(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
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Math 431: Final Exam
two minute period is again Poisson with rate 4, so the probability is
P
(no calls) +
P
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 Spring '12
 Miller
 Probability, Probability theory, continuity correction

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