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

As, we know that sum of all the probabilities $=1$ <br/><br/>$$ \begin{aligned} & \text { So, } \sum_{x=1}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\\\ & \Rightarrow k\left[1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+\ldots . \infty\right]=1 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Let } S=1+\frac{2}{3}+\frac{3}{3^2}+\ldots .+\infty \\\\ & \Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots .+\infty \end{aligned} $$ <br/><br/>On subtracting, we get <br/><br/>$$ \begin{aligned} & \frac{2 S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots .+\infty \\\\ & \Rightarrow \frac{2 S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}} \\\\ & \Rightarrow \frac{2 S}{3}=\frac{3}{2} \\\\ & \Rightarrow S=\frac{9}{4} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { So, } k \times \frac{9}{4}=1 \Rightarrow k=\frac{4}{9} \\\\ & \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{X}<2) \\\\ & =1-\mathrm{P}(\mathrm{X}=0)-\mathrm{P}(\mathrm{X}=1) \\\\ & =1-\frac{4}{9}(1)-\frac{4}{9} \times \frac{2}{3} \\\\ & =1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27} \end{aligned} $$