A random variable X takes values 0, 1, 2, 3, | vedclass

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(C) Given,$P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$.Since the sum of all probabilities in a probability distribution is $1$,we have $\sum_{x=0}^{\in

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$\frac{16}{25}$

Vedclass · Answer

(C) Given,$P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$.Since the sum of all probabilities in a probability distribution is $1$,we have $\sum_{x=0}^{\infty} P(X=x) = 1$.Substituting the given expression:$\sum_{x=0}^{\infty} K(x+1)\left(\frac{1}{5}\right)^x = 1$$K \left[ 1 + 2\left(\frac{1}{5}\right) + 3\left(\frac{1}{5}\right)^2 + 4\left(\frac{1}{5}\right)^3 + \ldots \right] = 1$.This is an arithmetico-geometric series of the form $\sum_{n=1}^{\infty} n r^{n-1} = (1-r)^{-2}$ where $r = \frac{1}{5}$.Thus,$K(1 - \frac{1}{5})^{-2} = 1$.$K(\frac{4}{5})^{-2} = 1 \Rightarrow K(\frac{5}{4})^2 = 1$.$K(\frac{25}{16}) = 1 \Rightarrow K = \frac{16}{25}$.Now,$P(X=0) = K(0+1)\left(\frac{1}{5}\right)^0 = K(1)(1) = K$.Therefore,$P(X=0) = \frac{16}{25}$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$1/6$	$k$	$1/4$	$k$	$1/6$

$A$ random variable $X$ takes 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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