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(A) The sum of all probabilities must be $1$: $\sum_{x=0}^{\infty} P(X = x) = 1$. $k \sum_{x=0}^{\infty} (x + 1)3^{-x} = 1$. Let $S = 1 + 2(3^{-1}) +

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$\frac{7}{27}$

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(A) The sum of all probabilities must be $1$: $\sum_{x=0}^{\infty} P(X = x) = 1$. $k \sum_{x=0}^{\infty} (x + 1)3^{-x} = 1$. Let $S = 1 + 2(3^{-1}) + 3(3^{-2}) + 4(3^{-3}) + \ldots = \sum_{x=0}^{\infty} (x + 1)3^{-x}$. This is an arithmetico-geometric series. $S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \ldots$ $\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \ldots$ Subtracting the two: $S - \frac{1}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots$ $\frac{2}{3}S = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2}$. $S = \frac{3}{2} 	imes \frac{3}{2} = \frac{9}{4}$. Since $kS = 1$,we have $k = \frac{4}{9}$. We need $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$. $P(X = 0) = k(0 + 1)3^0 = k = \frac{4}{9}$. $P(X = 1) = k(1 + 1)3^{-1} = 2k 	imes \frac{1}{3} = \frac{2}{3} 	imes \frac{4}{9} = \frac{8}{27}$. $P(X \geq 2) = 1 - (\frac{4}{9} + \frac{8}{27}) = 1 - (\frac{12 + 8}{27}) = 1 - \frac{20}{27} = \frac{7}{27}$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

If the probability that the random variable $X$ takes values $x$ is given by $P(X = x) = k(x + 1)3^{-x}$ for $x = 0, 1, 2, 3, \ldots$,where $k$ is a constant,then $P(X \geq 2)$ is equal to

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