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

Correct answer: (B) 3/4 and 1/8. We have, PX=0=1/2,PX=j=(13^jj=1,2,3,…,∞Mean =ΣX_iP_i=∑r=0∞r·13^r=3)/4 P(X is even) =(13^2+13^4+…∞=1)/91-1/9 =19/89=1/8.

Q45 / 90 · JEE Main 2021 (25 Jul Shift 2) · Mathematics

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

A
$\frac{3}{8}$ and $\frac{1}{8}$
Correct answer
B
$\frac{3}{4}$ and $\frac{1}{8}$
C
$\frac{3}{4}$ and $\frac{1}{9}$
D
$\frac{3}{4}$ and $\frac{1}{16}$

Solution

We have,

P (X = 0) = \frac{1}{2}, P (X = j) = \frac{1}{3^{j}} (j = 1, 2, 3, \dots, \infty)

Mean $= Σ X_{i} P_{i} = \sum_{r = 0}^{\infty} r \cdot \frac{1}{3^{r}} = \frac{3}{4}$

$P$ ( $X$ is even) $= \frac{1}{3^{2}} + \frac{1}{3^{4}} + \dots \infty$

$= \frac{\frac{1}{9}}{1 - \frac{1}{9}}$

$= \frac{\frac{1}{9}}{\frac{8}{9}} = \frac{1}{8} .$

Hence, the correct answer is option: 'B'.

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Solution

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