Customers arrive at a restaurant according to a Poisson distribution at a rate of 20 customers per hour. The restaurant opens for business at 11:00 am. (Assume that customers will stay long enough). Find the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am.

0.087

Step-by-step explanation:

Given that there were 17 customers at 11:07, probability of having 20 customers in the restaurant at 11:12 am could be computed as:

= Probability of having 3 customers in that 5 minute period. For every minute period, the number of customers coming can be modeled as:

X₅ ~ Poisson (20 (5/60))

X₅ ~ Poisson (1.6667)

Formula for computing probabilities for Poisson is as follows:

P (X=ₓ) = ((e^ (-λ)) λˣ) / ₓ!

P (X₅ = 3) = ((e^ (-λ)) λˣ) / ₓ! = (e^-1.6667) ((1.6667²) / 3!)

P (X₅ = 3) = (2.718^ (-1.6667)) ((2.78) / 6)

P (X₅ = 3) = (2.718^ (-1.6667)) 0.46

P (X₅ = 3) = 0.1889*0.46

P (X₅ = 3) = 0.086894

P (X₅ = 3) = 0.087

Therefore, the probability of having 20 customers in the restaurant at 11:12 am given that there were 17 customers at 11:07 am is 0.087.