A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{3}{4}\]
D) None of these
Correct Answer: C
Solution :
Required probability \[={}^{3}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{1}}.{{\left( \frac{1}{2} \right)}^{2}}+{}^{3}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{2}}.{{\left( \frac{1}{2} \right)}^{1}}=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\].
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