CUET · MATHS · PYQ PAPER 2025
Suppose \(X\) has Poisson distribution such that \(3 P(X=1)=2 P(X=2)\) then \(P(X>0)\) is :
- A \(1-e^3\)
- B \(\frac{2}{3} e^3-1\)
- C \(1-e^{-3}\)
- D \(\frac{2}{3} e^{-3}-1\)
Answer & Solution
Correct Answer
(A) \(1-e^3\)
Step-by-step Solution
Detailed explanation
\(3 P(X=1)=2 P(X=2)\) \(3 \frac{e^{-\lambda} \lambda^1}{1!} = 2 \frac{e^{-\lambda} \lambda^2}{2!}\) \(3 \lambda = \lambda^2 \implies \lambda = 3\) \(P(X>0) = 1 - P(X=0)\) \(P(X>0) = 1 - \frac{e^{-3} 3^0}{0!} = 1 - e^{-3}\)
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