For an alternative approach, note that the event of interest occurs if and only if the
time
Y
2
of the second arrival is less than or equal to 2. Hence, the desired probability
is
P
(
Y
2
≤
2) =
Z
2
0
f
Y
2
(
y
)
dy
=
Z
2
0
(0
.
6)
2
ye
−
0
.
6
y
dy.
This integral can be evaluated by integrating by parts, but this is more tedious than
the ±rst approach.
(d) The expected number of ±sh caught is equal to the expected number of ±sh caught
during the ±rst two hours (which is 2
λ
=2
·
0
.
6=1
.
2), plus the expectation of the
number
N
of ±sh caught after the ±rst two hours. We have
N
= 0 if he stops ±shing
at two hours, and
N
= 1, if he continues beyond the two hours. The event
{
N
=1
}
occurs if and only if no ±sh are caught in the ±rst two hours, so that
E
[
N
]=
P
(
N
=
1) =
P
(0
,
2) = 0
.
301. Thus, the expected number of ±sh caught is 1
.
2+0
.
301 = 1
.
501
.
(e) Given that he has been ±shing for 4 hours, the future ±shing time is the time until
the ±rst ±sh is caught. By the memorylessness property of the Poisson process, the
future time is exponential, with mean 1
/λ
. Hence, the expected total ±shing time is
4+(1
/
0
.
6) = 5
.
667.
Solution to Problem 6.11.
We note that the process of departures of customers who
have bought a book is obtained by splitting the Poisson process of customer departures,
and is itself a Poisson process, with rate
pλ
.
(a) This is the time until the ±rst customer departure in the split Poisson process. It
is therefore exponentially distributed with parameter
pλ
.
(b) This is the probability of no customers in the split Poisson process during an hour,
and using the result of part (a), equals
e
−
pλ
.
(c) This is the expected number of customers in the split Poisson process during an
hour, and is equal to
pλ
.
Solution to Problem 6.12.
Let
X
be the number of di²erent types of pizza ordered.
Let
X
i
be the random variable de±ned by
X
i
=
n
1
,
if a type
i
pizza is ordered by at least one customer,
0
,
otherwise.
We have
X
=
X
1
+
···
+
X
n
,and
E
[
X
n
E
[
X
1
].
We can think of the customers arriving as a Poisson process, and with each
customer independently choosing whether to order a type 1 pizza (this happens with
probability 1
/n
) or not. This is the situation encountered in splitting of Poisson pro-
cesses, and the number of type 1 pizza orders, denoted
Y
1
, is a Poisson random variable
with parameter
λ/n
.W
ehav
e
E
[
X
1
P
(
Y
1
>
0) = 1
−
P
(
Y
1
=0)=1
−
e
−
λ/n
,
so that
E
[
X
n
E
[
X
1
n
(
1
−
e
−
λ/n
)
.
71