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(B) Given that $P(X=x) = k(x+1) 5^{-x}$ for $x = 0, 1, 2, 3, \ldots$. Since the sum of all probabilities must be $1$,we have $\sum_{x=0}^{\infty} P(X=

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$\frac{16}{25}$

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(B) Given that $P(X=x) = k(x+1) 5^{-x}$ for $x = 0, 1, 2, 3, \ldots$. Since the sum of all probabilities must be $1$,we have $\sum_{x=0}^{\infty} P(X=x) = 1$. $k \sum_{x=0}^{\infty} (x+1) (\frac{1}{5})^x = 1$. 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 this series is given by $\frac{1}{(1-r)^2}$. So,$k 	imes \frac{1}{(1 - \frac{1}{5})^2} = 1$. $k 	imes \frac{1}{(\frac{4}{5})^2} = 1$. $k 	imes \frac{25}{16} = 1$,which gives $k = \frac{16}{25}$. Now,$P(X=0) = k(0+1) 5^{-0} = k(1)(1) = k$. Therefore,$P(X=0) = \frac{16}{25}$.

If a discrete random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = k(x+1) 5^{-x}$,where $k$ is a constant,then $P(X=0)$ is

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