The Practice of Statistics Third Edition Chapter 6: Probability and Simulation: The Study of...

The Practice of StatisticsThird Edition

Chapter 6:Probability and Simulation:The Study of Randomness

6.3 General Probability Rules

Copyright © 2008 by W. H. Freeman & Company

Daniel S. Yates

Essential Questions

• What is the addition rule for disjoint events?• What is the general addition rule for union of two

events?• How do you compute P(A U B)?• What is a joint event and joint probability?• What is the general multiplication rule for any

two events?• What is meant by the conditional probability P(A|

B)?• How do you define independent events in terms

of conditional probability?

Probability Rules from 6.2

Venn Diagrams: Disjoint Events

AB

S

( or ) ( ) ( )P A B P A P B

Union

P(A U B U C) = P(A) + P(B) + P(C)

Venn Diagrams: Non-disjoint Events

A

B

S

A and B

Venn Diagram, Non-Disjoint, in set notation

Suppose set A consists of the following

3,4,5,6

and set B consists of:

6,7,8,9

is the set 6A B

Notice that 6 is included in both sets.

So, in order to find the set (A U B) you must subtract one 6.

( ) is 3, 4,5,6,7,8,9A B

Venn Diagrams: Non-disjoint Events

A

B

S

( or ) ( ) ( ) ( and )P A B P A P B P A B

A and B

Joint Event

• Joint Event is the simultaneous occurrence of two events. For example, the outcome in set of A is six and the outcome in set of B is also six.

• The joint probability of a joint event is P(A and B). P(A and B) is the same as

P(A ∩ B).

General Rule for Union of Two Events

Example 1

• In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.

• A person selected at random from the town. What is the probability that the person will have brown hair or brown eyes? P( BH or BE)

• P( BH or BE) = P(BH) + P(BE) – P(BH ∩ BE) = .40 + .25 - .15 = 0.5

• What is the probability that person selected does not have brown hair or brown eyes?

• 1 – P( BH or BE) = 1 - .5 = 0.5

Two Methods for Picturing Probabilities

• In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.

Venn DiagramBrown Hair Brown Eyes

.25 .10.15

.5

TableBrown Hair

Yes No Total

Brown Yes 0.15 0.25

Eyes No

Total 0.40 1.00

Independent events

• The outcome of one trial does not influence or change the outcome of another trial.

Multiplication Rule

• For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events.

• P(A and B) = P(A) x P(B)

• P(A ∩ B) = P(A) x P(B)

A B

A∩B

Independence

• Please note, that we can use the multiplication rule for independent events, to verify if two events are independent.

• If P(A and B) ≠ P(A)·P(B), then the events are not independent.

Example 1 Continued•In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.

Brown Hair

Yes No Total

Brown Yes 0.15 0.25

Eyes No

Total 0.40 1.00

• What is the probability of selecting a person with brown hair? P(A) What is the probability of selecting a person with brown eyes? P(B)

• P(A) = 0.40 P(B) = 0.25

• What is the probability of selecting a person with brown hair and brown eyes? P( A and B)

• P(A and B) = 0.15

Conditional Probability

• Lets consider the situation of finding the probability of one event under the condition that we know the results of the other event.

• The notation for conditional probability is:P(A| B)

The bar means “given the information that…”So, P(A| B) reads, “The probability of event A given event B

occurs.”

Lets Consider Two Events That May or May Not Be Independent

If we take the equation P(A ∩ B) = P(A)P(B| A) and solve for P(B| A) we get the equation for calculating P(B| A).

Example 1 – Independence?

• Are Events A and B independent?• P(A)●P(B) = (0.40)(0.25) = 0.10 Since P(A and B) ≠

P(A)●P(B), Events A and B are not independent.• Second Method – Test P(B| A) = P(B)

•In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.

Brown Hair

Yes No Total

Brown Yes 0.15 0.10 0.25

Eyes No 0.25 0.50 0.75

Total 0.40 0.60 1.00

Event A = Brown Hair

Event B = Brown Eyes

375.040.0

15.0

)(

)()|(

AP

BAPABP

Since P(A| B) ≠ 0.40, then Events are not independent.

Example 1 – Conditional Probability

•In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.

Brown Hair

Yes No Total

Brown Yes 0.15 0.10 0.25

Eyes No 0.25 0.50 0.75

Total 0.40 0.60 1.00



What is the P(A ∩ B) ?

Ans: P(A ∩ B) = 0.15

What is the probability that the randomly selected person has brown hair given that he has brown eyes?

Ans: P(A| B) = P(A ∩ B) / P(B) = 0.15/0.25 = 0.6

Example 1 – Conditional Probability Continued

Brown Hair

Yes No Total

Brown Yes 0.15 0.10 0.25

Eyes No 0.25 0.50 0.75

Total 0.40 0.60 1.00



What is the probability of the randomly selected person has brown eyes given the person has brown hair?

