If a discrete random variable $X$ is defined as follows $\mathrm{P}[\mathrm{X}=x]=\left\{\begin{array}{cl} \frac{\mathrm{k}(x+1)}{5^x}, & ext { if } x=0,1,2 \ldots \ldots \ 0, & ext { otherwise } \end{array} ight.$ then $\mathrm{k}=$

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If a discrete random variable $X$ is defined as follows $\mathrm{P}[\mathrm{X}=x]=\left\{\begin{array}{cl} \frac{\mathrm{k}(x+1)}{5^x}, & 	ext { if } x=0,1,2 \ldots \ldots \ 0, & 	ext { otherwise } \end{array}ight.$ then $\mathrm{k}=$

Aicharya · Accepted Answer

We have, $\sum_{x=0}^{\infty} \mathrm{P}(\mathrm{X}=x)=1$ $\begin{aligned} & \Rightarrow \mathrm{k} \sum_{x=0}^{\infty}(x+1)\left(\frac{1}{5}ight)^x=1 \ & \Rightarrow \mathrm{k}\left[1+2\left(\frac{1}{5}ight)+3\left(\frac{1}{5}ight)^2+4\left(\frac{1}{5}ight)^3+\ldotsight]=1 \end{aligned}$ $\Rightarrow \mathrm{k}\left[\frac{1}{1-\frac{1}{5}}+\frac{1 	imes \frac{1}{5}}{\left(1-\frac{1}{5}ight)^2}ight]=1$ $\cdots\left[\begin{array}{r}\because a+(a+d) r+(a+2 d) r^2+\ldots . \ =\frac{a}{1-r}+\frac{d r}{(1-r)^2}\end{array}ight]$ $\begin{aligned} & \Rightarrow \mathrm{k}\left(\frac{5}{4}+\frac{5}{16}ight)=1 \ & \Rightarrow \frac{25 \mathrm{k}}{16}=1 \ & \Rightarrow \mathrm{k}=\frac{16}{25}\end{aligned}$

If a discrete random variable $X$ is defined as follows $\mathrm{P}[\mathrm{X}=x]=\left\{\begin{array}{cl}…

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