Question

The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 3.1 per week. Find the probability of the following events.

A. No accidents occur in one week.

Probability =

B. 5 or more accidents occur in a week.

Probability =

C. One accident occurs today.

Probability =

Answer 1

a)
Here, λ = 3.1 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 0)
P(X = 0) = 3.1^0 * e^-3.1/0!
P(X = 0) = 0.045
Ans: 0.045

b)
We need to calculate P(X > 4) = 1 - P(X <= 4).
P(X > 4) = 1 - (3.1^0 * e^-3.1/0!) + (3.1^1 * e^-3.1/1!) + (3.1^2 * e^-3.1/2!) + (3.1^3 * e^-3.1/3!) + (3.1^4 * e^-3.1/4!)
P(X > 4) = 1 - (0.045 + 0.1397 + 0.2165 + 0.2237 + 0.1733)
P(X > 4) = 1 - 0.7982
= 0.2018

c)
Here, λ = 3.1/7 = 0.4429 and x = 1
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 1)
P(X = 1) = 0.4429^1 * e^-0.4429/1!
P(X = 1) = 0.2844