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Ch5Probability Probability is a measure of the likelihood of a random phenomenon or chance behavior. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty. If we flip a coin 100 times and compute the proportion of heads observed after each toss of the coin, what will the proportion approach? The Law of Large Numbers As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome. Ch 5.1Probability Rule Objective A: Sample Spaces and Events Experiment – any activity that leads to well-defined results called outcomes. Outcome – the result of a single trial of a probability experiment. Sample space, S – the set of all possible outcomes of a probability experiment. Event, E – a subset of the sample space

Simple event, ie – an event with one outcome is called a simple event.

Compound event – consists of two or more outcomes. Example 1: A die is tossed one time. (a) List the elements of the sample space S.

S = {1, 2, 3, 4, 5, 6}

(b) List the elements of the event consisting of a number that is greater than 4. E = {5, 6} Example 2: A coin is tossed twice. List the elements of the sample space S, and list the elements of the event

consisting of at least one head. HH S = {HH, HT, TH, TT} TT

E = {HT, TH, HH}

H

H T

H

T T

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Objective B: Requirements for Probabilities 1. Each probability must lie on between 0 and 1. (0 ≤ P (E) ≤ 1) 2. The sum of the probabilities for all simple events in S equals 1. ( 𝑃(𝑒𝑖) = 1) If an event is impossible, the probability of the event is 0. If an event is a certainty, the probability of the event is 1. An unusual event is an event that has a low probability of occurring. Typically, an event with a probability less than 0.05 is considered as unusual. Probabilities should be expressed as reduced fractions or rounded to three decimal places. Example 1: A probability experiment is conducted. Which of these can be considered a probability of an outcome? (a) 2/5 (b) -0.28 (c) 1.09 Yes No No (0 ≤ P (E) ≤ 1) (0 ≤ P (E) ≤ 1) Example 2: Why is the following not a probability model?

Color Probability

Red 0.28

Green 0.56

Yellow 0.37

Condition 1: (0 ≤ P (E) ≤ 1)

Condition 2: ( 𝑃(𝑒𝑖) = 1)

Check: 0.28 + 0.56 + 0.37 = 1.21 ≠ 1

Condition 2 was not met.

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Example 3: Given: 𝑆 = {𝑒1, 𝑒2, 𝑒3, 𝑒4} 𝑃 𝑒1 = 𝑃 𝑒2 = 0.2 and 𝑃 𝑒3 = 0.5 Find: 𝑃 𝑒4 Condition 1: (0 ≤ P (E) ≤ 1) Condition 2: ( 𝑃(𝑒𝑖) = 1) P(𝑒1) + P(𝑒2) + P(𝑒31)+ P(𝑒4) = 1 0.2 + 0.2+ 0.5 + P(𝑒4) = 1 0.9 + P(𝑒4) = 1 P(𝑒4) = 1- 0.9 P(𝑒4) = 0.1

Objective C:Calculating Probabilities( )

( )( )

N EP E

N S

Example 1: The age distribution of employees for this college is shown below:

Age Number of Employees

Under 20 25

20 – 29 48

30 – 39 32

40 – 49 15

50 and over 10

= = 130

If an employee is selected at random, find the probability that he or she is in the following age groups

(a) Between 30 and 39 years of age

32

130=

16

65or 0.246 (3 decimal places)

(b) Under 20 or over 49 years of age

25+10

130 =

35

130 =

7

26 or 0.269 (3 decimal places)

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Example 2: Let the sample space be 1,2,3,4,5,6,7,8,9,10S . Suppose the outcomes are equally likely.

(a) Compute the probability of the event 5,9F .

N (F) = 2 N (S) = 10

P (F) = 𝑁 (𝐹)

𝑁 (𝑆) =

2

10 =

1

5 or 0.2

(b) Compute the probability of the event E = "an odd number."

