Question

The number of customers that order a pizza online is a Poisson variable with rate 55 per hour.

1) What is the probability that the operator will have to wait less than 3 minutes for the first order after 10:00 AM?

2) Between 10:00 and 10:30 have been ordered 25 pizzas. What is the probability that 10 pizzas have been ordered between 10:00 and 10:10?

3) What is the probability that the operator will have to wait less than 5 minutes for the first order after 10:00 AM, given that in the first 2 minutes there have been no orders?

4) What is the probability that an order was placed between 10:00 AM and 10:02 AM, given that only 1 order arrived between 10:00 AM and 10:05 AM?

Answer 1

P(x customer in k hours) = e^(-55*k)*(55*k)^x / (x!)

1.

P(T<=t) = 1 - e^(-55*t)

t = 3 min = 3/60 hours

P(T<=3/60) = 1 - e^(-55*3/60)

= 0.9361

2.

P(10 pizzas in 10 min | 25 pizzas in 30 min)

= P(10 pizzas in 10 min)*P(15 pizzas in remaining 20 min) / P(25 pizzas in 30 min)

= (e^(-55*10/60)*(55*10/60)^10 / (10!)) * (e^(-55*20/60)*(55*20/60)^15 / (15!)) / (e^(-55*30/60)*(55*30/60)^25 / (25!))

= 0.1264

3.

in first two min no order so, now time has to be less than 3 min

P(T<=3/60) = 1 - e^(-55*3/60)

= 0.9361

4.

probability that an order was placed between 10:00 AM and 10:02 AM, given that only 1 order arrived between 10:00 AM and 10:05 AM

= P(one order in 2 min)*P(0 order in next 3 min) / P(1 order in 5 min)

= (e^(-55*2/60)*(55*2/60)^1 / (1!)) * (e^(-55*3/60)*(55*3/60)^0 / (0!)) / (e^(-55*5/60)*(55*5/60)^1 / (1!))

= 0.4000

(please UPVOTE)