A) \[\frac{1}{4}\]
B) \[\frac{2}{3}\]
C) \[\frac{2}{5}\]
D) \[\frac{3}{5}\]
Correct Answer: C
Solution :
Total number of ways \[=5\,!\] Favourable number of ways \[2\,.\,4\,!\] Hence required probability \[=\frac{2\,.\,4\,!}{5\,!}=\frac{2}{5}\].
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