Question

5.1

A certain market has both an express checkout line and a superexpress checkout line. Let X, 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. 

$$ \begin{array}{cc|cccc}  & & \multicolumn{3}{|c} {x_{2}} \\ & & 0 & 1 & 2 & 3 \\ \hline & 0 & 0.08 & 0.07 & 0.04 & 0.00 \\ & 1 & 0.05 & 0.15 & 0.05 & 0.04 \\ x_{1} & 2 & 0.05 & 0.03 & 0.10 & 0.06 \\ & 3 & 0.01 & 0.02 & 0.04 & 0.07 \\ & 4 & 0.00 & 0.01 & 0.05 & 0.08 \end{array} $$

(a) What is X1 = 1 ' X2 = 1), that is, the probability that there is exactly one customer in each line? 

(b) What is XX2), that is, the probability that the numbers of customers in the two lines are identical? 

(c) Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of X1, and X2, Calculate the probability of this event. 

(d) What is the probability that the total number of customers in the two lines is exactiy four? 

Answer 1

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)

                      = 0.08 + 0.15 + 0.10 + 0.07

                      = 0.4

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

                    = 0.04 + 0.1 + 0.06 + 0.02 + 0.04 + 0.07 + 0 + 0.01 + 0.05 + 0.08

                    = 0.47