If a random variable $X$ takes the values $x_1, x_2, x_3, \ldots, x_{100}$ with probability $P\left(X=x_i\right)=$ $K i(i+1)$, then $200 K=$

Question

If a random variable $X$ takes the values $x_1, x_2, x_3, \ldots, x_{100}$ with probability $P\left(X=x_i\right)=$ $K i(i+1)$, then $200 K=$, $\frac{1}{1707}$, $\frac{1}{1717}$, $\frac{1}{1727}$, $\frac{1}{1777}$

MARKS App · Accepted Answer

It is given that for random variable $X$ which can take values $x_1, x_2, x_3, \ldots . ., x_{100}$ with probability $P\left(x=x_i\right)=K i(i+1)$ $\begin{aligned} & \because \quad \sum_{i=1}^{100} P\left(X=x_i\right)=1 \Rightarrow K \sum_{i=1}^{100} i(i+1)=1 \ & \Rightarrow \quad K \sum_{i=1}^{100}\left(i^2+i\right)=1 \ & \Rightarrow \quad K\left[\frac{100(100+1)(200+1)}{6}+\frac{100(100+1)}{2}\right]=1 \ & \Rightarrow \quad K \frac{100 \times 101}{6}[201+3]=1 \ & \Rightarrow \quad K=\frac{6}{100 \times 101 \times 204}[201+3]=1 \ & \therefore 200 K=200 \times \frac{6}{100 \times 101 \times 204}=\frac{1}{101 \times 17}=\frac{1}{1717} \end{aligned}$ Hence, option (b) is correct.

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