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 -Maths 9th

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 -Maths 9th
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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 truthNow, P(\(\bar{E}\)1) = \(rac{75}{100}\) = \(rac{3}{4}\) ⇒ P( \(\bar{E}\)1) = 1 - \(rac{3}{4}\) = \(rac{1}{4}\) P(\(\bar{E}\)2) = \(rac{80}{100}\) = \(rac{4}{5}\) ⇒ P(\(\bar{E}\)2) = 1 - \(rac{4}{5}\) = \(rac{1}{5}\)∴ Required probability = P(E1) x  P(\(\bar{E}\)2) +  P(\(\bar{E}\)1) x P(E2) = \(rac{3}{4}\)x \(rac{1}{5}\)+\(rac{1}{4}\)x \(rac{4}{5}\) = \(rac{7}{20}\).∴ % of cases in which A and B contradict each other = \(\bigg(rac{7}{20} imes100\bigg)\)% = 35%
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