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 a constant, then $P(X=0)$ is equal to

• A. $\frac{7}{25}$
• B. $\frac{18}{25}$
• C. $\frac{13}{25}$
• D. $\frac{19}{25}$
• E. $\frac{16}{25}$

Answer 1

The correct option is E

$\frac{16}{25}$

Explanation for the correct option:

Step 1. Find the value of $P(X=0)$:

Given that $P(X=x)=k(x+1){\left(\frac{1}{5}\right)}^x$, $x$ takes the values such as $0,1,2,3,\dots .$

As We know,

$\sum _{x=0}^{\infty }P(X=x)=1$

$\Rightarrow$ $k\sum _{x=0}^{\infty }(x+1){\left(\frac{1}{5}\right)}^x=1$

$\Rightarrow$$k\left[1+2\left(\frac{1}{5}\right)+3{\left(\frac{1}{5}\right)}^2+4{\left(\frac{1}{5}\right)}^3+\dots .\right]=1$

Step 2. On solving this equation, we get

$\Rightarrow$$k\left(\frac{25}{16}\right)=1$

$\Rightarrow$$k=\frac{16}{25}$

Now, $P(X=0)=\left(\frac{16}{25}\right)(0+1){\left(\frac{1}{5}\right)}^0$

$\Rightarrow$ $P(X=0)=\frac{16}{25}\times 1\times 1$

$\Rightarrow$ $P(X=0)=\frac{\mathbf{16}}{\mathbf{25}}$

Hence, Option ‘E’ is Correct.