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 ) respectively, are:

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 .

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MathematicsProbabilityRandom Variable and its Probability DistributionMedium2 minPYQ_2021

MathematicsMediumsingle choice

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:

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Answer:

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) $= \frac{1}{3^{2}} + \frac{1}{3^{4}} + \dots \infty$

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

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

Stream:JEESubject:MathematicsTopic:ProbabilitySubtopic:Random Variable and its Probability Distribution

⏱ 2mℹ️ Source: PYQ_2021

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