Let X be a random variable such that the probability function of a distribution is given by P(X=0)=12,P(X=j)=13j(j=1,2,3,.....,∞). Then the mean of the distribution and P(X is positive and even ) respectively are

Mean =∑j=0∞jP(X=j)=0(12)+∑j=1∞j(13j)=13+232+332+434+... =13[1+2(13)+3(12)2+...]=13(1−13)−2=13×94=34P(X is positive and even)=P(X=2)+P(X=4)+... =132+134+136+...=1/321−1/32=19×98=18

Let X be a random variable such that the probability function of a distribution is given by P(X=0)=12,P(X=j)=13j(j=1,2,3,.....,∞). Then the mean of the distribution and P(X is positive and even ) respectively are

Mean =∑j=0∞jP(X=j)=0(12)+∑j=1∞j(13j)=13+232+332+434+... =13[1+2(13)+3(12)2+...]=13(1−13)−2=13×94=34P(X is positive and even)=P(X=2)+P(X=4)+... =132+134+136+...=1/321−1/32=19×98=18

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

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

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

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

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

answer is B.

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Detailed Solution

Mean $= \sum_{j = 0}^{\infty} j P (X = j) = 0 (\frac{1}{2}) + \sum_{j = 1}^{\infty} j (\frac{1}{3^{j}}) = \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{2}} + \frac{4}{3^{4}} + ...$
$= \frac{1}{3} [1 + 2 (\frac{1}{3}) + 3 {(\frac{1}{2})}^{2} + ...] = \frac{1}{3} {(1 - \frac{1}{3})}^{- 2} = \frac{1}{3} \times \frac{9}{4} = \frac{3}{4}$
$P (X$ is positive and even $) = P (X = 2) + P (X = 4) + ...$
$= \frac{1}{3^{2}} + \frac{1}{3^{4}} + \frac{1}{3^{6}} + ... = \frac{1 / 3^{2}}{1 - 1 / 3^{2}} = \frac{1}{9} \times \frac{9}{8} = \frac{1}{8}$