Question

Suppose that in a week the number of accidents at a certain crossing has a Poisson distribution with an average of 0.6

a) What is the probability that there are at least 3 accidents at the crossing for two weeks?

b) What is the probability that the time between an accident and the next one is longer than 2 weeks?

Answer 1

a)
Here, λ = 0.6*2 = 1.2 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 2) = 1 - P(X <= 2).
P(X > 2) = 1 - (1.2^0 * e^-1.2/0!) + (1.2^1 * e^-1.2/1!) + (1.2^2 * e^-1.2/2!)
P(X > 2) = 1 - (0.3012 + 0.3614 + 0.2169)
P(X > 2) = 1 - 0.8795 = 0.1205

b)
We need to calculate P(X = 0)
P(X = 0) = 1.2^0 * e^-1.2/0!
P(X = 0) = 0.3012
Ans: 0.3012