JEE Mains · Maths · STD 12 - 13. probability
Let a random variable X take values \(0,1,2,3\) with \(\mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)=\mathrm{p}, \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{X}=3)\) and \(\mathrm{E}\left(\mathrm{X}^2\right)=2 \mathrm{E}(\mathrm{X})\). Then the value of \(8 \mathrm{p}-1\) is :
- A 0
- B 2
- C 1
- D 3
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\begin{aligned} & 2 p+2 q=\frac{1}{2} \\ & p+q \\ & E\left(x^2\right)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 \cdot p+4 \cdot q+9 q \\ & =p+13 q \\ & E(x)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 \cdot p+2 q+3 q=p+5 q \\ & p+13 q=2(p+5 q) \\ & p=3 q \\ & \text {…
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