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

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

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) The sum of all probabilities in a probability distribution must be equal to $1$. Thus,$\sum_{x=0}^{\infty} P(X=x) = 1$. Substituting the given expression: $\sum_{x=0}^{\infty} k(x+1)\left(\frac{1}{2}\right)^x = 1$. $k \sum_{x=0}^{\infty} (x+1)r^x = 1$,where $r = \frac{1}{2}$. We know that $\sum_{x=0}^{\infty} (x+1)r^x = 1 + 2r + 3r^2 + \dots = (1-r)^{-2}$. For $r = \frac{1}{2}$,the sum is $(1 - \frac{1}{2})^{-2} = (\frac{1}{2})^{-2} = 4$. Therefore,$k(4) = 1$,which gives $k = \frac{1}{4}$. Now,we need to find $P(X=1)$. $P(X=1) = k(1+1)\left(\frac{1}{2}\right)^1 = k(2)(\frac{1}{2}) = k$. Since $k = \frac{1}{4}$,$P(X=1) = \frac{1}{4}$.

$X=x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$2K$	$0$	$0.3$	$K$	$0.1$

$X=k$	$0$	$1$	$2$	$3$	$4$
$P(X=k)$	$0.1$	$0.4$	$0.3$	$0.2$	$0$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ random variable $X$ takes values $0, 1, 2, 3, \dots$ with probabilities $P(X=x) = k(x+1)\left(\frac{1}{2}\right)^x$. If $k$ is a constant,then $P(X=1) = $

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