Ans: P(B| A) = P(A ∩ B) / P(A) = 0.15/0.40 = 0.375

What is the probability that the randomly selected person has neither brown hair nor brown eyes?

Ans: P( Ac ∩ Bc ) = 0.50

Example 2

Age

18-29 30-64 65 and over Total

Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

Example – Question 1Age


Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• A= young (between 18 and 29)• P(A)=?

– 22512/103870

Example 2 – Question 2

Age


Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• B=married• P(B)=?

– 59920/103870

Example 2 – Question 3Age


Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• A=is young (between 18 and 29)• B=married• P(A and B)=?

– 7842/103870



Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• A=is young (between 18 and 29)• B=married• P(A | B)= (Read as “the probability of A given B”)

– 7842/59920• This is known as a “conditional probability”



Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• A=is young (between 18 and 29)• B=married• P(B | A)= (Read as “the probability of B given A”)

– 7842/22512

Example 2 Question 6Age


Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• P(A and B)= 7842/103870• P(A and B)= P(A)*P(B|A) 22512 7842

103870 22512



Married 7,842 43,808 8,270 59,920

Never Married

13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced 704 9,174 1,263 11,141

Total 22,512 62,689 18,669 103,870

• P(A and B)= 7842/103870• P(A and B)= P(B)*P(A|B) 59920 7842

103870 59920

Multiplication Rule Extended to Several Events

Recall the general multiplication rule for the intersection of two events.

Now, consider the intersection of three events. Remember

) B andA |P(C )A |P(B P(A) ) C B A P(

Example 6.29 page 448• Only 5% of male high school basketball, and football players go on

to play at the college level. Of these, only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years.

• Define– A = { competes in college} P(A) = 0.05– B = { competes professionally} P(B|A) = 0.017– C = { pro career longer than 3 years} P(C|A and B) = 0.40

• What is the probability that a high school athlete competes in college and then goes on to have a pro career of more than 3 years? P(A and B and C)

• P(A and B and C) = P(A)●P(B|A)●P(C|A and B) = (0.05)(0.017)(0.40) = 0.00034• SO, only 3 out of 10,000 high school athletes will have a pro-career

of more than 3 years.

Example 3 Using Tree Diagram

• Using the same probabilities from the previous problem: – A = { competes in college} P(A) = 0.05– B = { competes professionally} P(B|A) = 0.017

• Suppose the probability of a few high school athlete enter the pro’s directly from high school (does not compete in college) is 0.0001.P(B| Ac ) = 0.0001

• What is the probability that a high school athlete will go on to the professional sport?

Example 3Using the same probabilities from the previous problem:

A = { competes in college} P(A) = 0.05 B = { competes professionally} P(B|A) = 0.017

P(B| Ac ) = 0.0001

High School Athletes

AB

Bc

Ac

B

Bc

0.05

P(B|A) = 0.017

P(B|Ac) = 0.00010.95

0.983

0.9999

•What is the probability that a high school athlete will go on to the professional sport?• P(B) = P(A ∩ B|A) + P(Ac ∩ B|Ac) = 0.00085 + 0.000095 = 0.000945

P(A ∩ B|A) = 0.00085

P(A ∩ Bc|A) = 0.04915

P(Ac ∩ B|Ac) = 0.000095

P(Ac ∩ Bc|Ac) = 0.94991

Tree Diagrams

• Tree diagrams combine the addition and multiplication rules.

• The multiplication rule is use to reach the end of any complete branch

• The probability of any outcome is found by adding the probabilities of all branches that are part of the event.

A Simulation Problem

• Run a simulation for throwing two dice and finding the probability of rolling a sum of six. – Step 1: State the Problem or describe the

random event.– Step 2: State the assumptions.– Step 3: Assign digits to represent outcomes.– Step 4: Simulate many repetitions.– Step 5: State your conclusions.

Assigning Probability to Events1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

A 2 3 4 5 6 7 8 9 10 11 12

P(A) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 1/9 1/12 1/18 1/36

Solution – Step 1: State the Problem or describe the random event.

• Find the probability of having a sum of six after a toss of two die.– Step 2: State the assumptions.

• The toss of the dice are independent.• The outcome (the sum) are not equally likely.

– Step 3: Assign digits to represent outcomes.• Use 1-36 for the 36 combinations• Number correspondence: 1 = sum of 2; 2-3 = sum of 3; 4-6 = sum of 4; 7-10 = sum of 5; 11-15 = sum of 6

– Step 4: Simulate many repetitions.• Randint(1,36,20)→L1• L1≥11 and L1≤15→L2• Sum(L2)

– Step 5: State your conclusions• The simulation resulted in approximately 30% of the rolls having a

sum of six.

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