If the probability that the random variable X takes the value $x$ is given by $P(X=x)=k(x+1) 3^{-x}$, $\mathrm{x}=0,1,2,3 \ldots \ldots$, where k is a constant, then $\mathrm{P}(\mathrm{X} \geq 3)$ is equal to

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Aicharya · Accepted Answer

$\sum_{x=0}^{\infty} k(x+1) 3^{-x}=1$
$\Rightarrow \frac{1}{\mathrm{k}}=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+$ ...(i)
$\frac{1}{3 \mathrm{k}}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots$ ...(ii)
(i)- (ii) $\Rightarrow \frac{1}{\mathrm{k}}-\frac{1}{3 \mathrm{k}}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots$
$\begin{aligned} & \Rightarrow \mathrm{k}=\frac{4}{9} \\ & \mathrm{P}(\mathrm{x} \geq 3)=1-\mathrm{P}(\mathrm{x}=0)-\mathrm{P}(\mathrm{x}=1)-\mathrm{P}(\mathrm{x}=2) \\ & =1-\mathrm{k}\left(1+\frac{2}{3}+\frac{3}{9}\right)=\frac{1}{9}\end{aligned}$ ,

If the probability that the random variable X takes the value $x$ is given by $P(X=x)=k(x+1) 3^{-x}$,…

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