‘X’ speaks truth in 60% and &lsqu | vedclass

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Question

(c) Here $P(X) = \frac{3}{5}$, $P(Y) = \frac{1}{2}$

$	herefore $ Required probability $= P\,(X) \cdot P(\bar Y) + P(\bar X)P(Y)$

$=\left( {\fra

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(c) Here $P(X) = \frac{3}{5}$, $P(Y) = \frac{1}{2}$

$	herefore $ Required probability $= P\,(X) \cdot P(\bar Y) + P(\bar X)P(Y)$

$=\left( {\frac{3}{5}} ight)\,\left( {1 - \frac{1}{2}} ight) + \left( {1 - \frac{2}{5}} ight)\,\left( {\frac{1}{2}} ight)$

$=\frac{3}{5}\,.\,\frac{1}{2} + \frac{2}{5}.\frac{1}{2} = \frac{1}{2}$.

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

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