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

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(B) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.Given: $P(A) = 0.8$,$P(B) = 0.9$.T

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) Let $P(A)$ be the probability that $A$ speaks the truth and $P(B)$ be the probability that $B$ speaks the truth.Given: $P(A) = 0.8$,$P(B) = 0.9$.Therefore,the probability that $A$ lies is $P(A') = 1 - 0.8 = 0.2$,and the probability that $B$ lies is $P(B') = 1 - 0.9 = 0.1$.They contradict each other if $A$ speaks the truth and $B$ lies,$OR$ $A$ lies and $B$ speaks the truth.Required probability $= P(A) 	imes P(B') + P(A') 	imes P(B)$$= (0.8 	imes 0.1) + (0.2 	imes 0.9)$$= 0.08 + 0.18 = 0.26$.Converting to percentage: $0.26 	imes 100 = 26 \%$.

$A$ speaks truth in $80 \%$ of the cases and $B$ in $90 \%$ of the cases. Find the percentage when they contradict each other in describing the same facts. (in $\%$)

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