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(B) We know that the sum of all probabilities in a distribution is $1$,so $\sum_{k=0}^{\infty} P(X=k) = 1$. Given $P(X=k) = \frac{(k+1)c}{2^k}$,we hav

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$\frac{5}{16}$

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(B) We know that the sum of all probabilities in a distribution is $1$,so $\sum_{k=0}^{\infty} P(X=k) = 1$. Given $P(X=k) = \frac{(k+1)c}{2^k}$,we have $c \sum_{k=0}^{\infty} \frac{k+1}{2^k} = 1$. Let $S = \sum_{k=0}^{\infty} \frac{k+1}{2^k} = 1 + \frac{2}{2} + \frac{3}{4} + \frac{4}{8} + \ldots$. Then $\frac{1}{2}S = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \ldots$. Subtracting the two equations: $S - \frac{1}{2}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = \frac{1}{1 - 1/2} = 2$. Thus,$\frac{1}{2}S = 2$,which means $S = 4$. Since $c \cdot S = 1$,we get $4c = 1$,so $c = \frac{1}{4}$. Now,$P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$. $P(X=0) = \frac{(0+1)c}{2^0} = c = \frac{1}{4}$. $P(X=1) = \frac{(1+1)c}{2^1} = c = \frac{1}{4}$. $P(X=2) = \frac{(2+1)c}{2^2} = \frac{3c}{4} = \frac{3}{16}$. $P(X \geq 3) = 1 - [\frac{1}{4} + \frac{1}{4} + \frac{3}{16}] = 1 - [\frac{4+4+3}{16}] = 1 - \frac{11}{16} = \frac{5}{16}$.

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

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

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

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