Question

If the range of random variable $X=\{0,1,2,...\}$ and $P(X=k)=\frac{c(k+1)}{2^k}$ for $k=0,1,2,...$ then $c=$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

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

$P(X=0)=c$

$P(X=1)=\frac{2c}{2}$

$P(X=2)=\frac{3c}{2^2}$

$\sum P(X=r)=1$

$\Rightarrow \frac{c}{2^0}+\frac{2c}{2^1}+\frac{3c}{2^2}+........=1$

$\Rightarrow \frac{1c}{2}+\frac{2c}{2^2}+.......=\frac{1}{2}$

$\Rightarrow c+\frac{c}{2}+\frac{c}{2^2}+.......=\frac{1}{2}$

$\Rightarrow c[1+\frac{1}{2}+\frac{1}{2^2}+.....]$

$\Rightarrow c\left[\frac{1}{1-\frac{1}{2}}\right]=\frac{1}{2}$

$\Rightarrow c\left[2\right]=\frac{1}{2}$

$\Rightarrow c=\frac{1}{4}$