Rensselaer
Electrical, Computer, and Systems Engineering Department
ECSE 4500 Probability for Engineering Applications
PS#3 Solutions
February 19, 2010
1. (1.35) Use Eq. (1.107) from the text with
t
=0
,
t
+
τ
=10
,and
λ
(
τ
)
as given. This gives
Z
10
0
[1
−
e
−
u/
10
]
du
e
−
1
=3
.
68
.
Thus,
P
[
k
clicks in
10
seconds
]=
e
−
3
.
68
1
k
!
(3
.
68)
k
.
2. (1.39)
(a) Events associated with disjoint time intervals, under Poisson law, are independent. The
number of cars arriving at a tollbooth in the time interval
(0
,T
)
at a rate of
λ
per minute
is such that the
P
[
k
cars arrive in
(0
)] =
e
−
λT
[
λT
]
k
k
!
. Let us de
f
ne the events:
A
,
{
n
1
cars arrive in
(0
,t
1
)
}
,
B
,
{
n
2
cars arrive in
(
t
1
)
}
C
,
{
n
1
+
n
2
cars arrive in
(0
)
}
.
We are asked to
f
nd the
P
[
A

C
]
. From the de
f
nition of conditional probability, we
know that this equals
P
[
AC
]
P
[
C
]
. The event
AC
is the event that
n
1
cars arrive in
(0
1
)
and
n
1
+
n
2
cars arrive in
(0
)
. This is the same as saying that
n
1
cars arrive in
(0
1
)
and
n
2
cars arrive in
(
t
1
)
, which is nothing but the event
AB
. Therefore,
AC
=
AB
.Bu
t
from the Poisson law (given), we know that
P
[
AB
P
[
A
]
P
[
B
]
, because
A
and
B
are
events on disjoint time intervals. Therefore,
P
[
A

C
P
[
AC
]
P
[
C
]
=
P
[
AB
]
P
[
C
]
=
P
[
A
]
P
[
B
]
P
[
C
]
=
(
λt
1
)
n
1
e
−
λt
1
n
1
!
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 Spring '08
 WOODS
 Normal Distribution, Variance, Probability theory

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