Suppose p is the number of cars per minute passing through a certain road…

Concept

A Poisson distribution with mean λ gives the probability of exactly k events as P(X = k) = e−λ · λk/k!.

“Fewer than 3” means X < 3, i.e. X ∈ {0, 1, 2}, so we add the probabilities of these three disjoint outcomes: P(X < 3) = P(0) + P(1) + P(2).

Application (λ = 3)

1. P(0) = e−3 · 30/0! = e−3 · 1 = e−3.

2. P(1) = e−3 · 31/1! = 3e−3.

3. P(2) = e−3 · 32/2! = (9/2)e−3 (note the 2! = 2 in the denominator).

4. Add them: P(X < 3) = e−3(1 + 3 + 9/2) = e−3 · (17/2) = 17/(2e3).

Cross-check

Numerically 17/(2e3) ≈ 0.423, a valid probability in [0, 1]. Each term is computed with the correct factorial (1, 1, 2 for k = 0, 1, 2), and all three terms below k = 3 are included, confirming the cumulative value.