Let X have a Poisson distribution with parameter λ = 1. What is the…
2024
Let X have a Poisson distribution with parameter λ = 1. What is the probability that X ≥ 2 given that X ≤ 4?
- A.
0.2615
- B.
0.0812
- C.
0.0756
- D.
0.0671
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Correct answer: A
For X ~ Poisson(1), P(X = k) = e^(-1) / k!. We need P(X >= 2 | X <= 4) = P(2 <= X <= 4) / P(X <= 4). The numerator is e^(-1)(1/2! + 1/3! + 1/4!) = e^(-1)(17/24). The denominator is e^(-1)(1 + 1 + 1/2! + 1/3! + 1/4!) = e^(-1)(65/24). Therefore the probability is (17/24)/(65/24) = 17/65 ≈ 0.2615.