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 !!