InterviewSolution
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InterviewSolution
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A speaks truth in 75% and B in 80% of the cases. In what per cent of the cases are they likely to contradict each other in narrating the same incident ? |
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Answer» Let E1 : Event that A speak the truth E2 : Event that B speaks the truth Then, \(\bar{E}\)1 : Event that A tells a lie \(\bar{E}\)2 : Event that B tells a lie E1, E2 and so \(\bar{E}\)1, \(\bar{E}\)2, are independent events. ∴ A and B will contradict each other in the following two mutually exclusive ways (i) A speaks the truth , B tells a lie (ii) A tells a lie, B speaks the truth Now, P(\(\bar{E}\)1) = \(\frac{75}{100}\) = \(\frac{3}{4}\) ⇒ P( \(\bar{E}\)1) = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\) P(\(\bar{E}\)2) = \(\frac{80}{100}\) = \(\frac{4}{5}\) ⇒ P(\(\bar{E}\)2) = 1 - \(\frac{4}{5}\) = \(\frac{1}{5}\) ∴ Required probability = P(E1) x P(\(\bar{E}\)2) + P(\(\bar{E}\)1) x P(E2) = \(\frac{3}{4}\)x \(\frac{1}{5}\)+\(\frac{1}{4}\)x \(\frac{4}{5}\) = \(\frac{7}{20}\). ∴ % of cases in which A and B contradict each other = \(\bigg(\frac{7}{20}\times100\bigg)\)% = 35% |
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