X speaks the truth in 60% of cases and Y in 5 | vedclass

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(C) Let $P(X)$ be the probability that $X$ speaks the truth and $P(Y)$ be the probability that $Y$ speaks the truth.Given: $P(X) = 60\% = \frac{60}{10

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) Let $P(X)$ be the probability that $X$ speaks the truth and $P(Y)$ be the probability that $Y$ speaks the truth.Given: $P(X) = 60\% = \frac{60}{100} = \frac{3}{5}$ and $P(Y) = 50\% = \frac{50}{100} = \frac{1}{2}$.$X$ and $Y$ contradict each other if one speaks the truth and the other lies.This can happen in two ways: ($X$ speaks the truth $AND$ $Y$ lies) $OR$ ($X$ lies $AND$ $Y$ speaks the truth).Required probability $= P(X) \cdot P(\bar{Y}) + P(\bar{X}) \cdot P(Y)$.Here,$P(\bar{X}) = 1 - P(X) = 1 - \frac{3}{5} = \frac{2}{5}$ and $P(\bar{Y}) = 1 - P(Y) = 1 - \frac{1}{2} = \frac{1}{2}$.Required probability $= (\frac{3}{5} \cdot \frac{1}{2}) + (\frac{2}{5} \cdot \frac{1}{2}) = \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$.

$X$ speaks the truth in $60\%$ of cases and $Y$ in $50\%$ of cases. The probability that they contradict each other while narrating the same incident is

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