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(D) The mean $E(X)$ is given by $\sum_{j=0}^{\infty} j \cdot P(X=j)$.Since $P(X=0) = 0 \cdot \frac{1}{2} = 0$,we have $E(X) = \sum_{j=1}^{\infty} j \c

Vedclass · Accepted Answer

$\frac{3}{4}$ and $\frac{1}{8}$

Vedclass · Answer

(D) The mean $E(X)$ is given by $\sum_{j=0}^{\infty} j \cdot P(X=j)$.Since $P(X=0) = 0 \cdot \frac{1}{2} = 0$,we have $E(X) = \sum_{j=1}^{\infty} j \cdot \frac{1}{3^j}$.This is an arithmetico-geometric series of the form $\sum_{j=1}^{\infty} j r^j$ where $r = \frac{1}{3}$.The sum is given by $\frac{r}{(1-r)^2} = \frac{1/3}{(1-1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$.For $P(X 	ext{ is positive and even})$,we sum $P(X=j)$ for $j \in \{2, 4, 6, \ldots\}$.$P(X 	ext{ is positive and even}) = \sum_{k=1}^{\infty} P(X=2k) = \sum_{k=1}^{\infty} \frac{1}{3^{2k}} = \sum_{k=1}^{\infty} \left(\frac{1}{9}ight)^k$.This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{9}$.The sum is $\frac{a}{1-r} = \frac{1/9}{1-1/9} = \frac{1/9}{8/9} = \frac{1}{8}$.Thus,the mean is $\frac{3}{4}$ and the probability is $\frac{1}{8}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

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

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

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

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