A speaks the truth in 75% of cases and B spea | vedclass

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(A) Let $P(A)$ be the probability that $A$ speaks the truth,so $P(A) = 75/100 = 3/4$. Then $P(A') = 1 - 3/4 = 1/4$ is the probability that $A$ lies. L

Vedclass · Accepted Answer

$7/20$

Vedclass · Answer

(A) Let $P(A)$ be the probability that $A$ speaks the truth,so $P(A) = 75/100 = 3/4$. Then $P(A') = 1 - 3/4 = 1/4$ is the probability that $A$ lies. Let $P(B)$ be the probability that $B$ speaks the truth,so $P(B) = 80/100 = 4/5$. Then $P(B') = 1 - 4/5 = 1/5$ is the probability that $B$ lies. They contradict each other if one speaks the truth and the other lies. This can happen in two ways: $1$. $A$ speaks the truth and $B$ lies: $P(A) 	imes P(B') = (3/4) 	imes (1/5) = 3/20$. $2$. $A$ lies and $B$ speaks the truth: $P(A') 	imes P(B) = (1/4) 	imes (4/5) = 4/20$. The total probability of contradiction is the sum of these probabilities: $P(	ext{contradiction}) = 3/20 + 4/20 = 7/20$.

$A$ speaks the truth in $75\%$ of cases and $B$ speaks the truth in $80\%$ of cases. What is the probability that they contradict each other in stating the same fact?

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