Poisson Distribution w/ mean 3.6 - demand for car hire

Hi,

Question:

A car-hire firm finds that the daily demand for its cars follows a Poisson distribution with mean 3.6.

a) What is the probability that on a particular day the demand will be:

i) two or fewer,

Using a cumulative Poisson distribution table:

P(X is less than or equal to 2) = 0.3027

ii) between three and seven (inclusive),

P(X is between 3 and 7) = P(X is less than or equal to 7) - P(X is less than or equal to 2)
P(X is between 3 and 7) = 0.9692 - 0.3207
P(X is between 3 and 7) = 0.6665

iii) zero

P(X = 0) = 0.0273

b) What is the probability that 10 consecutive days will include two or more on which the demand is zero?

I really don't know how to do (b). Please could you give me some pointers as to how to start?

Many thanks.

Re: Poisson Distribution - demand for car hire

Part b) is a binomial calculation with Poisson probability. How many ways can you have 10 days with 2 of them containing a zero and 8 containing at least one?

Re: Poisson Distribution - demand for car hire

Hi,

This is what I've done:

X~B(15, 0.6973)

P(X=0) = (10C0)(0.0273)^0 (1-0.0273)^10 = 0.75820939...
P(X=1) = (10C1)(0.0273)^1 (1-0.0273)^9 = 0.21280062...

P(X=1) + P(X=2) = 0.9701

P(X is greater than or equal to 2) = 1 - 0.9701
P(X is greater than or equal to 2) = 0.290 (to 3 significant figures)

Is that correct?

Thanks again for helping.