We have
E
[stay of 2nd student] = 30, and, using the memorylessness property of the
exponential distribution,
E
[stay of 1st

stay of 1st
≥
5] = 5 +
E
[stay of 1st] = 35
.
Also
P
(1st student stays no more than 5 minutes) = 1
−
e
−
5
/
30
,
P
(1st student stays more than 5 minutes) =
e
−
5
/
30
.
By substitution we obtain
E
[Time] = (5 + 30)
·
(1
−
e
−
5
/
30
) + (35 + 30)
·
e
−
5
/
30
= 35 + 30
·
e
−
5
/
30
= 60
.
394
.
Solution to Problem 3.21.
(a) We have
f
Y
(
y
) = 1
/l
, for 0
≤
y
≤
l
. Furthermore,
given the value
y
of
Y
, the random variable
X
is uniform in the interval [0
, y
]. Therefore,
f
X

Y
(
x

y
) = 1
/y
, for 0
≤
x
≤
y
. We conclude that
f
X,Y
(
x, y
) =
f
Y
(
y
)
f
X

Y
(
x

y
) =
1
l
·
1
y
,
0
≤
x
≤
y
≤
l
,
0
,
otherwise.
(b) We have
f
X
(
x
) =
f
X,Y
(
x, y
)
dy
=
l
x
1
ly
dy
=
1
l
ln(
l/x
)
,
0
≤
x
≤
l.
(c) We have
E
[
X
] =
l
0
xf
X
(
x
)
dx
=
l
0
x
l
ln(
l/x
)
dx
=
l
4
.
(d) The fraction
Y/l
of the stick that is left after the first break, and the further fraction
X/Y
of the stick that is left after the second break are independent. Furthermore, the
random variables
Y
and
X/Y
are uniformly distributed over the sets [0
, l
] and [0
,
1],
respectively, so that
E
[
Y
] =
l/
2 and
E
[
X/Y
] = 1
/
2. Thus,
E
[
X
] =
E
[
Y
]
E
X
Y
=
l
2
·
1
2
=
l
4
.
Solution to Problem 3.22.
Define coordinates such that the stick extends from
position 0 (the left end) to position 1 (the right end). Denote the position of the first
break by
X
and the position of the second break by
Y
.
With method (ii), we have
X < Y
. With methods (i) and (iii), we assume that
X < Y
and we later account for
the case
Y < X
by using symmetry.
Under the assumption
X < Y
, the three pieces have lengths
X
,
Y
−
X
, and
1
−
Y
.
In order that they form a triangle, the sum of the lengths of any two pieces
must exceed the length of the third piece. Thus they form a triangle if
X <
(
Y
−
X
) + (1
−
Y
)
,
(
Y
−
X
)
< X
+ (1
−
Y
)
,
(1
−
Y
)
< X
+ (
Y
−
X
)
.
42