Chapter 4: Probability and Counting Rules

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Dr. Janet Winter, [email protected] Stat 200 Page 1

Chapter 4: Probability and Counting Rules

Learning Objectives

Upon successful completion of Chapter 4, you will be able to:

–

Determine sample spaces and find the probability of an event using classical probability or empirical probability.

–

Find the probability of compound events using the addition rules, multiplication rules, and the conditional probability equation.

I.

Basic Vocabulary

• Probability, the chance of an event occurring, is used in games of chance, the insurance industry, investments, weather forecasting, and in various other areas.

• A probability experiment is a chance process.

• An outcome is the result of a single trial of a probability experiment.

• Equally likely events have the same probability of happening.

II.

Sample Spaces

A. General Information

• A record of all possible outcomes of a probability experiment using lists, charts, trees or Venn Diagrams.

• Use lists, charts, and tree diagrams to answer questions about equally likely events.

• Use tree diagrams with probability recorded on the branches for events that are not equally likely.

B. Examples for Lists and Charts

I. Lists Activity

List the sample space for a 2 coin toss.

{(H, H), (, ), (, ), (, )}

Note: If you thought (H, T) and (T, H) were the same event and did not list them both, picture 1 coin was dipped in red paint and the other coin was not.

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Dr. Janet Winter, [email protected] Stat 200 Page 2 II. Chart Activity

Use a chart to list all events for tossing 1 die twice (or 2 dice once).

C. Tree Diagrams

I. A tree diagram uses line segments to list all possibilities of a sequence of events in a systematic way.

(1, 1) (1, 2)

(2, 1)

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Dr. Janet Winter, [email protected] Stat 200 Page 3 II. Example:

a) If both genders are equally likely the tree would be:

Gender for a family with 3 children

b) If there is an 80% chance the child is a girl, P(G)=.8 P(B)=1-P(G) P(B)=1-.8=.2

D. Venn Diagrams

I. Venn Diagrams represent probabilities using a diagram:

It is helpful to represent intersections.

B

B B

G

G B

G

B B

G

G B

G

B

B B

G

G B

G

B B

G

G B

G

P(A) P(B)

.2

.2 .8 .2

.8 .8

.8

S

P(A and B)

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Dr. Janet Winter, [email protected] Stat 200 Page 4 II. Example:

P(A) = .4 P(B) = .6 P(A and B) = .2

III.

Types of Probability

A. Empirical

I. Empirical probability uses actual experiences or experiments to determine the likelihood of outcomes.

II. Example:

B. Subjective Probability

I. Subjective probability is an estimate of probability based on opinions, past experiences, and any other available information.

II.

Example: P (rain tonight) = .30 or 30%

Example:

C. Classical Probability

I. Classical probability uses sample spaces and rules to determine the numerical probability that an event will happen.

For equally likely outcomes:

• Count the number of outcomes of interest and divide it by the total number of outcomes in the sample space.

• Counting, is used only for equally likely events.

• Using rules, classical probability can be used for other situations.

A .2

B .4

.2

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IV.

Probability Concepts

A. Recording Probability

I. Use fractions or reduced fractions OR use decimals rounded to 4 significant digits when possible.

II. For extremely small decimals, round to the first nonzero digit after the decimal point and use scientific notation.

B. General Probability Rules

I. The probability of an event E is between and including 0 and 1.

0 ≤ 𝑃 (𝐸) ≤ 1

This means that probabilities cannot be negative or greater than 1.

II. If the event E cannot happen, then P(E)=0 III. If an event E is certain, P(E)=1

IV. The sum of the probabilities of all outcomes in the sample space is 1.

� 𝑃 (𝑋) = 1.00

𝑛

𝑖=1

C. Complementary Events

I. The set of elements in the sample space that are not in E is called the complement of E and is denoted Ē.

P(Ē) = 1 – P(E)

II. Example

If P(win) = .8, the P(loose) = 1 – P(win) = 1 - .8 = .2 :

D. Law of Large Numbers

If an experiment is repeated a larger and larger number of times usually relative frequency approaches theoretical or classical probability. Experiment yields more accurate results with a larger number of repeats.

V.

Conjunctions for Two Events

A. Types of Conjunctions for 2 Events

I. A or B (means either A, B, or both A and B) P (A or B) II. A and B (means both A and B must happen) P (A and B)

III. A given B (means A happens knowing B has already happened) P (A | B)

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B. Characteristics of Events

I. Mutually Exclusive Events or Disjoint Events

• Cannot occur together

• Have no outcomes in common

• A∩B = ∅

Statisticians use the concept of mutually exclusive to find the P (A or B).

II. Independent and Dependent Events

• 2 events, A and B, are independent if A occurring does not affect the probability of B occurring.

• 2 events are dependent if A occurring does affect the probability of B occurring.

• Independent events result from situations that do not change in the experiment Eg. Dice, coins turning on a light.

• Independent events result when objects are replaced after they are selected for the first trial.

• Dependent events result from drawing a second object without replacing the first trial

Statisticians use the concept of independent to find the P (A and B).

