Let x be a random variable such that the probability function of a distribution is given by P(X=0)=(1/2), P(X=j)=(1/3j)(j=1,2,3, ldots ldots, ∞), Then the mean of the distribution and P(X is positive and even) respectively are:

Question

Let x be a random variable such that the probability function of a distribution is given by P(X=0)=(1/2), P(X=j)=(1/3j)(j=1,2,3, ldots ldots, ∞), Then the mean of the distribution and P(X is positive and even) respectively are: (A) (3/8) and (1/8) (B) (3/4) and (1/8) (C) (3/4) and (1/9) (D) (3/4) and (1/16)

Tardigrade Admin · Accepted Answer

mean =∑ x1 pi= displaystyle∑r=0∞ r ⋅ (1/3r)=(3/4) p(x is even )=(1/32)+(1/34)+ ldots . ∞ =((1/9)/1-(1)9)=(1 / 9/8 / 9)=(1/8)

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

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

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

$\frac{3}{4}$ and $\frac{1}{9}$

$\frac{3}{4}$ and $\frac{1}{16}$

mean $= \sum x_{1} p_{i} = r = 0 \sum \infty r \cdot \frac{1}{3 ^{r}} = \frac{3}{4}$
$p (x$ is even $) = \frac{1}{3 ^{2}} + \frac{1}{3 ^{4}} + \dots .\infty$
$= \frac{\frac{1}{9}}{1 - \frac{1}{9}} = \frac{1/9}{8/9} = \frac{1}{8}$

Q. Let x be a random variable such that the probability function of a distribution is given by P(X=0)=21​,P(X=j)=3j1​(j=1,2,3,……,∞), Then the mean of the distribution and P(X is positive and even) respectively are:

83​ and 81​

43​ and 81​

43​ and 91​

43​ and 161​