A) \[\frac{291}{364}\]
B) \[\frac{371}{464}\]
C) \[\frac{471}{502}\]
D) \[\frac{459}{512}\]
Correct Answer: D
Solution :
We have \[p=\frac{3}{4}\Rightarrow q=\frac{1}{4}\] and \[n=5\] Therefore required probability \[={}^{5}{{C}_{3}}{{\left( \frac{3}{4} \right)}^{3}}{{\left( \frac{1}{4} \right)}^{2}}+{}^{5}{{C}_{4}}{{\left( \frac{3}{4} \right)}^{4}}\left( \frac{1}{4} \right)+{}^{5}{{C}_{5}}{{\left( \frac{3}{4} \right)}^{5}}\] \[=\frac{10\,.\,27}{{{4}^{5}}}+\frac{5\,.\,81}{{{4}^{5}}}+\frac{243}{{{4}^{5}}}=\frac{270+405+243}{1024}=\frac{459}{512}.\]
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