E = {1, 3, 5, 7, 9} N (E) = 5

P (E) = 𝑁 (𝐸)

𝑁 (𝑆) =

5

10 =

1

2 or 0.5

Example 3: Two dice are tossed. Find the probability that the sum of two dice is greater than 8?

N (S) = 36 N (E) = 10

E = {(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)}

P(E) = 𝑁 (𝐸)

𝑁 (𝑆) =

10

36=

5

18 𝑜𝑟 ≈ 0.278

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Example 3:If one card is drawn from a deck, find the probability of getting (a) a club

P (a club) = 𝑁 (𝑎 𝑐𝑙𝑢𝑏 )

𝑁 (𝑆) =

13

52 =

1

4 or 0.25

(b) a 4 and a club

P (a 4 and a club) = 𝑁 (𝑎 4 𝑎𝑛𝑑 𝑎 𝑐𝑙𝑢𝑏 )

𝑁 (𝑆) =

1

52 ≈ 0.019

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Example 4:Three equally qualified runners, Mark, Bill, and Alan, run a 100-meter sprint, and the order of finish

is recorded. (a) Give a sample space S.

S = {MBA, MAB, BMA, BAM, AMB, ABM}

(b) What is the probability that Mark will finish last?

E = {BAM, ABM}

P (E) = 𝑁 (𝐸)

𝑁 (𝑆) =

2

6 =

1

3 or ≈ 0.333

Ch5.2 The Addition Rules and Complements Objective A: Addition Rule for Disjoint (Mutually Exclusive) Events Event A and B are disjoint (mutually exclusive) if they have no outcomes in common. Addition Rule for Disjoint Events

If E and F are disjoint events, then ( or ) ( ) ( )P E F P E P F .

Example 1: A standard deck of cards contains 52 cards. One card is randomly selected from the deck. Compute the probability of randomly selecting a two or three from a deck of cards.

M

M

M

M

B

B

B

B

A

A

A

A

M

A

B

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P (2 or 3) = P (2) + P (3) = 4

52 +

4

52 =

8

52 = 13 or ≈ 0.154

Objective B: General Addition Rule The General Addition Rule

For any two events E and F, ( or ) ( ) ( ) ( and )P E F P E P F P E F .

Example 1: A standard deck of cards contains 52 cards. One card is randomly selected from the deck. Compute the probability of randomly selecting a two or club from a deck of cards.

P (2 or club) = P (2) + P(club) – P(2 and club)

= 4

52 +

13

52 -

1

52 =

16

52 =

4

13 or ≈ 0.308

Objective C: Complement Rule Complement Rule

If E represents any event and cE represents the complement of E , then ( ) 1 ( )CP E P E .

Example 1: The chance of raining tomorrow is 70%. What is the probability that it will not rain tomorrow?

P (not raining) = 1 – P (raining) = 1- 0.7 = 0.3

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Example 2: In a large department store, there are 2 managers, 4 department heads, 16 clerks, and 4 stock persons. If a person is selected at random,

(a) find the probability that the person is a clerk or a manager;

P (clerk or manager) = P (clerk) +P (manager) –P (clerk and manager)

= 16

26 +

2

26 -

0

26 =

18

26 =

9

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(b) find the probability that the person is not a clerk.

P (not a clerk)= 1-P (clerk)

= 1- 16

26 =

26

26 -

16

26 =

10

26 =

5

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Example 3: A probability experiment is conducted in which the sample space of the experiment

is 1,2,3,4,5,6,7,8,9,10,11,12S .

Let event 2,3,5,6,7E , event 5,6,7,8F , and event 9,11G

(a) List the outcome in and E F . Are and E F mutually exclusive?

E = {2, 3, 5, 6, 7} F = {5, 6, 7, 8} Since there are common elements of 5, 6, and 7 between E and F,

E and F are not mutually exclusive. (b) Are and F G mutually exclusive? Explain. Yes, there is no common element between F and G.

