{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

I did not require continuity correction though it is

Info iconThis preview shows pages 3–5. Sign up to view the full content.

View Full Document Right Arrow Icon
(I did not require continuity correction, though it is applicable. Without conti- nuity correction, we would have gotten approximately 97.87%.) 4. [8 points] In a group of 50 people, how many pairs of people do you expect to share a birthday? (Note that a person can be in many such pairs.) Assume that there are 365 days in a year and that birthdays are independent and uniformly distributed. Solution: Let A 1 , . . . , A n be indicators, for each possible pair, that the pair shares a birthday. So n = ( 50 2 ) = 1225 and X = A 1 + · · · + A n is the number of pairs of people that share a birthday. Since E ( A i ) = P (the i th pair shares a birthday) = 1 365 , we have E ( X ) = 1225 365 3 . 356 . 5. Calls come into a call center at an average rate of 2 per minute. (a) [6 points] What is the probability that exactly three calls come in the first two minutes? Solution: We have a Poisson arrival process with rate λ = 2. So the number of calls in the first two minutes is Poisson distributed with rate 2 λ = 4. Therefore, the probability of exactly three calls is e - 4 4 3 3! 19 . 54% . (b) [6 points] What is the probability that there are at least two minutes between the 3rd and 5th calls? Solution: This is the probability that in the two minutes starting from the time of the 3rd call there is at most one more call. The number of calls in this Page 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Math 431: Final Exam two minute period is again Poisson with rate 4, so the probability is P (no calls) + P
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}