Example 21 - Chapter 13 Class 12 Probability
Last updated at August 13, 2026 by Teachoo
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Chapter 13 Class 12 Probability
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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) = (š(š_1 ).š(šø|š_1))/(š(š_1 ).š(šø|š_1)+š(š_2 ).š(šø|š_2)) P(S1) = Probability that man speaks truth = š/š P(E|S1) = Probability that six appears on the die, if the man speaks the truth = š/š P(S2) = Probability man lies = 1 ā P(E) = 1 ā 3/4 = š/š P(E|S2) = Probability that six appears on the die, if the man lies = P(6 does not appear) = 1 ā 1/6 = š/š Putting value in formula, P(S1|E) = (š(š_1 ).š(šø|š_1))/(š(š_1 ).š(šø|š_1)+š(š_2 ).š(šø|š_2)) = (š/š Ć š/š)/( š/š Ć š/š + š/š Ć š/š ) = (1/4 Ć 1/6 Ć 3)/( 1/4 Ć 1/6 [3 + 5] ) = 3/8 Therefore, required probability is š/š