(c) List the outcome in or E F . Find ( or )P E F by counting the

number of outcomes in or E F . E or F = {2, 3, 5, 6, 7, 8}

N (E or F) = 6

N (S)= 12

P (E or F) = 6

12 =

1

2 or 0.5

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(d) Determine ( or )P E F using the General Addition Rule.

P( E or F) = P (E) + P (F) – P (E and F)

= 5

12 +

4

12 -

3

12 =

6

12 =

1

2 or 0.5

(e) List the outcomes in CE . Find ( )CP E by counting the number of outcomes in CE .

𝐸𝐶 = {1, 4, 8, 9, 10, 11, 12}

P (𝐸𝐶)= 𝑁(𝐸𝐶)

𝑁 (𝑆) =

7

12

(f) Determine ( )CP E using the Complement Rule.

P (𝐸𝐶)= 1- P (E)

= 1 - 5

12 =

12

12 -

5

12 =

7

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Objective D: Contingency Table A contingency table relates two categories of data. It is also called a two-way table which consists of a row variable and a column variable. Each box inside the table is called a cell. Example 1: In a certain geographic region, newspapers are classified as being published daily morning, daily evening, and weekly. Some have a comics section and other do not. The distribution is shown here. Have comics (M) (E) (W) Section Morning Evening Weekly (CY) Yes 2 3 1 6 (CN) No 3 4 2 9 5 7 3 15 If a newspaper is selected at random, find these probabilities. (a) The newspaper is a weekly publication.

P(W) = 𝑁(𝑊)

𝑁(𝑆) =

3

15 =

1

5

(b) The newspaper is a daily morning publication or has comics.

P (M or CY) = P (M) + P (CY) – P (M and CY)

= 5

15 +

6

15 -

2

15 =

9

15 =

3

5

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Ch5.3 Independence and the Multiplication Rule Objective A: Independent Events Two events are independent if the occurrence of event E does not affect the probability of event F . Two events are dependent if the occurrence of event E affects the probability of event F . Example 1: Determine whether the events E and F are independent or dependent. Justify your answer. (a) E : The battery in your cell phone is dead. F : The battery in your calculator is dead. Independent (b) E : You are late to class. F :Your car runs out of gas. Dependent

Objective B: Multiplication Rule for Independent Events

If E and F are independent events, then ( and ) ( ) ( )P E F P E P F

Example 1: If 36% of college students are underweight, find the probability that if

three college students are selected at random, all will be underweight.

Independent case

P (1

st underweight and 2

nd underweight and 3

rd underweight)

P (1

st underweight) * P (2

nd underweight) * P (3

rd underweight)

= 0.36*0.36*0.36 ≈ 0.047

Example 2: If 25% of U.S. federal prison inmates are not U.S. citizens, find the probability that two randomly selected federal prison inmates will be U.S. citizens.

P (1st U. S. citizen and 2

nd U. S. citizen)

= P (1st U. S. citizen) * P (2

nd U. S. citizen)

= 0.75 * 0.75 ≈ 0.563

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Objective C: At-Least Probabilities Probabilities involving the phrase “at least” typically use the Complement Rule. The phrase at least means “greater than or equal to.” For example, a person must be at least 17 years old to see an R-rated movie. Example 1: If you make random guesses for four multiple-choice test questions (each with five possible answers), what is the probability of getting at least one correct? Direct method:

P (at least one correct)

= P (one correct) + P (two correct) + P (three correct) + P (four correct)

Indirect method:

What is the opposite of at least one correct? None correct.

P (none correct) =4

5 *

4

5 *

4

5 *

4

5= (

4

5)4

P (at least one correct) = 1- P (none correct)

= 1- (4

5)4 ≈ 0.590

Example 2: According to the Department of Health and Human Services, 30% of 18- to 25-year-olds have some form of mental illness. (a) What is the probability two randomly selected 18- to 25-year-olds have some form of mental illness?

