Question

A certain market has both an express checkout line and a superexpress checkout line. Let X₁ 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	1	2	3
x1	0	0.09	0.07	0.04	0.00
	1	0.05	0.15	0.04	0.04
	2	0.05	0.03	0.10	0.06
	3	0.01	0.03	0.04	0.07
	4	0.00	0.02	0.05	0.06

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

(b) What is P(X₁ = X₂), that is, the probability that the numbers of customers in the two lines are identical?
P(X₁ = X₂) =

(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.

A = {X₁ ≤ 2 + X₂ ∪ X₂ ≤ 2 + X₁}

A = {X₁ ≥ 2 + X₂ ∪ X₂ ≥ 2 + X₁}

    

A = {X₁ ≥ 2 + X₂ ∪ X₂ ≤ 2 + X₁}

A = {X₁ ≤ 2 + X₂ ∪ X₂ ≥ 2 + X₁}

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)	=

Answer 1

(a)

From the joint distribution table,

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 + 0.10 + 0.07

= 0.41

(c)

A = {X₁ ≥ 2 + X₂ ∪ 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 = 4, X₂ = 0) +

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

= 0.04 + 0.00 + 0.04 + 0.05 + 0.01 + 0.00 + 0.03 + 0.02 + 0.05

= 0.24

(d)

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

= 0.04 + 0.10 + 0.03 + 0.00

= 0.17

P(at least four) = 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 + 0.10 + 0.06 + 0.03 + 0.04 + 0.07 + 0.00 + 0.02 + 0.05 + 0.06

= 0.47