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

Given,

The probability that the random variable $X$ takes values $x$ is given by $P(X=x)=k(x+1)3^{-x}$, $x=0,1,2,3,\dots \dots$, where $k$ is a constant,

 Now we know that,

$P(X=0)+P(x=1)+P(x=2)+\dots \dots \dots =1$

$\Rightarrow \frac{k}{3^0}+\frac{2k}{3^1}+\frac{3k}{3^2}+\dots ..=1$

$\Rightarrow k\left(1+\frac{2}{3}+\frac{3}{3^2}+\dots .\right)=1$

Now finding,

$S=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\dots ....\left(1\right)$

$\Rightarrow \frac{S}{3}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+\dots \dots .....\left(2\right)$

Now on subtracting above two equation we get,

$\Rightarrow \frac{2S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\dots \dots$

$\Rightarrow S=\frac{9}{4}$

Hence, $k\left(1+\frac{2}{3}+\frac{3}{3^2}+\dots .\right)=1$

$\Rightarrow k=\frac{4}{9}$

Now finding,

 $P(X\ge 2)=P\left(2\right)+P\left(3\right)+\dots \dots$

$\Rightarrow P(X\ge 2)=1-P\left(0\right)-P\left(1\right)$

$\Rightarrow P(X\ge 2)=1-\left(\frac{k}{1}+\frac{2k}{3}\right)=1-\frac{20}{27}=\frac{7}{27}$