Q.11.The p.m.f of random variate X is
P(X) = \begin{cases} \frac{2x}{\mathrm{n}(\mathrm{n}+1)}, & x=1,2,3,\ldots, n \\ 0, & \text{otherwise} \end{cases} \text{Then } E(X) =A. \frac{n+1}{3}
B. \frac{2n+1}{3}
C. \frac{n+2}{3}
D. \frac{2n-1}{3}
Answer :- B. \frac{2n+1}{3}
Explanation :-
\mathrm{P(X)} = \begin{cases} \frac{2x}{\mathrm{n(n+1)}},x=1,2,3,\ldots,\mathrm{n} \\ 0\quad,\mathrm{~otherwise} \end{cases} \therefore \mathrm{E(X)} = \sum_{i=1}^n x_i p(x_i) =\frac{2}{\mathrm{n(n+1)}}+\frac{8}{\mathrm{n(n+1)}}+\ldots+\frac{2\mathrm{n^2}}{\mathrm{n(n+1)}} =\frac{2\left(1^2+2^2+\ldots+\mathrm{n}^2\right)}{\mathrm{n}(\mathrm{n}+1)} =\frac{\mathrm{2n(n+1)(2n+1)}}{\mathrm{6n(n+1)}} =\frac{2\mathrm{n}+1}{3}