A) \[\frac{37}{256}\]
B) \[\frac{219}{256}\]
C) \[\frac{128}{256}\]
D) \[\frac{28}{256}\]
Correct Answer: D
Solution :
Given that, mean\[=4\Rightarrow np=4\] and variance = 2 \[\Rightarrow \] \[npq=2\] \[\Rightarrow \] \[4q=2\] \[\Rightarrow \] \[q=\frac{1}{2}\] \[\therefore \] \[p=1-q=1-\frac{1}{2}=\frac{1}{2}\] Also, \[n=8\] Probability of 2 successes\[=P(X=2){{=}^{8}}{{C}_{2}}{{p}^{2}}{{q}^{6}}\] \[=\frac{8!}{2!6!}\times {{\left( \frac{1}{2} \right)}^{2}}\times {{\left( \frac{1}{2} \right)}^{6}}=28\times \frac{1}{{{2}^{8}}}=\frac{28}{256}\]
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