IOE 316 Fall 2007 – Homework 5 Solution
Due December 4, 2007
1. (25 points) Cars cross a certain point in the highway in accordance with a Poisson process
with rate
λ
= 4 per minute.
(a) (5 points) If Kate blindly runs across the highway, then what is the probability that she
will be uninjured if the amount of time that it takes her to cross the road is
s
seconds?
(Assume that if she is on the highway when a car passes by, then she will be injured.)
(b) (5 points) Compute part (a) for
s
= 2
,
5
,
10
,
20.
(c) (10 points) Now assume that Kate is agile enough to escape from a single car, but if she
encounters two or more cars while attempting to cross the road, then she will be injured.
What is the probability that she will be unhurt if it takes her
s
seconds to cross?
(d) (5 points) Compute part (c) for
s
= 5
,
10
,
20
,
30.
SOLUTION:
Let N(t) = the number of cars that cross the road by time
t
, where
t
is minutes.
(a) (5 points)
P
(Kate is uninjured) =
P
(
N
(
s
) = 0) =
e

λ
·
s
60
=
e

s
15
(b) (5 points)
s = 2: P(Kate is uninjured) =
e

4
·
s
60
=
e

2
15
= 0
.
87517
.
s = 5: P(Kate is uninjured) =
e

4
·
s
60
=
e

1
3
= 0
.
71653
.
s = 10: P(Kate is uninjured) =
e

4
·
s
60
=
e

2
3
= 0
.
513417
.
s = 20: P(Kate is uninjured) =
e

4
·
s
60
=
e

4
3
= 0
.
263597
.
(c) (10 points)
P
(Kate is uninjured)
=
P
(
N
(
t
) = 0 or
N
(
t
) = 1)
=
e

λt
+
λte

λt
=
e

λ
·
s
60
+
λ
s
60
e

λ
·
s
60
=
e

4
·
s
60
+ 4
s
60
e

4
·
s
60
(d) (5 points)
P(Kate is uninjured) =
e

4
·
s
60
+ 4
s
60
e

4
·
s
60
s = 5: P(Kate is uninjured) =
4
3
e

1
3
= 0
.
955375
s = 10: P(Kate is uninjured) =
5
3
e

2
3
= 0
.
855695
s = 20: P(Kate is uninjured) =
7
3
e

4
3
= 0
.
61506
s = 30: P(Kate is uninjured) = 3
·
e

2
= 0
.
406
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2. (10 points) A certain scientific theory supposes that mistakes in cell division occur according
to a Poisson process with rate 2.5 per year, and that an individual dies when 196 such mistakes
have occurred. Assuming this theory, find
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 Spring '08
 Dolinskaya
 Exponential distribution, Poisson process, Kate

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