2. A disk server receives requests from many client machines and requires 10 mil

Question

2. A disk server receives requests from many client machines and requires 10 milliseconds to respond to each request. Let N = the number of additional requests that arrive before the service interval is complete. The probability of exactly k additional requests in the 10-millisecond service interval is PlN = k) = e-0.9(0.9)k/k!, for k = 0.1.2. . . If 2 or more new calls arrive while the service interval is only partially complete, what is the probability that exactly 3 new calls will arrive before the server is ready to respond? Hint: Use rules for conditional probability. Look at Example 1 from the Lecture Notes to see how to calculate the denominator.

Explanation / Answer

here P(exaclty 3 calls given 2 or more calls) =P(N=3|N>=2) =P(N=3)/P(N>=2)

=P(N=3)/(1-P(N=0)-P(N=1)) =e-0.9*(0.9)3/3!/(1-e-0.9*(0.9)0/0!-e-0.9*(0.9)1/1!)

=0.0494/(1-0.4066-0.3569)=0.2171

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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