Suppose an online retailer inspected all shipments last month?

Question

There were for 4 or fewer books, to see if the proper books were in the shipment.
X is the number of books in the customer’s order,
Y is the number of correct books that were shipped.
We will assume that we always ship the correct NUMBER of books; it's the identities of the books that might be wrong.
Here are the results of the inspection:
THIS IS A TABLE COLUMNS AND ROWS
=Y
4 0 0 0 0.12
3 0 0 0.15 0.04
2 0 0.25 0.03 0.02
1 0.28 0.03 0.01 0.01
0 0.02 0.02 0.01 0.01
1 2 3 4 =X

(a) What is the average number of books (correct or not) per shipment in this data set?
(b) Given that a shipment had 3 books, what is the probability that they were all correct?
(c) Are X and Y independent? Give a quantitative reason for your answer.

Answer 1

The table format didn't come out very well. For this kind of data, I suggest using the same precision for each entry (0.00 rather than 0 when the other entries are to 2 decimal places) and using a space between entries.

a) First get the marginal probabilities for X by adding up the probabilities for each value of X:
P(X=1) = 0.28+0.02 = 0.30
P(X=2) = 0.25+0.03+0.02 = 0.30
P(X=3) = 0.15+0.03+0.01+0.01 = 0.20
P(X=4) = 0.12+0.04+0.02+0.01+0.01 = 0.20
So E(X) = 1(0.30) + 2(0.30) + 3(0.20) + 4(0.20) = 2.3.

b) P(Y=3 | X=3) = P(X=3 and Y=3) / P(X = 3)
= 0.15 / 0.20
= 0.75.

c) No, since P(X=1) and P(Y=4) are both nonzero but P(X=1 and Y=4) = 0.