Suppose X has Poisson distribution such that \(3P(X = 1) = 2P(X = 2)\) then \(P(X \gt 0)\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1 - e^{-3}\)
For a Poisson distribution with mean \(\lambda\),
\( \displaystyle P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!} \)
The given condition becomes
\( \displaystyle 3e^{-\lambda}\lambda = 2 \cdot \frac{e^{-\lambda}\lambda^2}{2} \)
\(3\lambda = \lambda^2\)
\(\lambda = 3 \quad (\lambda \gt 0)\)
Using the complement,
\(P(X \gt 0) = 1 - P(X = 0) = 1 - e^{-3}\)