Question

A random variable $X$ takes values $0,1,2,3,...$ with probability $p(X=x)=k(x+1){\left(\frac{1}{5}\right)}^x$ where $k$ is constant, then $P\{X=0)$ is

• A. $7/25$
• B. $18/25$
• C. $13/25$
• D. $19/25$
• E. $16/25$

Answer 1

Answer: (E)

$\therefore$ $P(X=0)=kp(X=1)=2k{\left(\frac{1}{5}\right)}^1$ $p(X=2)=3k{\left(\frac{1}{5}\right)}^1,....$
Since, $P(X=0)+P(X=1)+P(X=2)$ $+.....=1$ $\begin{array}{rl}& \underline{\begin{array}{rl}& \therefore k+2k\left(\frac{1}{5}\right)+3k{\left(\frac{1}{5}\right)}^2+.....=1\\ & +\frac{k}{5}+2k{\left(\frac{1}{5}\right)}^2+......=\frac{1}{5}\\ & ----\end{array}}\\ & k+k\frac{1}{5}+k{\left(\frac{1}{5}\right)}^2+.....=\frac{4}{5}\end{array}$
$\Rightarrow$ $\frac{k}{1-\frac{1}{5}}=\frac{4}{5}$
$\Rightarrow$ $k=\frac{16}{25}$
$\therefore$ $P(X=0)=\frac{16}{25}(0+1){\left(\frac{1}{5}\right)}^0$