The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and corresponding probabilities 0.1, 0.2, 0.3, 0.25, 0.15. A randomly selected customer will have 1, 2, or 3 packages for wrapping with probabilities 0.55, 0.25, and 0.2, respectively. Let Y = the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer). (a) Determine P (X = 3, Y = 3), i. e., p (3,3). (Round your answer to four decimal places.) P (X = 3, Y = 3) = (b) Determine p (4,11). (Round your answer to four decimal places.) p (4,11) =

(a) 0.0416 (b) 0.0012

Step-by-step explanation:

(a) We can separate the probabilities of the two events in a multiplication of probabilities:

P (X=3, Y=3) = P (X=3) * P (Y=3)

*It is assumed there is no correlation between the number of clients waiting and the number of packages they want to wrap.

There is only one combination possible of packages between the clients, because they all have to have at least one package to wrap.

P (X=3, Y=3) = P (X=3) * P (Y=3)

P (X=3, Y=3) = P (X=3) * P (Y1=1, Y2=1, Y3=1) = 0.25 * (0.55*0.55*0.55) = 0.0416

(b) To have 11 packages distributed in 4 clients, one needs to have 2 packages to wrap and the others 3 packages. There is 4 permutations of this, all with the same probability of ocurrence.

P (X=4, Y=11) = P (X=4) * P (Y1=3; Y2=3; Y3=3; Y4=2) * 4

P (X=4, Y=11) = 0.15 * (0.2*0.2*0.2*0.25) * 4=0.0012