Question

telephone calls arrive at an exchange have a poisson distribution at an average rate of one every second. Find the probabilities of the following;

a. no calls arriving in a given five-second period

b. between four and six calls arriving in the five-second period

c. there is atleast one call

d. there is at most one call

Answer 1

Here, lambda = 1 for every second

a)

Here, λ = 5 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) = 5^0 * e^-5/0!
P(X = 0) = 0.0067
Ans: 0.0067

b)

Here, λ = 5, x1 = 4 and x2 = 6.
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(4 <= X <= 6)
P(4 <= X <= 6) = (5^4 * e^-5/4!) + (5^5 * e^-5/5!) + (5^6 * e^-5/6!)
P(4 <= X <= 6) = 0.1755 + 0.1755 + 0.1462
P(4 <= X <= 6) = 0.4972

c)
Here, λ = 1 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) = 1 - P(x =0)

P(X >= 1) = 1 - 1^0 * e^-1/0!
P(X > = 1) = 1- 0.3679

P(x> =1) = 0.6321

d)

Here, λ = 1 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) = (1^0 * e^-1/0!) + (1^1 * e^-1/1!)
P(X <= 1) = 0.3679 + 0.3679
P(X <= 1) = 0.7358