# Poisson & Geometric

#### Wee G

Hello, I have a little problem finishing this question, I'll tell you how I started and if you could help me complete it will be most appreciated....

In a supermarket there is one queue for payment. The number of items that a customer buy is a random variable with Poisson distribution with a mean of 5.5 items. The customers come one after another to pay.
What is the probability that the 5th customer will be the first to collect 2 items at most, if we know that the 15th customer is the 3rd customer to collect 2 items at most. Assume that the customers are independent.

I have started by calculating the probability of collecting 2 items at most:
X~Poisson(5.5)
P(X<=2)=P(X=0)+P(X=1)+P(X=2)=0.00408+0.0224+0.06171=0.08819

I know that probably some Geometric random variable is involved now, but the condition confuse me, it doesn't sound like the lack of memory issue, or maybe I am wrong. How to solve it ?

Thanks !!

#### awkward

MHF Hall of Honor
Hint: The probability that X of any n customers all have 2 for fewer items has a Binomial distribution with p = whatever (I think you need to double check your arithmetic on that, I get a slightly different answer).

Think about what has to happen for customer 15 to be the third to have 2 or fewer items. Then what has to happen for customer 3 to be the first and customer 15 to be the third?

Finally, apply the definition of conditional probability.

#### Wee G

"Think about what has to happen for customer 15 to be the third to have 2 or fewer items. Then what has to happen for customer 3 to be the first and customer 15 to be the third?"

I am afraid I lost you here....

#### awkward

MHF Hall of Honor
OK, what I was getting at is that if customer 15 is the third to have two or fewer items, then
(1) Customer 15 must have two or fewer items, and
(2) Of customers 1-14, exactly 2 must have two or fewer items.

If in addition customer 3 is the first to have two or fewer items, then...