Here, μ = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:

P(X > 0) = 1 − P(x0)

Now `P(X)=(e^(-mu)mu^x)/(x!)` so `P(x_0)=(e^-3 3^0)/(0!)=4.9787xx10^-2`

Therefore the probability of `1` or more policies is given by:

`"Probability"=P(X>=0)`

`=1-P(x_0)`

`=1-4.9787xx10^-10`

`=0.95021`

(b) The probability of selling 2 or more, but less than 5 policies is:

`P(2<=X<5)`

`=P(x_2)+P(x_3)+P(x_4)`

`=(e^-3 3^2)/(2!)+(e^-3 3^3)/(3!)+(e^-3 3^4)/(4!)`

`=0.61611`

(c) Average number of policies sold per day: `3/5=0.6`

So on a given day, `P(X)=(e^-0.6(0.6)^1)/(1!)=0.32929`