A) \[\frac{9}{41}\]
B) \[\frac{25}{128}\]
C) \[\frac{1}{5}\]
D) \[\frac{27}{128}\]
Correct Answer: D
Solution :
Required probability\[={}^{4}{{C}_{2}}{{\left( \frac{3}{4} \right)}^{2}}{{\left( \frac{1}{4} \right)}^{2}}\] \[=\frac{4!}{2!2!}\times \frac{9}{16}\times \frac{1}{16}\] \[=\frac{24}{4}\times \frac{9}{16}\times \frac{1}{16}=\frac{27}{128}\]
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