A) \[\frac{11}{16}\]
B) \[\frac{15}{16}\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: A
Solution :
[a] \[\because E\left( X \right)=np=2\]and variance \[\left( X \right)=npq=1\] \[q=\frac{1}{2}.\] \[\therefore p=1-q=1-\frac{1}{2}=\frac{1}{2}\] Now, \[n\left( \frac{1}{2} \right)=2\] \[\Rightarrow n=4\]g \[\therefore p(X>1)=1-p(X\le 1)\] \[=1-\left[ ^{4}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{4}}{{+}^{4}}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{3}}.\left( \frac{1}{2} \right) \right]\] \[=1-\left[ 1{{\left( \frac{1}{2} \right)}^{4}}+4.{{\left( \frac{1}{2} \right)}^{4}}. \right]=1-\left( \frac{1}{16}+\frac{4}{16} \right)\] \[=1-\frac{5}{16}=\frac{16-5}{16}=\frac{11}{16}\] Hence option [a] is correct.
You need to login to perform this action.
You will be redirected in
3 sec
