JEE Main 2014 — Mathematics Probability & Statistics

Since $\mathrm{X}$X has a binomial distribution, $\mathrm{B}(n$B(n, p) $$\begin{aligned} &\therefore \mathrm{P}(X=2)={ }^n \mathrm{C}_2(p)^2(1-p)^{n-2} \\ &\text { and } \mathrm{P}(X=3)={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\ &\text { Given } \mathrm{P}(X=2)=\mathrm{P}(X=3) \\ &\Rightarrow{ }^n \mathrm{C}_2 p^2(1-p)^{n-2}={ }^n \mathrm{C}_3(p)^3(1-p)^{n-3} \\ &\Rightarrow \frac{n !}{2 !(n-2) !} \cdot \frac{p^2(1-p)^n}{(1-p)^2} \\ &=\frac{n !}{3 !(n-3) !} \cdot \frac{p^3(1-p)^n}{(1-p)^3} \end{aligned}$$​∴P(X=2)=nC2​(p)2(1−p)n−2 and P(X=3)=nC3​(p)3(1−p)n−3 Given P(X=2)=P(X=3)⇒nC2​p2(1−p)n−2=nC3​(p)3(1−p)n−3⇒2!(n−2)!n!​⋅(1−p)2p2(1−p)n​=3!(n−3)!n!​⋅(1−p)3p3(1−p)n​​ $$\begin{aligned} &\Rightarrow \frac{1}{n-2}=\frac{1}{3} \cdot \frac{p}{1-p} \\ &\Rightarrow 3(1-p)=p(n-2) \\ &\Rightarrow 3-3 p=n p-2 p \\ &\Rightarrow n p=3-p \\ &\Rightarrow \mathrm{E}(X)=\text { mean }=3-p \\ &(\because \text { mean of } \mathrm{B}(n, p)=n p) \end{aligned}$$​⇒n−21​=31​⋅1−pp​⇒3(1−p)=p(n−2)⇒3−3p=np−2p⇒np=3−p⇒E(X)= mean =3−p(∵ mean of B(n,p)=np)​