IEOR 4106 Midterm Exam. Open text book and class notes; 1.5 hours. 100 Points total
1. (35 points) Voice messages are made from a cell phone according to a Poisson process at
rate 8 per hour, and independent of this, text messages are sent from the phone according
to a Poisson process at rate 2 per hour. (Time is in hours.)
(a) (10 points) Given that 4 text messages were sent during the hours of 10:00AM to
11:30AM, what is the probability that exactly two of these four were made between
11:00AM and 11:30AM?
SOLUTION:
We can treat the (unordered) four arrival times
U
1
, . . . U
4
as iid uni
formly distributed over the 90 minute interval, which for simplicity we denote by
(0
,
90).
Each one, independently, would arrive between 11:00AM and 11:30AM
with “success” probability
p
=
P
(
U
∈
(60
,
90)) = 30
/
90 = 1
/
3.
So, letting
N
denote the number of these 4 that do so yields a
binomial
(
p,
4) distribution;
P
(
N
= 2) =
(
4
2
)
p
2
(1

p
)
2
= 6(1
/
3)
2
(2
/
3)
2
= 8
/
27.
(b) (10 points, 5 each)
i. What is the probability that the first two messages are both voice, and the third
is text?
SOLUTION:
lambda
1
= 8
, λ
2
= 2.
Letting
p
=
λ
1
/
(
λ
1
+
λ
2
) = 0
.
80, we get
p
2
(1

p
) =
(0
.
8)
2
(0
.
20) = 0
.
128.
ii. Starting from now (time 0) what is the expected length of time until at least
one voice and one text message will been made?
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 Spring '08
 Whitt
 Probability theory, Exponential distribution, Poisson process, Text messages

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