Question

[-14 Points] DETAILS DEVORESTAT9 5.E.005.MI. 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 0.5, 0.35, and 0.15, 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) =

Answer 1

it can bee happen in the following combinations ANEWER a) P(x=3, y=3) is possible only if all the three Customers in the line have got each callage to be wrapped P(X= 3, y = 3) = P(X=3) x6:9)3 = 0.95x65) 3 = 0.03125 03125 reauired probability b) Now Similarly P (X=4, 9=11) is computed •3,3,3,2 6073,3,2,3 (08) 3, 2, 3,3 Coa) 9,3 13,3 P(x=41Y=11) = 4X2 CX= 4) *0.16°40-35 = 4x015 x 15) 3x0.35 -0.00070875 CS Scanned with CamScanner