P (1st with some form of mental illness and 2

nd with some form of mental illness)

= P (1

st with some form of mental illness) * P (2

nd with some form of mental illness)

= 0.30*0.30 = 0.09

(b) What is the probability six randomly selected 18- to 25-year-olds have some form of mental illness?

P (1st with mental illness and 2nd with mental illness and 3rd with mental illness and 4th with mental illness and 5th with mental illness and 6th with mental illness) = P (1st with mental illness) * P (2nd with mental illness) * P (3rd with mental illness) * P (4th with mental illness) * P (5th with mental illness) * P (6th with mental illness) = 0.30 * 0.30 * 0.30 * 0.30 * 0.30 * 0.30 = 0.000729

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(c) What is the probability at least one of six randomly selected 18- to 25-year-olds has some form of mental illness? 30% of 18- to 25-year-olds have some form of mental illness therefore 1-30% do not have some form of mental illness. 1-0.30 = 0.70 P (at least one of 6 has mental illness)

= 1 – P (none with mental illness)

= 1 - 0.706 = 1-0.117649 = 0.882351 ≈ 0.882

(d) Would it be unusual that among four randomly selected 18- to 25-year-olds, none has

some form of mental illness?

P (all four have no mental illness) = (0.70)4 = 0.2401 Since 0.2401 ≤ 0.05, it would not be unusual for none of the four to have mental illness.

Ch 5.4 Conditional Probability and the General Multiplication Rule Objective A: Conditional Probability and the General Multiplication Rule A1. Multiplication Rule for Dependent Events If E and F are dependent events, then ( and ) ( ) ( | )P E F P E P F E .

The probability of E and F is the probability of event E occurring times the probability of event F occurring, given the occurrence of event E . Example 1: A box has 5 red balls and 2 white balls. If two balls are randomly selected (one after the other), what is the probability that they both are red? (a) With replacement

P (1st red and 2nd red)

= P (1st red) * P (2nd red)

= 5

7 *

5

7 =

25

49

(b) Without replacement

P (1st red and 2nd red)

= P (1st red) * P (2

nd red│1

st red)

= 5

7 *

4

6 =

20

42 =

10

21

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Example 2: Three cards are drawn from a deck without replacement. Find the probability that all are jacks.

Without replacement Dependent case P (1st jack and 2nd jack and 3rd jack)

= 4

52 *

3

51 *

2

50 =

1

13 *

14

17 *

1

25 =

1

5525 ≈ 0.00018

(Almost zero percent of a chance)

A2. Conditional Probability

If E and F are any two events, then( and ) ( and )

( )( ) ( )

P E F N E FP F E

P E N E .

The probability of event F occurring, given the occurrence of event E , is found by dividing the probability of E and F by the probability of E . Example 1:At a local Country Club, 65% of the members play bridge and swim, and 72% play bridge. If a member is selected at random, find the probability that the member swims, given that the member plays bridge.

Let B be the event of playing bridge.

Let S be the event of swimming.

Given: P (B and S) = 0.65 P (B) = 0.72

Find: P (S│ B)

P (S │B) = 𝑃 (𝑆 and 𝐵)

𝑃 (𝐵) =

0.65

0.72 ≈ 0.903

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Objective B: Application Example 1: Eighty students in a school cafeteria were asked if they favored a ban on smoking in the cafeteria. The results of the survey are shown in the table.

(Fa) (O) (N) Class Favor Oppose No opinion (F) Freshman 15 27 8 50 (S) Sophomore 23 5 2 30 38 32 10 80 If a student is selected at random, find these probabilities. (a) The student is a freshman or favors the ban.

P(F or Fa) = P (F) + P (Fa) – P (F and Fa)

50

80 +

38

80 -

15

80 =

73

80

(b) Given that the student favors the ban, the student is a sophomore.

P (S│Fa) = 𝑃 (𝑆 𝑎𝑛𝑑 𝐹𝑎)

𝑃 (𝐹𝑎)

= 23/80

38/80 =

23

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