Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Edition 2
Roy D. Yates and David J. Goodman
Problem Solutions
: Yates and Goodman,
1.5.5 1.6.4 1.7.5 1.8.1 1.8.3 1.8.4 1.8.6 1.9.2
1.9.4 1.10.1 and 1.10.2
Problem 1.5.5 Solution
The sample outcomes can be written
ijk
where the first card drawn is
i
, the second is
j
and the third is
k
. The sample space is
S
=
{
234
,
243
,
324
,
342
,
423
,
432
}
.
(1)
and each of the six outcomes has probability 1
/
6. The events
E
1
, E
2
, E
3
, O
1
, O
2
, O
3
are
E
1
=
{
234
,
243
,
423
,
432
}
,
O
1
=
{
324
,
342
}
,
(2)
E
2
=
{
243
,
324
,
342
,
423
}
,
O
2
=
{
234
,
432
}
,
(3)
E
3
=
{
234
,
324
,
342
,
432
}
,
O
3
=
{
243
,
423
}
.
(4)
(a) The conditional probability the second card is even given that the first card is even is
P
[
E
2

E
1
] =
P
[
E
2
E
1
]
P
[
E
1
]
=
P
[243
,
423]
P
[234
,
243
,
423
,
432]
=
2
/
6
4
/
6
= 1
/
2
.
(5)
(b) The conditional probability the first card is even given that the second card is even is
P
[
E
1

E
2
] =
P
[
E
1
E
2
]
P
[
E
2
]
=
P
[243
,
423]
P
[243
,
324
,
342
,
423]
=
2
/
6
4
/
6
= 1
/
2
.
(6)
(c) The probability the first two cards are even given the third card is even is
P
[
E
1
E
2

E
3
] =
P
[
E
1
E
2
E
3
]
P
[
E
3
]
= 0
.
(7)
(d) The conditional probabilities the second card is even given that the first card is odd
is
P
[
E
2

O
1
] =
P
[
O
1
E
2
]
P
[
O
1
]
=
P
[
O
1
]
P
[
O
1
]
= 1
.
(8)
(e) The conditional probability the second card is odd given that the first card is odd is
P
[
O
2

O
1
] =
P
[
O
1
O
2
]
P
[
O
1
]
= 0
.
(9)
1