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

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(A) The sum of all probabilities in a probability distribution must be equal to $1$. Therefore,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the give

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$\frac{16}{25}$

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(A) The sum of all probabilities in a probability distribution must be equal to $1$. Therefore,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the given expression: $\sum_{x=0}^{\infty} k(x+1)\left(\frac{1}{5}ight)^x = 1$. This is an arithmetico-geometric series of the form $\sum_{x=0}^{\infty} (x+1)r^x$ where $r = \frac{1}{5}$. The sum of the series $\sum_{x=0}^{\infty} (x+1)r^x = 1 + 2r + 3r^2 + \dots = \frac{1}{(1-r)^2}$. Substituting $r = \frac{1}{5}$: $\frac{1}{(1 - 1/5)^2} = \frac{1}{(4/5)^2} = \frac{1}{16/25} = \frac{25}{16}$. Thus,$k 	imes \frac{25}{16} = 1$,which gives $k = \frac{16}{25}$. We need to find $P(X=0)$. $P(X=0) = k(0+1)\left(\frac{1}{5}ight)^0 = k 	imes 1 	imes 1 = k$. Therefore,$P(X=0) = \frac{16}{25}$.

$A$ random variable $X$ takes the values $0, 1, 2, 3, \dots$ 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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