Mathematics - Probability Question with Solution | TestHub

Since $\mathrm{X}$ has a binomial distribution, $\mathrm{B}(n$, p) $$\begin{array}{rl}& \therefore \mathrm{P}(X=2)={}^n{\mathrm{C}}_2(p{)}^2(1-p{)}^{n-2}\\ & \text{ and }\mathrm{P}(X=3)={}^n{\mathrm{C}}_3(p{)}^3(1-p{)}^{n-3}\\ & \text{ Given }\mathrm{P}(X=2)=\mathrm{P}(X=3)\\ & \Rightarrow {}^n{\mathrm{C}}_2{p}^2(1-p{)}^{n-2}={}^n{\mathrm{C}}_3(p{)}^3(1-p{)}^{n-3}\\ & \Rightarrow \frac{n!}{2!(n-2)!}\cdot \frac{p^2(1-p{)}^n}{(1-p{)}^2}\\ & =\frac{n!}{3!(n-3)!}\cdot \frac{p^3(1-p{)}^n}{(1-p{)}^3}\end{array}$$ $$\begin{array}{rl}& \Rightarrow \frac{1}{n-2}=\frac{1}{3}\cdot \frac{p}{1-p}\\ & \Rightarrow 3(1-p)=p(n-2)\\ & \Rightarrow 3-3p=np-2p\\ & \Rightarrow np=3-p\\ & \Rightarrow \mathrm{E}(X)=\text{ mean }=3-p\\ & (\because \text{ mean of }\mathrm{B}(n,p)=np)\end{array}$$