A) 2 ? p
B) 3 ? p
C) \[\frac{p}{2}\]
D) \[\frac{p}{3}\]
Correct Answer: B
Solution :
Since X has a binomial distribution, B (n, p) \[\therefore \]\[P(X=2){{=}^{n}}{{C}_{2}}{{(p)}^{2}}{{(1-p)}^{n-2}}\] and\[P(X=3){{=}^{n}}{{C}_{3}}{{(p)}^{3}}{{(1-p)}^{n-3}}\] Given \[P(X=2)=P(X-3)\] \[\Rightarrow \]\[^{n}{{C}_{2}}{{p}^{2}}{{(1-p)}^{n-2}}{{=}^{n}}{{C}_{3}}{{(p)}^{3}}{{(1-p)}^{n-3}}\] \[\Rightarrow \]\[\frac{n!}{2!(n-2)!}.\frac{{{p}^{2}}{{(1-p)}^{n}}}{{{(1-p)}^{2}}}=\frac{n!}{3!(n-3)!}.\frac{{{p}^{3}}{{(1-p)}^{n}}}{{{(1-p)}^{3}}}\] \[\Rightarrow \]\[\frac{1}{n-2}=\frac{1}{3}.\frac{p}{1-p}\]\[\Rightarrow \]\[3(1-p)=p(n-2)\] \[\Rightarrow \]\[3-3p=np-2p\]\[\Rightarrow \]\[np=3-p\] \[\Rightarrow \]\[E(X)=mean\,=3-p\] (\[\because \] mean of B (n, p) = np)
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