Question

Let $X$ be a random variable such that the probability function of a distribution is given by$P(X=0)=\frac{1}{2},P(X=j)=\frac{1}{3^j}(j=1,2,3,\dots ,\infty )$. Then the mean of the distribution and $P\left(X\text{is positive and even}\right)$ respectively are

• A. $\frac{3}{4}$ and $\frac{1}{9}$
• B. $\frac{3}{4}$ and $\frac{1}{16}$
• C. $\frac{3}{4}$ and $\frac{1}{8}$
• D. $\frac{3}{8}$ and $\frac{1}{8}$

Answer 1

The correct option is C $\frac{3}{4}$ and $\frac{1}{8}$

Mean of distribution $\left(M\right)$
$=\sum x_i{P}_i$
$\Rightarrow M=0\left(\frac{1}{2}\right)+1\left(\frac{1}{3}\right)+2\left(\frac{1}{3^2}\right)+3\left(\frac{1}{3^3}\right)+\dots \infty$
$\Rightarrow M=\frac{1}{3}+2\left(\frac{1}{3^2}\right)+3\left(\frac{1}{3^3}\right)+\dots \infty$
$\Rightarrow \frac{M}{3}=\frac{1}{3^2}+2\left(\frac{1}{3^3}\right)+\dots \infty$
$\therefore M-\frac{M}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \infty$
$\Rightarrow \frac{2M}{3}=\frac{\frac{1}{3}}{\frac{2}{3}}\Rightarrow \frac{2M}{3}=\frac{1}{2}$
$\therefore$ Mean $=\frac{3}{4}$

$P(X\text{is positive and even)}$
$=\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\dots$
$=\frac{\frac{1}{3^2}}{1-\frac{1}{3^2}}=\frac{1}{8}$