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

Question

Dr. Abubacker · Accepted Answer

(d)
$P(X = 0) = k,\ P(X = 1)2k\left( \frac{1}{5} ight)^{1}$
$P(X = 2) = 3k\left( \frac{1}{5} ight)^{2},\ldots$
Since, $P(X = 0) + P(X = 1) + P(X = 2) + ... = 1$
$	herefore k + 2k\left( \frac{1}{5} ight) + 3k\left( \frac{1}{5} ight)^{2} + ... = 1$
and $+ \frac{k}{5} + 2k\left( \frac{1}{5} ight)^{2} + ... = \frac{1}{5}$
$\ \ \  - \ \ \  - \ \ \ \ \ \  - \ \ \ \ \ \ \  - \ \ \ \ \ \ \ \ \ \ \ \  -$
­­­­­­­­­­­­­____________________________________
$k + k\left( \frac{1}{5} ight) + k\left( \frac{1}{5} ight)^{2} + ... = \frac{4}{5}$
$\Rightarrow \ \frac{k}{1 - \frac{1}{5}} = \frac{4}{5}$
$\Rightarrow \ k = \frac{16}{25}$
$	herefore\ P(X = 0) = \frac{16}{25}(0 + 1)\left( \frac{1}{5} ight)^{0} = \frac{16}{25}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

Want more practice?