Question

10. Canada’s Department of Transportation maintains statistics for involuntary denial of boarding. In February 2018, Canadian Airlines rate of involuntarily denying boarding was 0.74 per 10,000 passengers. What is the probability that in the next 10,000 passengers, there will be

a. No one involuntarily denied boarding?

b. At least one person involuntarily denied boarding?

c. At least two persons involuntarily denied boarding?

Answer 1

a)

Here, λ = 0.74 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) = 0.74^0 * e^-0.74/0!
P(X = 0) = 0.4771

b)

Here, λ = 0.74 and x = 1
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X >=1) = 1 - P(X <= 0).
P(X > =1) = 1 - (0.74^0 * e^-0.74/0!)
P(X > =1 ) = 1 - (0.4771)
P(X >=1 ) = 1 - 0.4771 = 0.5229

2)

Here, λ = 0.74 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 <= 1).
P(X >=2) = 1 - (0.74^0 * e^-0.74/0!) + (0.74^1 * e^-0.74/1!)
P(X >=2) = 1 - (0.4771 + 0.3531)
P(X >=2) = 1 - 0.8302 = 0.1698