Question

A random variable $X$ takes values $0,1,2,3,...$ with probability proportional to $(x+1){\left(\frac{1}{5}\right)}^x$. Then

• A. $P(X=0)=\frac{16}{25}$
• B. $P(X\ge 1)=\frac{9}{25}$
• C. $P(x\ge 1)=\frac{1}{25}$
• D. $E\left(X\right)=\frac{1}{2}$

Answer 1

Answer: (A), (B), (D)

$P(X=0)=\frac{16}{25}$
$P(X\ge 1)=\frac{9}{25}$
$E\left(X\right)=\frac{1}{2}$

Let $P(X=x)=\alpha (x+1){\left(\frac{1}{5}\right)}^x,x\le 0$
We have $\sum P\left(X\right)=1$
$\Rightarrow \alpha [1+2\left(\frac{1}{5}\right)+3{\left(\frac{1}{5}\right)}^2+...]=1$
$\Rightarrow \alpha \frac{1}{{(1-\frac{1}{5})}^2}=1\Rightarrow \frac{25\alpha }{16}=1\Rightarrow \alpha =\frac{16}{25}$.
Now, $P(X=0)=\alpha (0+1){\left(\frac{1}{5}\right)}^0=\alpha =\frac{16}{25}$.
$\Rightarrow P(X\ge 1)=1-P(X=0)=1-\frac{16}{25}=\frac{9}{25}$.
Also, $E\left(X\right)=\sum _{x=0}^{\infty }xP(X=x)=\alpha \sum _{x=0}^{\infty }x(x+1){\left(\frac{1}{5}\right)}^x$
$\Rightarrow \frac{25}{16}E\left(X\right)=\left(1\right)\left(2\right)\left(\frac{1}{5}\right)+\left(2\right)\left(3\right){\left(\frac{1}{5}\right)}^2+\left(3\right)\left(4\right){\left(\frac{1}{5}\right)}^3+.....$
$\Rightarrow \frac{5}{16}E\left(X\right)=\left(1\right)\left(2\right){\left(\frac{1}{5}\right)}^2+\left(2\right)\left(3\right){\left(\frac{1}{5}\right)}^3+.....$
Subtracting, we get
$\Rightarrow \frac{20}{16}E\left(X\right)=2\left(\frac{1}{5}\right)+4{\left(\frac{1}{5}\right)}^2+6{\left(\frac{1}{5}\right)}^3+.....$
$=\frac{2/5}{(1-1/5{)}^2}=\frac{2}{5}\times \frac{25}{16}=\frac{5}{8}$ $\Rightarrow E\left(X\right)=\frac{1}{2}$.