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Last updated at Feb. 15, 2020 by Teachoo

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Example 21 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.Let S1 : man speaks the truth S2 : man lies E : six on the die We need to find the Probability that it is actually a six, if the man reports that it a six i.e. P(S1|E) P(S1|E) = (๐(๐_1 ).๐(๐ธ|๐_1))/(๐(๐_1 ).๐(๐ธ|๐_1)+๐(๐_2 ).๐(๐ธ|๐_2)) P(S1) = Probability that man speaks truth = ๐/๐ P(E|S1) = Probability that six appears on the die, if the man speaks the truth = ๐/๐ P(S2) = Probability man lies = 1 โ P(E) = 1 โ 3/4 = ๐/๐ P(E|S2) = Probability that six appears on the die, if the man lies = P(6 does not appear) = 1 โ 1/6 = ๐/๐ Putting value in formula, P(S1|E) = (๐(๐_1 ).๐(๐ธ|๐_1))/(๐(๐_1 ).๐(๐ธ|๐_1)+๐(๐_2 ).๐(๐ธ|๐_2)) = (3/4 ร 1/6)/( 3/4 ร 1/6 + 1/4 ร 5/6 ) = (1/4 ร 1/6 ร 3)/( 1/4 ร 1/6 [3 + 5] ) = 3/8 Therefore, required probability is ๐/๐

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Chapter 13 Class 12 Probability

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.