If a discrete random variable \(X\) takes values \(0,1,2,3, \ldots \ldots .\). with probability \(\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 5^{-x}\), where k is a constant, then \(\mathrm{P}(\mathrm{X}=0)\) is

Given that \(\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 5^{-x}\), where \(\mathrm{X}=0,1,2,3, \ldots\)
...(i)
\(\begin{aligned}
& \text { Since, } \sum_{x=0}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\
& \Rightarrow \mathrm{k} \sum_{x=0}^{\infty}(x+1) 5^{-x}=1 \\
& \Rightarrow \mathrm{k}\left[1+2(5)^{-1}+3(5)^{-2}+4(5)^{-3}+\ldots\right]=1 \\
& \Rightarrow \mathrm{k}\left[1+2\left(\frac{1}{5}\right)+3\left(\frac{1}{5}\right)^2+4\left(\frac{1}{5}\right)^3+\ldots\right]=1
\end{aligned}\)
\(\begin{gathered}
\Rightarrow \mathrm{k} \times\left[\frac{1}{1-\frac{1}{5}}+\frac{1 \times \frac{1}{5}}{\left(1-\frac{1}{5}\right)^2}\right]=1 \\
\Rightarrow \mathrm{k} \times \frac{25}{16}=1 \\
\Rightarrow \mathrm{k}=\frac{16}{25} \\
\therefore \quad \mathrm{P}(\mathrm{X}=0)=\frac{16}{25}(0+1)\left(\frac{1}{5}\right)^{\circ} \\...[From (i)]
\quad=\frac{16}{25}
\end{gathered}\)