MCQ
The p.m.f. of a random variable $X$ is
$
P(x)=\left\{\begin{array}{cc}
\frac{2 x}{n(n+1)}, & x=1,2,3, \ldots . . n \\
0, & \text { otherwise, }
\end{array}\right.
$
then $E(X)$ is
  • A
    $\frac{n+1}{6}$
  • B
    $\frac{2 n+1}{6}$
  • C
    $\frac{n+1}{3}$
  • D
    $\frac{2 n+1}{3}$

Answer

(d) : We have, $E(X)=\Sigma x P(X=x)$
$
\begin{aligned}
& =1 \times P(X=1)+2 \times P(X=2)+\ldots \ldots+n \times P(X=n) \\
& =\frac{2}{n(n+1)}+2 \times \frac{2 \times 2}{n(n+1)}+\ldots \ldots .+n \frac{2 \times n}{n(n+1)} \\
& =\frac{2}{n(n+1)}\left[1^2+2^2+\ldots . .+n^2\right] \\
& =\frac{2}{n(n+1)}\left[\frac{n(n+1)(2 n+1)}{6}\right]=\frac{2 n+1}{3}
\end{aligned}
$

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