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(A) Given the probability mass function $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \{0, 1, 2, \dots\}$.Since the sum of all probabilities must be $1$, w

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$\frac{20}{27}$

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(A) Given the probability mass function $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \{0, 1, 2, \dots\}$.Since the sum of all probabilities must be $1$, we have $\sum_{n=0}^{\infty} P(X=n) = 1$.$k \sum_{n=0}^{\infty} \frac{n+1}{3^n} = k \left( \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots \right) = 1$.This is an arithmetico-geometric series with $a=1$, $d=1$, and $r=\frac{1}{3}$.The sum of the series is $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{1}{1-1/3} + \frac{1 \cdot (1/3)}{(1-1/3)^2} = \frac{3}{2} + \frac{1/3}{4/9} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$.Thus, $k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}$.We need to find $P(X $P(X=0) = k \cdot \frac{0+1}{3^0} = k \cdot 1 = \frac{4}{9}$.$P(X=1) = k \cdot \frac{1+1}{3^1} = k \cdot \frac{2}{3} = \frac{4}{9} \cdot \frac{2}{3} = \frac{8}{27}$.$P(X < 2) = \frac{4}{9} + \frac{8}{27} = \frac{12+8}{27} = \frac{20}{27}$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

If the probability function of a random variable $X$ is given by $P(X=n) = \frac{k(n+1)}{3^n}$ for $n \in \mathbb{N} \cup \{0\}$ where $k$ is a constant, then $P(X < 2) = $

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