2. 0/1 points | Previous Answers DevoreStat9 5.E.005 My Notes Ask Your Teacher T

Question

2. 0/1 points | Previous Answers DevoreStat9 5.E.005 My Notes Ask Your Teacher 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 the total number of packages to be wrapped for the customers waiting in line (assume that the numberof packages submitted by one customer is independent of the number submitted by any other customer) (a) Determine P(X-3,Y-3),i.e, (3,3). (Round your answer to four decimal places.) P(X- 3, Y- 3) 0.054 (b) Determine p(4,11). (Round your answer to four decimal places.) P(4,11) Need Help?Read It Talk to a Tutor

Explanation / Answer

Question 2:

a) P(X = 3, Y = 3) is possible only if all the three customers in the line have got 1 each package to be wrapped.

Therefore, we get:

P(X = 3, Y = 3) = P(X = 3) * 0.53 = 0.25*0.53 = 0.03125

Therefore 0.03125 is the required probability here.

b) Now similarly, here

P(X = 4, Y =11 ) is computed as:

It can happen in the following combinations:

Therefore the required probability here would be computed as:

P(X = 4, Y =11 ) = 4*P(X = 4) * 0.153*0.35 = 4*0.15*0.153*0.35 = 0.00070875

Therefore the required probability here is 0.00070875

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