MATH 471 HW 9 Solutions
Pengsheng Ji
November 7, 2006
1
(a) Let N be the number of customers arriving in the store during a typical
hour. It has a Poisson distribution with parameter
λ
. Let K be the number of
female customers arriving in a given hour. We know that, conditioning on
N
=
n
,
the distribution of K is
Binomial
(
n,p
).
By the law of total probability, the probability of
K
=
k
is given by
P
(
K
=
k
) =
∞
X
n
=
k
P
(
K
=
k,N
=
n
) =
∞
X
n
=
k
P
(
K
=
k

N
=
n
)
P
(
N
=
n
)
=
∞
X
n
=
k
n
!
k
!(
n

k
)!
p
k
(1

p
)
n

k
·
e

λ
λ
n
n
!
=
e

λ
p
k
λ
k
k
!
∞
X
n
=
k
[
λ
(1

p
)]
n

k
(
n

k
)!
.
Actually the summation is the Taylor series of
e
λ
(1

p
)
. Therefore,
P
(
K
=
k
) =
e

λ
p
k
λ
k
k
!
·
e
λ
(1

p
)
=
e

λp
(
λp
)
k
k
!
,
which means K has a Poisson distribution.
(b) Let A=
¨
No females arriving
¨
, B=
¨
No males arriving
¨
, q=1p. Then,
P
(
A
) =
P
(
K
= 0) =
e

λp
. Similarly to Part(a), the number of males in a typical hour has
a Poisson distribution with parameter
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 '08
 PROTSAK
 Poisson Distribution, Probability theory, Exponential distribution, Poisson process, Pengsheng Ji

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