A) \[\frac{1}{4}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{2}\]
D) \[\frac{2}{3}\]
Correct Answer: C
Solution :
Here \[P(X)=\frac{3}{5}\], \[P(Y)=\frac{1}{2}\] \[\therefore \] Required probability = \[P\,(X)\].\[P(\bar{Y})+P(\bar{X})P(Y)\] =\[\left( \frac{3}{5} \right)\,\left( 1-\frac{1}{2} \right)+\left( 1-\frac{2}{5} \right)\,\left( \frac{1}{2} \right)\]=\[\frac{3}{5}\,.\,\frac{1}{2}+\frac{2}{5}.\frac{1}{2}=\frac{1}{2}\].
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