If the range of a random variable $X$ is $\{0,1,2,3,4, \ldots \ldots\}$ with $P(X=k)=\frac{(k+1) a}{3^k}$ for $k \geq 0$, then $a$ is equal to

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Aicharya · Accepted Answer

Given that $P(X=k)=\frac{(k+1) a}{3^k} 	ext { for } x \in(0,1,2, \ldots \infty)$ As we know that $P(0)+P(1)+P(2)+\ldots \infty=1$ $\begin{aligned} S & =a\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\ldots \inftyight) \ \frac{1}{3} S & =a\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots \inftyight) \ \hline S-\frac{1}{3} S & =a\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \inftyight) \ \Rightarrow \quad \frac{2}{3} S & =a\left(\frac{1}{1-\frac{1}{3}}ight) \ \Rightarrow \quad \frac{2}{3} S & =\frac{3 a}{2} \ \Rightarrow \quad S & =\frac{9 a}{4} \end{aligned}$ From equation (i) $\begin{array}{llrl} \Rightarrow & & \frac{9 a}{4} & =1 \ \Rightarrow & a & =\frac{4}{9} \end{array}$

If the range of a random variable $X$ is $\{0,1,2,3,4, \ldots \ldots\}$ with $P(X=k)=\frac{(k+1) a}{3^k}$…

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