For a random variable X,if P(X=k) = frac{(k+1 | vedclass

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(B) The sum of probabilities for all possible values of a random variable must be equal to $1$. Thus,$\sum_{k=0}^{\infty} P(X=k) = 1$. Substituting th

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$4/9$

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(B) The sum of probabilities for all possible values of a random variable must be equal to $1$. Thus,$\sum_{k=0}^{\infty} P(X=k) = 1$. Substituting the given expression: $\sum_{k=0}^{\infty} \frac{(k+1)a}{3^k} = 1$. Taking $a$ as a constant: $a \sum_{k=0}^{\infty} (k+1) \left(\frac{1}{3}ight)^k = 1$. Let $S = \sum_{k=0}^{\infty} (k+1) x^k$ where $x = 1/3$. This is an arithmetico-geometric series: $S = 1 + 2x + 3x^2 + 4x^3 + \ldots$. We know that $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$. Differentiating both sides with respect to $x$: $\sum_{k=1}^{\infty} k x^{k-1} = \frac{1}{(1-x)^2}$. Multiplying by $x$: $\sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$. Then $S = \sum_{k=0}^{\infty} (k+1) x^k = \sum_{k=0}^{\infty} k x^k + \sum_{k=0}^{\infty} x^k = \frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{x + 1 - x}{(1-x)^2} = \frac{1}{(1-x)^2}$. For $x = 1/3$,$S = \frac{1}{(1 - 1/3)^2} = \frac{1}{(2/3)^2} = \frac{1}{4/9} = 9/4$. Therefore,$a 	imes (9/4) = 1$,which gives $a = 4/9$.

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

For a random variable $X$,if $P(X=k) = \frac{(k+1)a}{3^k}$ for $k=0, 1, 2, \ldots$,then $a = $

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