**5.21 Assume that the number of new visitors to a website in one hour is distributed as a Poisson variable. The mean number of new visitors to the website is 4.0 per hour. What is the probability that in any given hour**

a. zero new visitors will arrive at the website?

b. exactly one new visitor will arrive at the website?

c. two or more new visitors will arrive at the website?

d. fewer than three new visitors will arrive at the website?

**Answer:**

Here the mean number of visitors λ=4

Probability in Poisson distribution is given by

P(x; λ) = (e^-λ) (λ^x) / x!

where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.

**Part (a)**

when zero new visitors will arrive at the website i.e x=0

P(x=0,4)=(e^-4)(4^0)/0! ;;

p(x=0,4)=0.01831563888.

**Part(b)**

exactly one new visitor will arrive at the website i.e x=1

P(x=1,4)=(e^-4)(4^1)/1!

P(x=1,4)=0.07326255

**Part©**

two or more new visitors will arrive at the website x≥2

P(x≥2,4)=1-P(x=1,4)-P(x=0,4)

P(x≥2,4)=1-0.01831563888-0.07326255

P(x≥2,4)=0.90842181112

**Part(d)**

fewer than three new visitors will arrive at the website i.e.x≤3

P(x<3,4)=P(x=0,4)+P(x=1,4)+P(x=2,4)

P(x<3,4)=0.01831563888+0.07326255+0.14652511111

P(x<3,4)=0.23810329999