If the probability that the random variable X takes values x is given by P(X=x)=k(x+1)3^-x…

Question

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

Accepted Answer

As, we know that sum of all the probabilities $=1$

$$
\begin{aligned}
& \text { So, } \sum_{x=1}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\
& \Rightarrow k\left[1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+\ldots . \infty\right]=1
\end{aligned}
$$

$$
\begin{aligned}
& \text { Let } S=1+\frac{2}{3}+\frac{3}{3^2}+\ldots .+\infty \\
& \Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots .+\infty
\end{aligned}
$$

On subtracting, we get

$$
\begin{aligned}
& \frac{2 S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots .+\infty \\
& \Rightarrow \frac{2 S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}} \\
& \Rightarrow \frac{2 S}{3}=\frac{3}{2} \\
& \Rightarrow S=\frac{9}{4}
\end{aligned}
$$

$$
\begin{aligned}
& \text { So, } k \times \frac{9}{4}=1 \Rightarrow k=\frac{4}{9} \\
& \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{X}<2) \\
& =1-\mathrm{P}(\mathrm{X}=0)-\mathrm{P}(\mathrm{X}=1) \\
& =1-\frac{4}{9}(1)-\frac{4}{9} \times \frac{2}{3} \\
& =1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27}
\end{aligned}
$$

Solutions · 1