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

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