A person P speaks the truth in 75% of cases a | vedclass

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(A) Let $P(P)$ be the probability that person $P$ speaks the truth and $P(R)$ be the probability that person $R$ speaks the truth. Given: $P(P) = \fra

Vedclass · Accepted Answer

$\frac{7}{20}$

Vedclass · Answer

(A) Let $P(P)$ be the probability that person $P$ speaks the truth and $P(R)$ be the probability that person $R$ speaks the truth. Given: $P(P) = \frac{75}{100} = \frac{3}{4}$ and $P(R) = \frac{80}{100} = \frac{4}{5}$. Consequently,the probabilities of them lying are $P(P') = 1 - \frac{3}{4} = \frac{1}{4}$ and $P(R') = 1 - \frac{4}{5} = \frac{1}{5}$. They contradict each other if one speaks the truth and the other lies. Case $I$: $P$ speaks the truth and $R$ lies: $P(P) 	imes P(R') = \frac{3}{4} 	imes \frac{1}{5} = \frac{3}{20}$. Case $II$: $R$ speaks the truth and $P$ lies: $P(R) 	imes P(P') = \frac{4}{5} 	imes \frac{1}{4} = \frac{4}{20}$. Total probability of contradiction $= \frac{3}{20} + \frac{4}{20} = \frac{7}{20}$.

$A$ person $P$ speaks the truth in $75\%$ of cases and another person $R$ in $80\%$ of cases. What is the probability that they are likely to contradict each other in narrating the same event?

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