C. Addition Rules for: P (A or B)

I. Event A or event B means A,B or both A and B occur II. Rules

a) If A and B are mutually exclusive, the probability that A or B will occur is:

P (A or B) = P (A) + P (B)

Note: A ∩ B = ∅

A B

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Dr. Janet Winter, [email protected] Stat 200 Page 7 b) If A and B are not mutually exclusive, then:

P (A or B) = P (A) + P (B) - P(A and B)

-

Note: A ∩ B ≠ ∅

Note: The intersection or (A and B) is included in A and also included in B.

III.

Examples

1. P(A) = .4 P(B) = .1 A and B are mutually exclusive

a) P(A) = .4 P(B) = .20 P(A and B) = .15

P(A or B) = .4 + .2 - .15 = .45 - or-

P(A or B) = .25 + .15 + .05 = .45

Question 1

To find the number of elements in A or B, the number in “A” plus the number in “B” is too many because:

a) the number in “A” is too large.

b) the number in “B” is too large.

c) the elements in “A” and “B” are counted twice.

d) I do not know.

P(A) P(B)

A .4

B .1

.25 A B .05

S

P(A and B)

.5

.15 .55

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Question 2

In a group of apartments for rent, 8 have 2 bedrooms and 4 have fireplaces. If 2 apartments have both 2 bedrooms and a fireplace, then how many apartments have 2 bedrooms or a fireplace?

a) 4 – because it’s 8 with 2 bedrooms minus 4 with fireplaces: 8 – 4 = 4.

b) 10 – because it’s 8 with 2 bedrooms plus 4 with fireplaces minus 2 with both 2 bedrooms and a fireplace: 8 + 4 – 2 = 10.

c) 12 – because it’s 8 with 2 bedrooms plus 4 with fireplaces: 8 + 4 = 12.

d) I do not know.

D. Multiplication Rules

I. Multiplication Rule #1 for independent events.

Multiplication Rule #1 states when 2 events are independent, the probability of both occurring is:

P(A and B) = P(A) x P(B)

Since the occurrence of either A or B does not affect the probability of the occurrence of the other

Examples

Assume the chance a couple has a boy or a girl is equally likely and the gender of a child is independent of the gender of any brothers or sisters.

:

a) What is the probability a couple has exactly one girl in their family of 3 children?

The girl could be the oldest, the middle, or the youngest child.

Thus, there are three ways this can occur each P (1 girl in a family of 3 children) =

with probability or

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Dr. Janet Winter, [email protected] Stat 200 Page 9 b) What is the probability a couple has exactly two girls in a family of three children?

Since the family has two girls and one boy, the one boy can be the oldest, middle, or youngest child. There are three different ways a family can have two girls each with probability or P (2 girls in a family of 3 children) =

c) What is the probability a family with 3 children has exactly three girls?

There is only one way this can happen GGG with probability or P (3 girls in a family of 3 children) =

d) What is the probability a family with 3 children has at least 1 girl?

P (at least 1 girl) = P (1 girl or 2 girls or 3 girls)

= P (1 girl) + P (2 girls) + P (3 girls) =

Another way to complete this problem is P (at least one girl) = 1 – P (no girls) The only way a family of 3 children would have no girls is to have all boys which can only occur in one way BBB with probability

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Dr. Janet Winter, [email protected] Stat 200 Page 10 e) P(A) = .3 P(B) = .4 A and B are independent. Construct a Venn Diagram.

P(A and B) = P(A) × P(B) since A and B are independent P(A and B) = . 3 × .4 = .12

Start with the intersection

Question 3

At a university, 16% of students are engineering majors. Of those engineering majors, 50% have completed a statistics course. The percent of students in that university who are engineering majors and have completed a statistics course is:

f) 16%.

g) 50% minus 16%, 34%.

h) 50% times 16%, or 8%.

i) I do not know.

Question 4

There are 10 boys and 10 girls in a second grade, and 8 boys and 12 girls in a third grade.

If a random second grade student and a random third grade student were chosen independently, the probability that both are girls is:

a) 22/18, because there are 22 girls and 18 boys altogether.

b) 10/20 times 12/20, or 1/2 times 3/5, because there are 10 girls out of 20 and 12 girls out of 20, chosen independently.

c) 22/40, or 11/20, because there are 22 girls and 18 boys altogether.

d) I do not know.

Question 5

The Gallup Poll reported that 53% of Americans wear seat belts when they are driving in a car. If 4 people are selected at random, what is the probability that they all used a safety belt the last time they got into a car? Round your answer to 4 decimal places.

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Dr. Janet Winter, [email protected] Stat 200 Page 11 III. Conditional Probability P (B|A)

a) The conditional probability of event B in relationship to event A is the probability that event B occurs after event A has already occurred.

b) Examples

1. Find the probability of drawing an ace from a well-shuffled deck.

:

4/52

2. Find the probability of drawing a second ace from a well-shuffled deck if the first ace is not returned to the deck.

3/51 IV. Multiplication Rule #2

Multiplication Rule #2 states when 2 events are dependent, the probability of both occurring is:

P(A and B) = P(A) x P(B|A)

Example

Find the probability of dealing 4 jacks in a row.

