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(B) For a probability distribution, the sum of all probabilities must be $1$: $\sum_{k=0}^{\infty} P(X=k) = 1$. Substituting the given expression: $c

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$\frac{1}{18}$

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(B) For a probability distribution, the sum of all probabilities must be $1$: $\sum_{k=0}^{\infty} P(X=k) = 1$. Substituting the given expression: $c \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 1$. Let $S = \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 3 \sum_{k=0}^{\infty} \frac{1}{3^k} + 2 \sum_{k=0}^{\infty} \frac{k}{3^k}$. The first part is a geometric series: $3 	imes \frac{1}{1 - 1/3} = 3 	imes \frac{3}{2} = \frac{9}{2}$. The second part is an arithmetico-geometric series: $\sum_{k=0}^{\infty} k x^k = \frac{x}{(1-x)^2}$. For $x = 1/3$, this is $\frac{1/3}{(1-1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$. So, $S = \frac{9}{2} + 2 	imes \frac{3}{4} = \frac{9}{2} + \frac{3}{2} = 6$. Since $c 	imes S = 1$, we have $c = 1/6$. Now, $P(X=3) = \frac{(2(3)+3)c}{3^3} = \frac{9c}{27} = \frac{c}{3}$. Substituting $c = 1/6$, we get $P(X=3) = \frac{1/6}{3} = \frac{1}{18}$.

If $X$ is a random variable with probability distribution $P(X=k) = \frac{(2k+3)c}{3^k}$, $k=0, 1, 2, \ldots, \infty$, then $P(X=3) =$

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