Question

Assume that the number of networks errors experienced in a day on a local area network (LAN) is distributed as a Poisson random variable. The mean number of network errors experienced in a day is 1.6. What is the probability that in any given day

a) zero network errors will occur

b) exactly one network error will occur

c) two or more network errors will occur

d) fewer than three network errors will occur

Answer 1

X ~ Poi ( )

Where = 1.6

Poisson probability distribution is

P(X) = e-X / X!

a)

P(X = 0) = e-1.6

= 0.2019

b)

P(X = 1) = e-1.6 * 1.6

= 0.3230

c)

P(X >= 2) = 1 - P(X <= 1)

= 1 - [ P(X = 0) + P(X = 1) ]

= 1 - [ e-1.6 + e-1.6 * 1.6 ]

= 0.4751

d)

P(X < 3) = P(X <= 2)

= e-1.6 + e-1.6 * 1.6 + e-1.6 * 1.62 / 2!

= 0.7834