A random variable X has the range {0, 1, 2, l | vedclass

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(B) The sum of all probabilities in a probability distribution is $1$. Therefore, $\sum_{r=0}^{\infty} P(X=r) = 1$. Given $P(X=r) = k(1+r) 3^{-r}$, we

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$\frac{8}{9}$

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(B) The sum of all probabilities in a probability distribution is $1$. Therefore, $\sum_{r=0}^{\infty} P(X=r) = 1$. Given $P(X=r) = k(1+r) 3^{-r}$, we have $k \sum_{r=0}^{\infty} (1+r) \left(\frac{1}{3}ight)^r = 1$. Let $S = \sum_{r=0}^{\infty} (1+r) x^r$ where $x = \frac{1}{3}$. This is an arithmetico-geometric series: $S = 1 + 2x + 3x^2 + 4x^3 + \ldots$. Multiplying by $x$: $xS = x + 2x^2 + 3x^3 + \ldots$. Subtracting the two: $S(1-x) = 1 + x + x^2 + x^3 + \ldots = \frac{1}{1-x}$. Thus, $S = \frac{1}{(1-x)^2}$. For $x = \frac{1}{3}$, $S = \frac{1}{(1 - 1/3)^2} = \frac{1}{(2/3)^2} = \frac{9}{4}$. So, $k 	imes \frac{9}{4} = 1 \implies k = \frac{4}{9}$. Now, $P(X=0) + P(X=1) + P(X=2) = k \left[ (1+0)3^0 + (1+1)3^{-1} + (1+2)3^{-2} ight]$. $= \frac{4}{9} \left[ 1 + \frac{2}{3} + \frac{3}{9} ight] = \frac{4}{9} \left[ 1 + \frac{2}{3} + \frac{1}{3} ight] = \frac{4}{9} 	imes 2 = \frac{8}{9}$.

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

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0$	$2k$	$k$	$3k$	$2k^2$	$2k$	$k^2+k$	$7k^2$

$A$ random variable $X$ has the range $\{0, 1, 2, \ldots\}$. If $P(X=r) = k(1+r) 3^{-r}$ for $r=0, 1, 2, \ldots$, where $k > 0$ is a real number, then $P(X=0) + P(X=1) + P(X=2) =$

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