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 3 ^ j (j = 1, 2, 3, , ∞) . Then the mean of the distribution and P ( X is positive and even )

Correct answer: B. (3)/(4) and (1)/(8) We have, P(X = 0) = (1)/(2), P(X = j) = 1 3 ^ j (j = 1, 2, 3, , ∞) Mean = Σ X _ i P _ i = ∑ _ r = 0 ^ ∞ r · 1 3 ^ r = (3)/(4) P ( X is even) = 1 3 ^ 2 + 1 3 ^ 4 + ∞ = ((1)/(9))/(1 - (1)/(9)) = ((1)/(9))/((8)/(9)) = (1)/(8) .

JEE Main 202125 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2021 · 25 July · Evening Shift · Mathematics · Probability

Let X be a random variable such that the probability function of a...

Let

X

be a random variable such that the probability function of a distribution is given by

P\left(X = 0\right) = \dfrac{1}{2}, P\left(X = j\right) = \dfrac{1}{{3}^{j}}\left(j = 1, 2, 3, \ldots , \infty\right) .

Then the mean of the distribution and

P ( X

is positive and even

)

respectively, are:

Options

A $\dfrac{3}{8}$ and $\dfrac{1}{8}$
B $\dfrac{3}{4}$ and $\dfrac{1}{8}$
C $\dfrac{3}{4}$ and $\dfrac{1}{9}$
D $\dfrac{3}{4}$ and $\dfrac{1}{16}$

Correct answer

B. $\dfrac{3}{4}$ and $\dfrac{1}{8}$

Step-by-step solution

We have,

P\left(X = 0\right) = \dfrac{1}{2}, P\left(X = j\right) = \dfrac{1}{{3}^{j}}\left(j = 1, 2, 3, \ldots , \infty\right)

Mean $= Σ{X}_{i}{P}_{i} = \mathop{∑}_{r = 0}^{\infty}r \cdot \dfrac{1}{{3}^{r}} = \dfrac{3}{4}$

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

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

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

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