Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability.

Chapter 4Probability and Counting Rules

Section 4-2

Sample Spaces and Probability

If two dice are rolled one time, find the probability of getting these results.

a. A sum of 6b. Doublesc. A sum of 7 or 11d. A sum greater than 9e. A sum less than or equal to 4

Section 4-2 Exercise #13

a. A sum of 6

There are 62 or 36 outcomes.

There are 5 ways to get a sum of 6.

They are (1,5), (2,4), (3,3), (4,2), and (5,1).

The probability then is

5

36 .

Total of 36 outcomes

b. Doubles

There are six ways to get doubles. They are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6).

The probability is then

6

36 =

1

6 .

c. A sum of 7 or 11

There are six ways to get a sum of 7. They are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).


8

36 =

2

9 .

Total of 36 outcomes

There are two ways to get a sum of 11. They are (5,6) and (6,5).

d. A sum of greater than 9

To get a sum greater than nine, one must roll a 10, 11, or 12.


6

36 =

1

6 .

There are six ways to get a 10, 11, or 12. They are (4,6), (5,5), (6,4), (6,5), (5,6), and (6,6).

e. The patient has had 1 or 2 tests done.

P (1 or 2 tests) =

8 + 2

30

= 10

30 = 1

3

Number of Tests

Performed

Number of

Patients0 121 82 23 3

4 or more 5


Section 4-3Exercise #23

a. A king or a queen or a jack.b. A club or a heart or a spade.c. A king or a queen or a diamond.d. An ace or a diamond or a heart.e. A 9 or a 10 or a spade or a club.

If one card is drawn from an ordinary deck of cards, find the probability of getting the following:

There are 4 kings, 4 queens, and 4 jacks,

hence:

P (king or queen or jack) =

12

52 =

3

13

If one card is drawn from an ordinary deck of cards, find the probability of getting the following:a. A king or a queen or a jack.

b. A club or a heart or a spade.

There are 13 clubs, 13 hearts, and 13

spades, hence:

P(club or heart or spade)

P =

13 + 13 + 13

52 =

39

52 =

3

4


c. A king or a queen or a diamond.There are 4 kings, 4 queens, and 13 diamonds but the king and queen of diamonds were counted twice, hence:

P =

4

52 +

4

52 +

13

52–

2

52

P(king or queen or diamond)

P(king) + P(queen) + P(diamond) – P(king or queen of diamonds)

=

=

19

52


d. An ace or a diamond or a heart.

There are 4 aces, 13 diamonds and 13 hearts. There is one ace of diamonds and one ace of hearts, hence:

– P (ace of hearts or ace of diamonds)

= P (ace) + P (diamond) + P (heart)P(ace or diamond or heart)

P =

4

52 +

13

52 +

13

52–

2

52 =

28

52 =

7

13


e. A 9 or a 10 or a spade or a club.

There are 4 nines, 4 tens, 13 spades, and 13 clubs. There is one nine of spades, one ten of spades, one nine of clubs, and one ten of clubs, hence:

P ( 9 or 10 or spade or club)

– P(9 of spades or 9 of clubs)


= P(9) + P(10) + P(spade) + P(club)


e. A 9 or a 10 or a spade or a club.

P ( 9 or 10 or spade or club)



= P(9) + P(10) + P(spade) + P(club)

P =

4

52 +

4

52 +

13

52 +

13

52–

2

52–

2

52

=

30

52 =

15

26



Section 4-4

The Multiplication Rules and Conditional Probability



At a local university 54.3% of incoming first-year students have computers. If three students are selected at random, find the following probabilities.a. None have computersb. At least one has a computerc. All have computers

P(no computer) = 1 – 0.543

P(none of the three) = (0.457)3

= 0.457

= 0.0954

At a local university 54.3% of incoming first-year students have computers. If three students are selected at random, find the following probabilities.a. None have computers

b. At least one has a computer

P (at least one) = 1 – P (none of the three)

= 1– 0.0954

= 0.9046

At a local university 54.3% of incoming first-year students have computers. If three students are selected at random, find the following probabilities.

c. All have computers

P(all three) = (0.543)3

= 0.1601

At a local university 54.3% of incoming first-year students have computers. If three students are selected at random, find the following probabilities.



A manufacturer makes two models of an item: model I, which accounts for 80% of unit sales, and model II, which accounts for 20% of unit sales. Because of defects, the manufacturer has to replace (or exchange) 10% of its model I and 18% of its model II. If a model is selected at random, find the probability that it will be defective. 0.1 D (0.8)(0.1) = 0.080.8

0.9 ND

0.18 D (0.2)(0.18) = 0.0360.2

0.82 ND

Finally, use the addition rule, since the item is chosen at random from model I or model II.

P(defective) = 0.08 + 0.036 = 0.116

0.1 D (0.8)(0.1) = 0.080.8

0.9 ND

0.18 D (0.2)(0.18) = 0.0360.2

0.82 ND



In Rolling Acres Housing Plan, 42% of the houses have a deck and a garage; 60% have a deck. Find the probability that a home has a garage, given that it has a deck.

P(garage|deck) = 0.42

0.60

= 0.7 or 70%



Consider this table concerning utility patents granted for a specific year.Select one patent at random.a. What is the probability that it is a foreign patent, given that it was issued to a corporation?b. What is the probability that it was issued to an individual, given that it was a U.S. patent?

