A random variable $X$ takes values $0,1,2,3, \ldots$ with probability…
A random variable $X$ takes values $0,1,2,3, \ldots$ with probability $P(X=x)=K(x+1)\left(\frac{1}{5}\right)^x$, where $K$ is constant, then $P(X=0)$ is
$\frac{7}{25}$
$\frac{18}{25}$
$\frac{16}{25}$
$\frac{13}{25}$
Solution
Given, $P(X=x)=K(x+1)\left(\frac{1}{5}\right)^x$
To find, $P(X=0)$
Since sum of all the probabilities in a probability distribution
is 1 .
$\therefore \quad P(X=0)+P(X=1)+P(X=2)+\ldots=1$
$\Rightarrow K(0+1)(1 / b)^0+K(1+1)\left(\frac{1}{5}\right)^1+K(2+1)\left(\frac{1}{5}\right)^2+\ldots=1$
$\begin{array}{ll}\Rightarrow & K+2 K\left(\frac{1}{5}\right)+3 K\left(\frac{1}{5}\right)^2+\ldots=1 \\ \Rightarrow & K\left[1+2\left(\frac{1}{5}\right)+3\left(\frac{1}{5}\right)^2+\ldots\right]=1\end{array}$
Let $r=1 / 5$
$K\left[1+2 r+3 r^2+\ldots\right]=1$
$\begin{aligned} \Rightarrow & K(1-r)^{-2} & =1 \\ \Rightarrow & K\left(1-\frac{1}{5}\right)^{-2} & =1\end{aligned}$
$\Rightarrow \quad K=\frac{16}{25}$
$\therefore \quad P(X=0)=\frac{16}{25}(0+1)\left(\frac{1}{5}\right)^0=\frac{16}{25}$