A certain market has both an express checkout line and a superexpress checkout l

Question

A certain market has both an express checkout line and a superexpress checkout line. LetX1 denote the number of customers in line at the express checkout at a particular time of day, and let x2 denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of X1 and X2 is as given in the accompanying table. x2 0 0.09 0.07 0.04 0.00 10.06 0.15 0.04 0.04 x 2 3 0.01 0.02 0.04 0.07 4 0.00. 0.05 0.06 xi 20.05 0.04 0.10 0.06 (a) What is PX1, X2 -1), that is, the probability that there is exactly one customer in each line? P(X11, X21) (b) What is P(XX2), that is, the probability that the numbers of customers in the two lines are identical? P(Xi-X2)- (c) Let A denote the event that there are at least two more customers in one line than in the otherline. Express A in terms of Xi and Xz Calculate the probability of this event. P(A) (d) What is the probability that the total number of customers in the two lines is exactly four? At least four? P(exactly four)- P(at least four)-

Explanation / Answer

a) P(X1 = 1, X2 = 1) = 0.15

b) P(X1 = X2) = P(X1 = 0, X2 = 0) + P(X1 = 1, X2 = 1) + P(X1 = 2, X2 = 2) + P(X1 = 3, X2 = 3)

                      = 0.09 + 0.15 + 0.1 + 0.07

                      = 0.41

c) Option-B) A = {X1> 2 + X2 U X2> 2 + X1}

P(A) = P(X1 = 0, X2 = 2) + P(X1 = 0, X2 = 3) + P(X1 = 1, X2 = 3) + P(X1 = 2, X2 = 0) + P(X1 = 3, X2 = 0) + P(X1 = 3, X2 = 1) + P(X1 = 4, X2 = 0) + P(X1 = 4, X2 = 1) + P(X1 = 4, X2 = 2)

        = 0.04 + 0 + 0.04 + 0.05 + 0.01 + 0.02 + 0 + 0.01 + 0.05

        = 0.22

d) P(exactly 4) = P(X1 = 1, X2 = 3) + P(X1 = 2, X2 = 2) + P(X1 = 3, X2 = 1) + P(X1 = 4, X2 = 0)

                        = 0.04 + 0.1 + 0.02 + 0

                        = 0.16

P(at least 4) = P(exactly 4) + P(X1 = 2, X2 = 3) + P(X1 = 3, X2 = 2) + P(X1 = 3, X2 = 3) + P(X1 = 4, X2 = 1) + P(X1 = 4, X2 = 2) + P(X1 = 4, X2 = 3)

                    = 0.16 + 0.06 + 0.04 + 0.07 + 0.01 + 0.05 + 0.06

                    = 0.45

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