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14.Probability
medium
‘$X$’ speaks truth in $60\%$ and ‘$Y$’ in $50\%$ of the cases. The probability that they contradict each other narrating the same incident is
A
$\frac{1}{4}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$\frac{2}{3}$
Solution
(c) Here $P(X) = \frac{3}{5}$, $P(Y) = \frac{1}{2}$
$\therefore $ Required probability $= P\,(X) \cdot P(\bar Y) + P(\bar X)P(Y)$
$=\left( {\frac{3}{5}} \right)\,\left( {1 – \frac{1}{2}} \right) + \left( {1 – \frac{2}{5}} \right)\,\left( {\frac{1}{2}} \right)$
$=\frac{3}{5}\,.\,\frac{1}{2} + \frac{2}{5}.\frac{1}{2} = \frac{1}{2}$.
Standard 11
Mathematics
