X speaks truth in 60% and Y in 50% of the cas | 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}{100

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(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 $P(Y) = 50\% = \frac{50}{100} = \frac{1}{2}$.Then $P(\bar{X}) = 1 - \frac{3}{5} = \frac{2}{5}$ and $P(\bar{Y}) = 1 - \frac{1}{2} = \frac{1}{2}$.They contradict each other if one speaks the truth and the other lies.Required probability $= P(X) \cdot P(\bar{Y}) + P(\bar{X}) \cdot P(Y)$$= (\frac{3}{5} \cdot \frac{1}{2}) + (\frac{2}{5} \cdot \frac{1}{2})$$= \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$.

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

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