Section 4.6 Probability
303
Because events are sets, they can be combined using set operations. Suppose
that E 1 and E 2 are two events from the same sample space S. If we are interested
in the outcomes in either E 1 or E 2 or both, this will be the event E 1 c E 2 . If we are
interested in the outcomes in both E 1 and E 2 , this will be the event E 1 d E 2 . And if
we are interested in all the outcomes that are not in E 1 , this will be E 1 ′.
example 67
Employees from testing, development, and marketing participate in a drawing in
which one employee name is chosen. There are 5 employees in testing (2 men and
3 women), 23 in development (16 men and 7 women), and 14 in marketing (6 men
and 8 women).
The sample space has 42 names, that is, 0 S 0 = 42. Let W be the event that a
name drawn belongs to a woman. Then 0 W 0 = 3 + 7 + 8 = 18. Therefore, the
probability P(W ) that the name drawn belongs to a woman is 0 W 0 / 0 S 0 = 18/42 =
3/7. Let M be the event that a name drawn belongs to someone from marketing. Then
0 M 0 = 14. Thus, the probability P(M ) that the name drawn belongs to someone from
marketing is 0 M 0 / 0 S 0 = 14/42 = 1/3. The event that the name drawn belongs to a
woman in marketing is W d M. Because there are 8 women in marketing, 0 W d M 0
= 8, and the probability P(W d M ) that the name drawn belongs to a woman from
marketing is 8/42 = 4/21. Finally, the event that a name drawn belongs to either a
woman or to someone from marketing is W c M, and 0 W c M 0 = 3 + 7 + 14 = 24.
Hence P(W c M ) = 24/42 = 4/7.
Probability involves finding the size of sets, either of the sample space or of
the event of interest. Therefore many of our previous counting techniques come
into play. We may need to use the addition or multiplication principles, the principle of inclusion and exclusion, or the formula for the number of combinations of
r things from n objects. (In Example 67, we could have found the size of the union
of the women and the marketing people by using the principle of inclusion and
exclusion: 0 W c M 0 = 0 W 0 + 0 M 0 − 0 W d M 0 = 18 + 14 − 8 = 24.)
PraCtiCe 43 In Example 67, what is the probability of drawing the name of a male from development?
Of drawing a name from testing or development?
■
example 68
At a party, each card in a standard deck is torn in half and both halves are placed in
a box. Two guests each draw a half-card from the box. What is the probability that
they draw two halves of the same card?
There are 52 # 2 = 104 half-cards in the box. The size of the sample space is
the number of ways to pick two objects from 104, that is, 0 S 0 = C(104,2). Let H be
the event that the halves match. There are 52 ways that the halves can match, so
0 H 0 = 52. The probability is therefore
P(H) =
0 H 0
0 S 0
=
52
C(104,2)
=
52
104!
2!102!
=
52
104 # 103
2
=
52
52 # 103
=
1
103
> 0.0097
303
Because events are sets, they can be combined using set operations. Suppose
that E 1 and E 2 are two events from the same sample space S. If we are interested
in the outcomes in either E 1 or E 2 or both, this will be the event E 1 c E 2 . If we are
interested in the outcomes in both E 1 and E 2 , this will be the event E 1 d E 2 . And if
we are interested in all the outcomes that are not in E 1 , this will be E 1 ′.
example 67
Employees from testing, development, and marketing participate in a drawing in
which one employee name is chosen. There are 5 employees in testing (2 men and
3 women), 23 in development (16 men and 7 women), and 14 in marketing (6 men
and 8 women).
The sample space has 42 names, that is, 0 S 0 = 42. Let W be the event that a
name drawn belongs to a woman. Then 0 W 0 = 3 + 7 + 8 = 18. Therefore, the
probability P(W ) that the name drawn belongs to a woman is 0 W 0 / 0 S 0 = 18/42 =
3/7. Let M be the event that a name drawn belongs to someone from marketing. Then
0 M 0 = 14. Thus, the probability P(M ) that the name drawn belongs to someone from
marketing is 0 M 0 / 0 S 0 = 14/42 = 1/3. The event that the name drawn belongs to a
woman in marketing is W d M. Because there are 8 women in marketing, 0 W d M 0
= 8, and the probability P(W d M ) that the name drawn belongs to a woman from
marketing is 8/42 = 4/21. Finally, the event that a name drawn belongs to either a
woman or to someone from marketing is W c M, and 0 W c M 0 = 3 + 7 + 14 = 24.
Hence P(W c M ) = 24/42 = 4/7.
Probability involves finding the size of sets, either of the sample space or of
the event of interest. Therefore many of our previous counting techniques come
into play. We may need to use the addition or multiplication principles, the principle of inclusion and exclusion, or the formula for the number of combinations of
r things from n objects. (In Example 67, we could have found the size of the union
of the women and the marketing people by using the principle of inclusion and
exclusion: 0 W c M 0 = 0 W 0 + 0 M 0 − 0 W d M 0 = 18 + 14 − 8 = 24.)
PraCtiCe 43 In Example 67, what is the probability of drawing the name of a male from development?
Of drawing a name from testing or development?
■
example 68
At a party, each card in a standard deck is torn in half and both halves are placed in
a box. Two guests each draw a half-card from the box. What is the probability that
they draw two halves of the same card?
There are 52 # 2 = 104 half-cards in the box. The size of the sample space is
the number of ways to pick two objects from 104, that is, 0 S 0 = C(104,2). Let H be
the event that the halves match. There are 52 ways that the halves can match, so
0 H 0 = 52. The probability is therefore
P(H) =
0 H 0
0 S 0
=
52
C(104,2)
=
52
104!
2!102!
=
52
104 # 103
2
=
52
52 # 103
=
1
103
> 0.0097