Clients arrive at a law firm according to a Poisson process with a rate of = 3 c

Question

Clients arrive at a law firm according to a Poisson process with a rate of = 3 clients per hour.

(a) What is the probability that no clients arrive to the firm from 10:00 to 12:30?

(b) Assume each client arriving needs interview which takes an exponentially distributed amount of time with a mean of 30 minutes. Let client A arrive at time 10:30. Given that client A is still present in the firm by 11:00, what is the probability that client A will still be in the firm by 12:00?

(c) It is known that the firm was visited by 10 clients in the afternoon, i.e. between 13:30 and 17:30. What is the probability that there were no clients during the first two hours?

Explanation / Answer

The poisson formula is as follows:-

e^-*^x/x!

a)

=3 per hour. Hence from 10 to 12:30, it would be =7.5. With x=0 customers, we get the probability of 0.000553

b)

=30. From 10:30 to 12 would be 90 minutes. Hence x=90. Probability is 5.49721442782145E-19.

c)

=3 per hour. Hence from 13:30 to 15:30, it would be =6. With x=0 customers, we get the probability of 0.00248. From 15:30 to 17:30, it would be =6 but with x=10, we get the probability of 0.0413. Hence, the final probability is 0.00248*0.0413=0.06

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.