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.
I have been asked this question in homework.
My query is from Bayes Theorem topic in chapter Probability of Mathematics – Class 12
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Correct answer is (d) 3/8
Easiest explanation: Let E = event that the man reports that six in the throwing of the die and let, S1 = event that six occurs and S2 = event that six does not occur.
P(S1) = Probability that six occurs = 1/6.
P(S2) = Probability that six does not occur = 5/6.
Also, P(E|S1) = P(Probability that man reports six occurs when six actually has occurred on the die) = 3/4.
P(E|S2) = P(Probability that man reports six occurs when six not actually occurred on the die) =
1 – 3/4 = 1/4.
By using Bayes’ theorem,
P(S2|E) = P(S1)P(E|S1)/(P(S1)P(E│S1)+P(S2)P(E|S2))
= (1/6 × 3/4) / ((1/6 × 3/4)) + (5/6 × 1/4)) = 3/8.