Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval? A. 8/(2e^3) B. 9/(2e^3) C. 17/(2e^3) D. 26/(2e^3)

GATE 2013 · Engineering Mathematics · Poisson Distribution · medium

Answer: P(p < 3) = 17 / (2e^3). Answer: C.

1. Poisson PMF at k = 0, 1, 2: P(p=0) = e^(-3) * 1 / 1 = e^(-3). P(p=1) = e^(-3) * 3 / 1 = 3*e^(-3). P(p=2) = e^(-3) * 9 / 2 = (9/2)*e^(-3).
2. Sum to get P(p < 3): P(p < 3) = e^(-3) * (1 + 3 + 9/2) = e^(-3) * (2/2 + 6/2 + 9/2) = e^(-3) * 17/2 = 17 / (2 * e^3).