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Probability question

2021 · 25 Jul · Shift 2 · Q38
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  5. /2021 · 25 Jul · Shift 2 · Q38

Probability question

2021 · 25 Jul · Shift 2 · Q38

JEE MainMathematicsProbabilityMCQ+4 / −1
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,...,∞)P(X = 0) = {1 \over 2},P(X = j) = {1 \over {{3^j}}}(j = 1,2,3,...,\infty )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 :
  1. A
    38{3 \over 8}83​ and 18{1 \over 8}81​
  2. B
    34{3 \over 4}43​ and 18{1 \over 8}81​
  3. C
    34{3 \over 4}43​ and 19{1 \over 9}91​
  4. D
    34{3 \over 4}43​ and 116{1 \over 16}161​
View written solutionFree

Correct answer: B

  1. Given probability distribution

We have P(X=0)=12,P(X=0)=\frac12,P(X=0)=21​, and for j=1,2,3,…j=1,2,3,\dotsj=1,2,3,…, P(X=j)=13j.P(X=j)=\frac{1}{3^j}.P(X=j)=3j1​.

First, verify that this is a valid distribution: ∑j=1∞13j=131−13=12.\sum_{j=1}^{\infty} \frac{1}{3^j} = \frac{\frac13}{1-\frac13} = \frac12.∑j=1∞​3j1​=1−31​31​​=21​. So total probability is P(X=0)+∑j=1∞P(X=j)=12+12=1.P(X=0)+\sum_{j=1}^{\infty}P(X=j)=\frac12+\frac12=1.P(X=0)+∑j=1∞​P(X=j)=21​+21​=1. Hence it is valid.

  1. Find the mean E(X)E(X)E(X)

By definition, E(X)=∑xxP(X=x).E(X)=\sum_x xP(X=x).E(X)=∑x​xP(X=x). So,

=\sum_{j=1}^{\infty} \frac{j}{3^j}.$$ Now use the standard result $$\sum_{j=1}^{\infty} jx^j=\frac{x}{(1-x)^2}, \quad |x|<1.$$ Here $x=\frac13$, so $$E(X)=\frac{\frac13}{\left(1-\frac13\right)^2} =\frac{\frac13}{\left(\frac23\right)^2} =\frac{\frac13}{\frac49} =\frac13\cdot \frac94 =\frac34.$$ Thus, $$\boxed{E(X)=\frac34}.$$ 3. **Find $P(X \text{ is positive and even})$** Positive even values are $2,4,6,\dots$. Therefore, $$P(X \text{ is positive and even})=\sum_{k=1}^{\infty} P(X=2k) =\sum_{k=1}^{\infty} \frac{1}{3^{2k}}.$$ This is a geometric series with first term $\frac{1}{9}$ and common ratio $\frac{1}{9}$: $$\sum_{k=1}^{\infty} \frac{1}{3^{2k}}=\sum_{k=1}^{\infty} \left(\frac19\right)^k =\frac{\frac19}{1-\frac19} =\frac{\frac19}{\frac89} =\frac18.$$ Thus, $$\boxed{P(X \text{ is positive and even})=\frac18}.$$ 4. **Match with options** We obtained: - Mean $= \frac34$ - $P(X$ is positive and even$)=\frac18$ This matches **Option B**.
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