)
A man is known to speak truth 3 out of 4 times. He throws a die and report that it is a 6. Find the probability that it is actually 6.

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Let E be the event that the man reports that six occurs in the throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur.

Then P (S1) = Probability that six occurs = 1/6
P (S2 ) = Probability that six does not occur = 5/6
P (E/S1) = Probability that the man reports that six occurs when six has actually occurred on the die

= Probability that the man speaks the truth = 3/4
P (E/S2 ) = Probability that the man reports that six occurs when six has not actually occurred on the die

= Probability that the man does not speak the truth=1-3/4=1/4.

Thus, by Bayes’ theorem, we get
P (S1/E) = Probability that the report of the man that six has occurred is actually a six

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