Question

pls explain as much as possible, thanks!!

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 have 1. 2. or 3 packages for wrapping with probabilities 0.55,0.25, and 0.2, respectively. Let the total number of packages to be wrapped for the customers waiting in line (ume that the number of packages submitted by one customer is independent of the number submitted by any other customer). (a) Determine PX, Y = 3). Lep(3.3). (Round your answer to four decimal places.) PX - 3, Y-3) - (b) Determine (4. 11). Round your answer to four decimal places)

Answer 1

a) We are given here that:
P(X = 0) = 0.1,
P(X = 1) = 0.2,
P(X = 2) = 0.3,
P(X = 3) = 0.25,
P(X = 4) = 0.15

For each customer, the number of packages here is defined with the following PDF:
P(K = 1) = 0.55,
P(K = 2) = 0.25,
P(K = 3) = 0.2

a) The probability here is computed as:
P(X = 3, Y = 3)

= P(X = 3)*P(K = 1)*P(K = 1)*P(K = 1)

= 0.25*0.55³ = 0.04159375

Therefore 0.0416 is the required probability here.

b) The probability here is computed as:

P(X = 4) P(Y = 11)

For X = 4 ,Y = 11 can be obtained as:

• 3 + 3 + 3 + 2
  Therefore, probability here is computed as:
  P(Y = 11 | X = 4) = (4c1)*[P(K = 3)]³*P(K = 2) = 4*0.2³*0.25 = 0.008

Therefore, p(4, 11) = P(X = 4)P(Y = 11 | X = 4) = 0.15*0.008 = 0.0012

Therefore 0.0012 is the required probability here.