Corporation Government Individual

U.S. 70,894 921 6129

Foreign 63,182 104 6267

a. What is the probability that it is a foreign patent, given that it was issued to a corporation?P(foreign patent | corporation)

=

P(corporation and foreign patent)

P(corporation)


U.S. 70,894 921 6129

Foreign 63,182 104 6267

P(foreign patent | corporation)

=


P(corporation)


U.S. 70,894 921 6129

Foreign 63,182 104 6267

=

63,182

147,497

70,894 + 63,182

147,497

= 63,182

134,076

P(foreign patent | corporation)

=


P(corporation)


U.S. 70,894 921 6129

Foreign 63,182 104 6267

= 0.4712

=

63,182

147,497

70,894 + 63,182

147,497

b. What is the probability that it was issued to an individual, given that it was a U.S. patent?P (individual | U.S.)

=

P(U.S. & individual)P(U.S.)


U.S. 70,894 921 6129

Foreign 63,182 104 6267

P(individual | U.S.) = P(U.S. & individual)

P(U.S.)


U.S. 70,894 921 6129

Foreign 63,182 104 6267

=

6129

147,497

70,894 + 921 + 6129

147,497

=

612977,944

P(individual | U.S.) = P(U.S. & individual)

P(U.S.)


U.S. 70,894 921 6129

Foreign 63,182 104 6267

=

6129

147,497

70,894 + 921 + 6129

147,497

= 0.0786


Section 4-5

Counting Rules



How many different 3 - digit identification tags can be made if the digits can be used more than once? If the first digit must be a 5 and repetitions are not permitted?

If digits can be used more than once: Since there are three spaces to fill and 10 choices for each space, the solution is:

10 10 10 = 1000

If the first digit must be a 5 and repetitions are not permitted: There is only one way to assign the first digit, 9 ways to assign the second, and 8 ways to assign the third:

1 9 8 = 72

How many different 3 - digit identification tags can be made if the digits can be used more than once? If the first digit must be a 5 and repetitions are not permitted?



How many different ID cards can be made if there are 6 digits on a card and no digit can be used more than once?

Since order is important, the solution is:

10P6 =

10!

(10–6)!

9 8 7 6 5 4 3 2 1 10 =

4 3 2 1

= 151,200



How many ways can a committee of 4 people be selected from a group of 10 people?

Since order is not important, the solution is:

10C4 = 210 = 10!

6!4!



How many ways can a foursome of 2 men and 2 women

be selected from 10 men and 12 women in a golf club?

10C12 12C2 = 10!

8!2! 12!

10!2!

= 45 66

= 2970


Section 4-6Probability and Counting Rules



In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.a. All 3 selected will be women.b. All 3 selected will be men.c. 2 men and 1 woman will be selected.d. 1 man and 2 women will be selected.

There are 4C3 ways of selecting

3 women and 7C3 total ways to

select 3 people; hence,

4 3

7 3all women =

CP

C =

4

35

In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.a. All 3 selected will be women.

b. All 3 selected will be men.


3 men; hence,

3 3

7 3all men =

CP

C

1 =

35

In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.

c. 2 men and 1 woman will be selected.


2 men and 4C1 ways of selecting

1 woman; hence,

4 13 22 men and 1 woman = 7 3

C CP

C

=

12

35


d. 1 man and 2 women will be selected.


1 man and 4C2 ways of selecting

2 women; hence,

3 1 4 21 man and two women =

7 3

C CP

C

=

18

35




A committee of 4 people is to be formed from 6 doctors and 8 dentists. Find the probability that the committee will consist of:a. All dentists.b. 2 dentists and 2 doctors.c. All doctors.d. 3 doctors and 1 dentist.e. 1 doctor and 3 dentists.

8C4

14C4 =

701001

= 10143

A committee of 4 people is to be formed from 6 doctors and 8 dentists. Find the probability that the committee will consist of:a. All dentists.

b. 2 dentists and 2 doctors.

6C2 8C2

14C4 =

4201001

= 60

143

A committee of 4 people is to be formed from 6 doctors and 8 dentists. Find the probability that the committee will consist of:

c. All doctors.

6C4

14C4 =

151001


d. 3 doctors and 1 dentist.

6C3 8C1

14C4 =

1601001


e. 1 doctor and 3 dentists.

6C1 8C3

14C4 =

3361001

=

48143




A drawer contains 11 identical red socks and 8 identical black socks. Suppose that you choose 2 socks at random in the dark.a. What is the probability that you

get a pair of red socks?

b. What is the probability that you get a pair of black socks?

c. What is the probability that you get 2 unmatched socks?

d. Where did the other red sock go?

11C2

19C2 =

55

171 = 0.3216

A drawer contains 11 identical red socks and 8 identical black socks. Suppose that you choose 2 socks at random in the dark.a. What is the probability that you

get a pair of red socks?

b. What is the probability that you get a pair of black socks?

8C2

19C2 =

28

171 = 0.1637

A drawer contains 11 identical red socks and 8 identical black socks. Suppose that you choose 2 socks at random in the dark.

c. What is the probability that you get 2 unmatched socks?

11C1 8C1

19C2 =

88

171 = 0.5146


d. Where did the other red sock go?

It probably got lost in the wash!




Find the probability that if 5 different- sized washers are arranged in a row, they will be arranged in order of size.

There are 5! = 120 ways to arrange 5 washers in a row and 2 ways to have them in correct order, small to large or large to small; hence, the probability is:

2

120 =

1

60

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Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability.

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