Question

On your way to work you usually stop by your favorite coffee shop. You can either walk-in or drive-through. The service time is exponentially distributed, with an average time of 5 minutes if you order inside and 7 minutes if you drive-through. Upon arrival at the shop, there is a 40% chance that the parking lot is full, so you would need to order at the drive-through.

(a)	What is the overall average service time (in minutes) at the coffee shop?

(b)	Today you have only 6 minutes to spare on your way to work. What is the probability that you arrive on time for work if you stop by the coffee shop?

(c)	Suppose that today the parking lot is full. Given that you have been in line at the drive-through for 4 minutes, what is the probability that you will be served in the next 2 minutes?

Answer 1

Let X1 be service time for walk-in. X1 ~ exp(1/5)

Let X2 be service time for drive-through. X2 ~ exp(1/7)

P(parking lot is full or drive-through) = 0.40

P(parking lot is not full or walk in) = 0.60

a)Overall average service time = P(walk-in)*average time for walk-in + P(drive-through)*average time for drive-through

= 0.60*5 + 0.40*7 = 5.8 minutes

b) P(arrive on time) = P(service time < 6)

= P(walk-in)*P(X1<6) + P(drive-through)*P(X2<6)

= 0.60*(1-exp(-6/5) ) + 0.40*( 1-exp(-6/7) ) = 0.42+0.23 = 0.65

c) P(X2 < 6 / X2 > 4 )

= P(4<X2<6) / P(X2 > 4)

=>

= = 0.248