Question

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).

Answer 1

P[X=3,Y=3] = 0.0416

Explanation:

Solution:

- X is the RV denoting the no. of customers in line.

- Y is the sum of Customers C.

- Where no. of Customers C's to be summed is equal to the X value.

- Since both events are independent we have:

P[X=3,Y=3] = P[X=3]*P[Y=3/X=3]

P[X=3].P[Y=3/X=3] = P[X=3]*P[C1+C2+C3=3/X=3]

P[X=3]*P[C1+C2+C3=3/X=3] = P[X=3]*P[C1=1,C2=1,C3=1]

P[X=3]*P[C1=1,C2=1,C3=1] = P[X=3]*(P[C=1]^3)

- Thus, we have:

P[X=3,Y=3] = P[X=3]*(P[C=1]^3) = 0.25*(0.55)^3

P[X=3,Y=3] = 0.0416