ISyE 2027
Probability with Applications
Fall 2011
R. D. Foley
Solution to Homework 9
December 12, 2011
1.
Solution: (a)The c.d.f of
Z
is
F
Z
(
t
) =
Pr
{
Z
6
t
}
=
1

Pr
{
Z > t
}
=
1

Pr
{
X > t
,
Y > t
}
=
1

Pr
{
X > t
}
Pr
{
Y > t
}
=
1

e

λt
e

μt
=
1

e
(
λ
+
μ
)
t
where
t
>
0.
(b)The p.d.f of
Z
is
(
λ
+
μ
)
e
(
λ
+
μ
)
t
.
(c)
Z
has exponential distribution with parameter
λ
+
μ
.
(d)The expected value of
Z
is
1
λ
+
μ
=
1
3
+
7
=
1
10
, so the expected value of
Z
is 6 minutes.
2.
Solution: Suppose
X
denotes the length of time until A finishes the race.
Y
denotes the length of time until B finishes the race.
Z
denotes the
length of time until C finishes the race. Then
X
,
Y
and
Z
are independent,
exponentially distributed random variables with parameters 2 per hour, 3
per hour and 4 per hour.
(a)The probability that A takes longer than 1 hour is Pr
{
X >
1
}
=
e

2
.
(b)Let
V
=
min
{
Y
,
Z
}
, so
V
has exponential distribution with parameter
3
+
4
=
7 per hour. Thus, the probability that A wins is
Pr
{
X < V
}
=
Z
∞
0
Pr
{
X < V

V
=
t
}
7
e

7
t
dt
=
Z
∞
0
Pr
{
X < t
}
7
e

7
t
dt
=
Z
∞
0
(
1

e

2
t
)
7
e

7
t
dt
=
2
9
(c)Let
W
denote the length of time until the winning turtle finishes the
race. Then
W
=
min
{
X
,
Y
,
Z
}
, so the c.d.f of
W
is 1

e

9
t
. Thus, the
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 Fall '08
 Zahrn
 Normal Distribution, Poisson Distribution, Probability theory, R. D. Foley

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