CUET · MATHS · PYQ PAPER 2023
The random variable X can take only the values 0, 1, 2. Given that \(P(X = 0) = P(X = 1) = p\) and \(E(X^2) = E(X) + 1\), then the value of p is:
- A \(\frac{1}{4}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{2}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(P(X=2) = 1 - P(X=0) - P(X=1) = 1 - p - p = 1 - 2p\) \(E(X) = 0(p) + 1(p) + 2(1-2p) = p + 2 - 4p = 2 - 3p\) \(E(X^2) = 0^2(p) + 1^2(p) + 2^2(1-2p) = p + 4(1-2p) = p + 4 - 8p = 4 - 7p\) \(E(X^2) = E(X) + 1 \Rightarrow 4 - 7p = (2 - 3p) + 1\) \(4 - 7p = 3 - 3p\) \(1 = 4p\)…
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