:

This example has an implied “and” or it is “Find the probability the first card is a jack and the second card is a jack and the third card is a jack and

4 52 ×

3 51 ×

2 50 ×

1

49 = 3.69 × 10⁻⁶ 𝑜𝑟 .000004

the fourth is a jack.” The

“and” means multiply. The “dealing” means without replacing the cards.

E. Formula for Conditional Probability

Use Multiplication Rule #2 to find the formula for conditional probability.

P(𝐴 and 𝐵)

P(𝐴) = P(𝐵|𝐴)

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F. Intuitive Approach to Conditional Probability

P (B|A) is found by assuming that event A has occurred and then calculating the probability that B will occur.

• Example: This is the data from the medical experiment with the Salk vaccine.

Polio No Polio Totals

Salk Vaccine Group 33 200712 200745

Placebo Group 115 201114 200229

Totals 148 401826 400974

a) Find the probability a person contracted polio given that they received the placebo using the intuitive methods.

𝐏 (𝐩𝐨𝐥𝐢𝐨|placebo)= = .011

Notice the sample space is reduced to the 200229 children in the plabebo group.

Within the placebo group, 115 children were diagnosed with polio.

b) Redo the question with the formula.

Question 6

20 students and 5 professors are walking across campus. 15 of the students and 4 of the professors are carrying a textbook. What is the probability that the person is carrying a textbook, given that the person is a professor?

a) 20% because 4 out of 20 = 4 / 20 = 20%.

b) 50% because the professor either is carrying a textbook or isn’t.

c) 80% because 4 out of 5 = 4 / 5 = 80%.

d) I do not know.

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G. Summary of Multiplication Rules

To find the probability of 2 or more events occurring in sequence:

I. For 2 independent events, use Multiplication Rule #1:

P(A and B) = P(A) x P(B)

II. For 2 dependent events, use Multiplication Rule #2:

Question 7

4 cards are randomly selected from a 52-card deck and not replaced. What is the probability all 4 cards are clubs?

a) 13/5 x 12/51 x 11/50 x 10/49.

b) 4/52 x 3/51 x 2/50 x 1/49.

c) 4/52 x 3/52 x 2/52 x 1/52.

d) I do not know.

Question 8

If medication is 75% effective against a bacterial infection, find the probability that all 4 patients given the medication will improve.

a) .75 x .74 x .73 x .72.

b) .75 x .75 x .75 x .75.

c) 4 x 3 x 2 x 1.

d) I do not know.

VI.

Conclusion

A tree diagram, chart or list can be used when a list of all possible outcomes is necessary.

However, when only the number of outcomes is needed, rules are more efficient.

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Answer: Activity (Lists)

List the sample space for a 2 coin toss.

{(H, H), (H, T), (T, H), (T, T)}

Note: If you thought (H, T) and (T, H) were the same event and did not list them both, picture 1 coin was dipped in red paint and the other coin was not.

Answer: Activity (Charts)

Use a chart to list all events for tossing 1 die twice (or 2 dice once).

Answer: Question 1

To find the number of elements in A or B, the number in “A” plus the number in “B” is too many because:

C – the elements in “A” and “B” are counted twice.

Answer: Question 2

In a group of apartments for rent, 8 have 2 bedrooms and 4 have fireplaces. If 2 apartments have both 2 bedrooms and a fireplace, then how many apartments have 2 bedrooms or a fireplace?

B – 10 – because it’s 8 with 2 bedrooms plus 4 with fireplaces minus 2 with both 2 bedrooms and a fireplace: 8 + 4 – 2 = 10.

Answer: Question 3

At a university, 16% of students are engineering majors. Of those engineering majors, 50% have completed a statistics course. The percent of students in that university who are engineering majors and have completed a statistics course is

C – 50% times 16%, or 8%.

Answer: Question 4

There are 10 boys and 10 girls in a second grade, and 8 boys and 12 girls in a third grade. If a random second grade student and a random third grade student were chosen independently, the probability that both are girls is:

B – 10/20 times 12/20, or 1/2 times 3/5, because there are 10 girls out of 20 and 12 girls out of 20, chosen independently.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

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Answer: Question 5

The Gallup Poll reported that 53% of Americans wear seat belts when they are driving in a car.

If 4 people are selected at random, what is the probability that they all used a safety belt the last time they got into a car? Round your answer to 4 decimal places.

.53⁴ = .0789

Answer: Question 6

20 students and 5 professors are walking across campus. 15 of the students and 4 of the professors are carrying a textbook. What is the probability that the person is carrying a textbook, given that the person is a professor?

C – 80% because 4 out of 5 = 4 / 5 = 80%.

Answer: Question 7

4 cards are randomly selected from a 52 card deck and not replaced. What is the probability all 4 cards are clubs?

A – 13/5 x 12/51 x 11/50 x 10/49

Answer: Question 8

If medication is 75% effective against a bacterial infection, find the probability that all 4 patients given the medication will improve.

B – .75 x .75 x .75 x .75

Chapter 4: Probability and Counting Rules