Assume that the number of network errors experienced in a day on a local area network (LAN) is distributed as a Poisson random variable.

Question

The average number of network errors experienced in a day is 2.4. 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

For a Poisson random variable distribution

P(k exact events in interval) = λ^k e ^(-λ)/k!, here e Euler constant e = 2.71828... and λ average no. of events.

To find answers a, b, c, and let us first calculate probability for exact events k = 0, 1, 2, 3, and 4. The values are indicated in a table below. In this case λ = 2.4 and e ^ (-λ) = .09072

k P (x=k)

0 0.0907

1 0.2177

2 0.2613

3 0.2090

4 0.1254

From table answer a. Probability exactly 0 error occur 0.0907

b. Probability exactly 1 error will occur 0.2177

c. Probability 2 or more error will occur = probability less than 2 error won't occur

P (k≥ 2) = 1- P (k<2) = 1 - 0.3084 = 0.6916

d. Probability fewer than 3 error occur P (k<3) = 0.0907 + 0.2177 + 0.261= 0.5697

I hope this helps.

Shailesh Kadakia, Senior Math & Physics Teacher

www.Skylativity.com