Let Xdenote the number of customers in line, and let Y denote the total number of items purchased by all the customers in line. Assume the number of items purchased by one customer is independent of the number of items purchased by any other customer.

Question

The number of customers in line at a supermarket express checkout counter is a random variable whose probability mass function is given in the following table.







For each customer, the number of items to be purchased is a random variable with probability mass function







a. Find P(X=2 and Y=2).

b. Find P(X=2 and Y=6).

c. Find P(Y=2).

Answer 1

(a) ?(? = 2and Y = 2) = ?(? = 2) ?(Y= 2 | ? = 2). Now ?(? = 2) = 0.30. Given that ? = 2, Y= 2 if and only if each of the two customers purchases one item. Therefore P(Y = 2 | X = 2) = 0.052 = 0.0025, so P(X = 2 and Y = 2) = (0.30)(0.0025) = 0.00075.



(b) P(X = 2 and Y = 6) = P(X = 2)P(Y = 6 | X = 2). Now P(X = 2) = 0.30. Given that X = 2, Let be the number of items purchased by the first customer, and let be the number of items purchased by the second customer. Then the event Y = 6 is equivalent to the event { = 5 and = 1} or { = 4 and = 2} or {= 3 and = 3} or {= 2and = 4} or { = 1 and = 5}.



P(Y = 6|X = 2)=P( = 5 and = 1) + P( = 4 and = 2)+P( = 3 and = 3)

+ P ( = 2 and = 4)+P ( = 1 and = 5)

=(0.15)(0.05) + (0.30)(0.15) + (0.25)(0.25) + (0.15)(0.30) + (0.05)(0.15)

=0.1675



P(X = 2 and Y = 6) = (0.30)(0.1675) = 0.05025



c) P (Y = 2) = P(X = 1 and Y = 2)+P(X = 2 and Y = 2).

P(X = 1 and Y = 2) = P(X = 1)P(Y = 2|X = 1) = (0.25)(0.15) = 0.0375.

From part (a), P(X = 2 and Y = 2) = 0.00075.

P(Y = 2) = 0.0375 + 0.00075 = 0.03825.