A random variable X takes the values 0, 1, 2, | vedclass

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(A) We know that the sum of all probabilities in a probability distribution is $1$,so $\sum_{x=0}^{\infty} P(X=x) = 1$.Substituting the given expressi

Vedclass · Accepted Answer

$\frac{16}{25}$

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(A) We know that the sum of all probabilities in a probability distribution is $1$,so $\sum_{x=0}^{\infty} P(X=x) = 1$.Substituting the given expression: $k \sum_{x=0}^{\infty} (x+1) \left(\frac{1}{5}ight)^x = 1$.Expanding the series: $k \left[ 1 + 2\left(\frac{1}{5}ight) + 3\left(\frac{1}{5}ight)^2 + 4\left(\frac{1}{5}ight)^3 + \ldots ight] = 1$.This is an arithmetico-geometric series of the form $\sum_{n=0}^{\infty} (a+nd)r^n = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ where $a=1, d=1, r=\frac{1}{5}$.Calculating the sum: $k \left[ \frac{1}{1-\frac{1}{5}} + \frac{1 	imes \frac{1}{5}}{(1-\frac{1}{5})^2} ight] = 1$.$k \left[ \frac{5}{4} + \frac{1/5}{16/25} ight] = 1 \Rightarrow k \left[ \frac{5}{4} + \frac{5}{16} ight] = 1$.$k \left[ \frac{20+5}{16} ight] = 1 \Rightarrow k \left( \frac{25}{16} ight) = 1 \Rightarrow k = \frac{16}{25}$.Now,$P(X=0) = k(0+1)\left(\frac{1}{5}ight)^0 = k(1)(1) = k$.Therefore,$P(X=0) = \frac{16}{25}$.

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

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