If the range of a random variable X is {0, 1, | vedclass

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(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.As we know that the sum of all probabilities in a probability dis

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$\frac{4}{9}$

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(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.As we know that the sum of all probabilities in a probability distribution is $1$,so $\sum_{k=0}^{\infty} P(X=k) = 1$.Substituting the given expression:$a \left( 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty ight) = 1 \quad \dots (i)$Let $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty$.Then $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \ldots \infty$.Subtracting the two equations:$S - \frac{1}{3}S = 1 + \left( \frac{2}{3} - \frac{1}{3} ight) + \left( \frac{3}{3^2} - \frac{2}{3^2} ight) + \ldots \infty$$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \infty$This is an infinite geometric series with first term $1$ and common ratio $r = \frac{1}{3}$.$\frac{2}{3}S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{2/3} = \frac{3}{2}$.$S = \frac{3}{2} 	imes \frac{3}{2} = \frac{9}{4}$.From equation $(i)$,$a 	imes S = 1 \implies a 	imes \frac{9}{4} = 1$.Therefore,$a = \frac{4}{9}$.

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

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

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$,then $a$ is equal to

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