A man speaks truth in 80 % of the cases and a | vedclass

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(B) Let the two men be $A$ and $B$. Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.Gi

Vedclass · Accepted Answer

$\frac{13}{50}$

Vedclass · Answer

(B) Let the two men be $A$ and $B$. Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.Given: $P(A) = \frac{80}{100} = 0.8$ and $P(B) = \frac{90}{100} = 0.9$.The probability that $A$ lies is $P(\bar{A}) = 1 - 0.8 = 0.2$.The probability that $B$ lies is $P(\bar{B}) = 1 - 0.9 = 0.1$.They contradict each other if one speaks the truth and the other lies.This can happen in two mutually exclusive ways: ($A$ speaks truth $AND$ $B$ lies) $OR$ ($B$ speaks truth $AND$ $A$ lies).Required Probability $= P(A) 	imes P(\bar{B}) + P(B) 	imes P(\bar{A})$.$= (0.8 	imes 0.1) + (0.9 	imes 0.2)$.$= 0.08 + 0.18 = 0.26$.$= \frac{26}{100} = \frac{13}{50}$.

$A$ man speaks truth in $80 \%$ of the cases and another in $90 \%$ of the cases. While stating the same fact,what is the probability that they contradict each other?

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