Question

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

• A. $\frac{2}{3}$
• B. $\frac{4}{9}$
• C. $\frac{8}{27}$
• D. $\frac{16}{81}$

Answer 1

The correct option is B

$\frac{4}{9}$

Explanation of the correct option.

Given : Random variable $X$ has a range $\left\{0,1,2,3,4,..\right\}$.

$P(X=K)=\left[\left(\frac{\left(k+1\right)a}{3^k}\right)\right]$ for $k\ge 0$

We know that the sum of all the probabilities is equal to $1$.

$\sum _{k=0}^{\infty }\frac{\left(k+1\right)a}{3^k}$

$\Rightarrow a\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+............\infty \right)=1$

We can see that above series is the arithmetic geometric series with common difference $1$ and common ratio $\frac{1}{3}$.

We know that sum to infinity in AGP with the first term $\text{'}a\text{'}$ common difference $\text{'}d\text{'}$ and common ratio $\text{'}r\text{'}$ is,

$S_{\infty }=\frac{a}{\left(1-r\right)}+\frac{dr}{{\left(1-r\right)}^2}$ for $\left|r\right|<1$

$$\begin{gathered} \Rightarrow a\left(\frac{1}{\left(1-{\displaystyle \frac{1}{3}}\right)}+\frac{1\left({\displaystyle \frac{1}{3}}\right)}{{\left(1-{\displaystyle \frac{1}{3}}\right)}^2}\right)=1 \\ \Rightarrow a\left(\frac{1}{{\displaystyle \frac{2}{3}}}+\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{4}{9}}}\right)=1 \\ \Rightarrow a\left(\frac{3}{2}+\frac{3}{4}\right)=1 \\ \Rightarrow a\left(\frac{9}{4}\right)=1 \\ \Rightarrow a=\frac{4}{9} \end{gathered}$$

Hence, option $\left(B\right)$ is the correct option.