A) \[\frac{1}{256}\]
B) \[\frac{3}{256}\]
C) \[\frac{9}{256}\]
D) \[\frac{7}{256}\]
Correct Answer: C
Solution :
Given mean of Binomial distribution \[=np=4\] ?..(i) and \[\text{variance = npq = 2}\] ??(ii) From Eqs. (i) and (ii), \[4q=2\] \[\Rightarrow \] \[q=\frac{2}{4}=\frac{1}{2}\] \[\therefore \] \[p=(1-q)=1-\frac{1}{2}=\frac{1}{2}\] From Eq. (i), \[n=4\times 2\] \[\Rightarrow \] \[n=8\] Now, \[P(X>6){{=}^{8}}{{C}_{7}}\left( \frac{1}{2} \right){{\left( \frac{1}{2} \right)}^{7}}+{{\left( \frac{1}{2} \right)}^{8}}\] \[=\frac{8}{256}+\frac{1}{256}=\frac{9}{256}\]
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