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Probability question

2025 · 7 Apr · Shift 2 · Q43
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Probability question

2025 · 7 Apr · Shift 2 · Q43

JEE MainMathematicsProbabilityMCQ+4 / −1
Let a random variable X take values 0, 1, 2, 3 with P(X=0)=P(X=1)=p, P(X=2)=P(X=3) and E(X2)=2E(X). Then the value of 8p−1 is :
  1. A
    2
  2. B
    0
  3. C
    3
  4. D
    1
View written solutionFree

Correct answer: A

  1. Set up the probabilities

Given:

  • P(X=0)=P(X=1)=pP(X=0)=P(X=1)=pP(X=0)=P(X=1)=p
  • P(X=2)=P(X=3)P(X=2)=P(X=3)P(X=2)=P(X=3)

Let P(X=2)=P(X=3)=q.P(X=2)=P(X=3)=q.P(X=2)=P(X=3)=q.

Since total probability is 111, p+p+q+q=1p+p+q+q=1p+p+q+q=1 2p+2q=12p+2q=12p+2q=1 p+q=12  ⟹  q=12−p.p+q=\frac12 \implies q=\frac12-p.p+q=21​⟹q=21​−p.

  1. Use the condition E(X2)=2E(X)E(X^2)=2E(X)E(X2)=2E(X)

First compute E(X)E(X)E(X): E(X)=0⋅p+1⋅p+2q+3q=p+5q.E(X)=0\cdot p+1\cdot p+2q+3q=p+5q.E(X)=0⋅p+1⋅p+2q+3q=p+5q.

Now compute E(X2)E(X^2)E(X2): E(X2)=02⋅p+12⋅p+22q+32q=p+4q+9q=p+13q.E(X^2)=0^2\cdot p+1^2\cdot p+2^2q+3^2q=p+4q+9q=p+13q.E(X2)=02⋅p+12⋅p+22q+32q=p+4q+9q=p+13q.

Given: E(X2)=2E(X)E(X^2)=2E(X)E(X2)=2E(X) So, p+13q=2(p+5q).p+13q=2(p+5q).p+13q=2(p+5q).

Simplify: p+13q=2p+10qp+13q=2p+10qp+13q=2p+10q 3q=p.3q=p.3q=p.

  1. Solve for ppp

From p+q=12p+q=\frac12p+q=21​ and p=3qp=3qp=3q, 3q+q=123q+q=\frac123q+q=21​ 4q=124q=\frac124q=21​ q=18.q=\frac18.q=81​. Thus, p=3q=38.p=3q=\frac38.p=3q=83​.

  1. Find 8p−18p-18p−1

8p−1=8(38)−1=3−1=2.8p-1=8\left(\frac38\right)-1=3-1=2.8p−1=8(83​)−1=3−1=2.

  1. Evaluate options
  • A: 222 ✅
  • B: 000
  • C: 333
  • D: 111

Therefore, the correct option is A.

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