Question

Assume that the monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What is the probability that there will be

(a) more than 3 such accidents in the next month? 0.1807

(b) less than 2 such accidents in the next 3 months?

(c) exactly 6 such accidents in the next 4 months?

Answer 1

a)

Here, λ = 2.2 and x = 3
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 3) = 1 - P(X <= 3).
P(X > 3) = 1 - (2.2^0 * e^-2.2/0!) + (2.2^1 * e^-2.2/1!) + (2.2^2 * e^-2.2/2!) + (2.2^3 * e^-2.2/3!)
P(X > 3) = 1 - (0.1108 + 0.2438 + 0.2681 + 0.1966)
P(X > 3) = 1 - 0.8193 = 0.1807

b)

Here, λ = 6.6 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X < 2).
P(X < 2) = (6.6^0 * e^-6.6/0!) + (6.6^1 * e^-6.6/1!)
P(X < 2) = 0.0014 + 0.009
P(X <2) = 0.0104

c)

Here, λ = 8.8 and x = 6
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 6)
P(X = 6) = 8.8^6 * e^-8.8/6!
P(X = 6) = 0.0972
Ans: 0.0972