chapter 4 probability and counting rules section 4-2 sample spaces and probability
TRANSCRIPT
![Page 1: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/1.jpg)
Chapter 4Probability and Counting Rules
Section 4-2
Sample Spaces and Probability
![Page 2: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/2.jpg)
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
![Page 3: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/3.jpg)
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 .
![Page 4: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/4.jpg)
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).
The probability then is
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.
The probability then is
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).
![Page 5: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/5.jpg)
![Page 6: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/6.jpg)
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
![Page 7: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/7.jpg)
Chapter 4Probability and Counting Rules
Section 4-3Exercise #23
![Page 8: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/8.jpg)
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:
![Page 9: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/9.jpg)
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.
![Page 10: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/10.jpg)
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
If one card is drawn from an ordinary deck of cards, find the probability of getting the following:
![Page 11: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/11.jpg)
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
If one card is drawn from an ordinary deck of cards, find the probability of getting the following:
![Page 12: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/12.jpg)
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
If one card is drawn from an ordinary deck of cards, find the probability of getting the following:
![Page 13: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/13.jpg)
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(10 of spades or 10 of clubs)
= P(9) + P(10) + P(spade) + P(club)
If one card is drawn from an ordinary deck of cards, find the probability of getting the following:
![Page 14: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/14.jpg)
e. A 9 or a 10 or a spade or a club.
P ( 9 or 10 or spade or club)
– P(9 of spades or 9 of clubs)
– P(10 of spades or 10 of clubs)
= 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
If one card is drawn from an ordinary deck of cards, find the probability of getting the following:
![Page 15: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/15.jpg)
Chapter 4Probability and Counting Rules
Section 4-4
The Multiplication Rules and Conditional Probability
![Page 16: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/16.jpg)
Chapter 4Probability and Counting Rules
Section 4-4Exercise #7
![Page 17: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/17.jpg)
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
![Page 18: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/18.jpg)
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
![Page 19: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/19.jpg)
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.
![Page 20: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/20.jpg)
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.
![Page 21: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/21.jpg)
Chapter 4Probability and Counting Rules
Section 4-4Exercise #21
![Page 22: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/22.jpg)
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
![Page 23: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/23.jpg)
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
![Page 24: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/24.jpg)
Chapter 4Probability and Counting Rules
Section 4-4Exercise #31
![Page 25: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/25.jpg)
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%
![Page 26: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/26.jpg)
Chapter 4Probability and Counting Rules
Section 4-4Exercise #35
![Page 27: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/27.jpg)
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
![Page 28: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/28.jpg)
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)
Corporation Government Individual
U.S. 70,894 921 6129
Foreign 63,182 104 6267
![Page 29: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/29.jpg)
P(foreign patent | corporation)
=
P(corporation and foreign patent)
P(corporation)
Corporation Government Individual
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
![Page 30: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/30.jpg)
P(foreign patent | corporation)
=
P(corporation and foreign patent)
P(corporation)
Corporation Government Individual
U.S. 70,894 921 6129
Foreign 63,182 104 6267
= 0.4712
=
63,182
147,497
70,894 + 63,182
147,497
![Page 31: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/31.jpg)
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.)
Corporation Government Individual
U.S. 70,894 921 6129
Foreign 63,182 104 6267
![Page 32: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/32.jpg)
P(individual | U.S.) = P(U.S. & individual)
P(U.S.)
Corporation Government Individual
U.S. 70,894 921 6129
Foreign 63,182 104 6267
=
6129
147,497
70,894 + 921 + 6129
147,497
=
612977,944
![Page 33: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/33.jpg)
P(individual | U.S.) = P(U.S. & individual)
P(U.S.)
Corporation Government Individual
U.S. 70,894 921 6129
Foreign 63,182 104 6267
=
6129
147,497
70,894 + 921 + 6129
147,497
= 0.0786
![Page 34: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/34.jpg)
Chapter 4Probability and Counting Rules
Section 4-5
Counting Rules
![Page 35: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/35.jpg)
Chapter 4Probability and Counting Rules
Section 4-5Exercise #9
![Page 36: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/36.jpg)
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
![Page 37: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/37.jpg)
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?
![Page 38: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/38.jpg)
Chapter 4Probability and Counting Rules
Section 4-5Exercise #21
![Page 39: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/39.jpg)
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
![Page 40: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/40.jpg)
Chapter 4Probability and Counting Rules
Section 4-5Exercise #31
![Page 41: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/41.jpg)
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!
![Page 42: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/42.jpg)
Chapter 4Probability and Counting Rules
Section 4-5Exercise #41
![Page 43: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/43.jpg)
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
![Page 44: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/44.jpg)
Chapter 4Probability and Counting Rules
Section 4-6Probability and Counting Rules
![Page 45: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/45.jpg)
Chapter 4Probability and Counting Rules
Section 4-6Exercise #3
![Page 46: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/46.jpg)
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.
![Page 47: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/47.jpg)
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.
![Page 48: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/48.jpg)
b. All 3 selected will be men.
There are 3C3 ways of selecting
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.
![Page 49: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/49.jpg)
c. 2 men and 1 woman will be selected.
There are 3C2 ways of selecting
2 men and 4C1 ways of selecting
1 woman; hence,
4 13 22 men and 1 woman = 7 3
C CP
C
=
12
35
In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.
![Page 50: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/50.jpg)
d. 1 man and 2 women will be selected.
There are 3C1 ways of selecting
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
In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.
![Page 51: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/51.jpg)
Chapter 4Probability and Counting Rules
Section 4-6Exercise #9
![Page 52: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/52.jpg)
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.
![Page 53: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/53.jpg)
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.
![Page 54: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/54.jpg)
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:
![Page 55: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/55.jpg)
c. All doctors.
6C4
14C4 =
151001
A committee of 4 people is to be formed from 6 doctors and 8 dentists. Find the probability that the committee will consist of:
![Page 56: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/56.jpg)
d. 3 doctors and 1 dentist.
6C3 8C1
14C4 =
1601001
A committee of 4 people is to be formed from 6 doctors and 8 dentists. Find the probability that the committee will consist of:
![Page 57: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/57.jpg)
e. 1 doctor and 3 dentists.
6C1 8C3
14C4 =
3361001
=
48143
A committee of 4 people is to be formed from 6 doctors and 8 dentists. Find the probability that the committee will consist of:
![Page 58: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/58.jpg)
Chapter 4Probability and Counting Rules
Section 4-6Exercise #11
![Page 59: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/59.jpg)
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?
![Page 60: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/60.jpg)
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?
![Page 61: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/61.jpg)
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.
![Page 62: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/62.jpg)
c. What is the probability that you get 2 unmatched socks?
11C1 8C1
19C2 =
88
171 = 0.5146
A drawer contains 11 identical red socks and 8 identical black socks. Suppose that you choose 2 socks at random in the dark.
![Page 63: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/63.jpg)
d. Where did the other red sock go?
It probably got lost in the wash!
A drawer contains 11 identical red socks and 8 identical black socks. Suppose that you choose 2 socks at random in the dark.
![Page 64: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/64.jpg)
Chapter 4Probability and Counting Rules
Section 4-6Exercise #15
![Page 65: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/65.jpg)
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
![Page 66: Chapter 4 Probability and Counting Rules Section 4-2 Sample Spaces and Probability](https://reader033.vdocument.in/reader033/viewer/2022061614/56649d205503460f949f58e9/html5/thumbnails/66.jpg)