section 6.3 probability models

Section 6.3Probability Models

Statistics APMrs. Skaff

Today you will learn how to… Construct Venn Diagrams, Tables, and

Tree Diagrams and use them to calculate probabilities

Calculate probabilities for nondisjoint events

Modify the multiplication rule to accommodate non-independent events

Calculate conditional probabilities

AP Statistics, Section 6.3, Part 1 2

Non-Independent Events

AP Statistics, Section 6.3, Part 1 3

You draw a card from a deck and then draw another one without replacing the first card. What is the probability that you draw a diamond and then another diamond?

GENERAL MULTIPLICATION RULEThe joint probability that both of two events

A and B happen together can be found by

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

Here P(B|A) is the conditional probability that B occurs given the information that A occurs.

4

5

Venn Diagrams: Disjoint Events

AB

S

6

Venn Diagrams: Disjoint Events

AB

S

Rule #3 (addition rule for disjoint events!)

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

7

Venn Diagrams: Non-disjoint Events

A

B

S

A and B

P(A or B) = P(A) + P(B) – P(A and B)

8

Example Claire and Alex are awaiting the decision

about a promotion. Claire guesses her probability of her getting a promotion at .7 and Alex’s probability at .5. Claire also thinks the probability of both getting promoted is .3

9

Example What’s the probability of either

Claire or Alex getting promotedP(C or A)? P(C) = 0.7 P(A) = 0.5 P(C and A) = 0.3

10

Example What’s the probability of either

Claire or Alex getting promotedP(C or A)?

P(C) = 0.7 P(A) = 0.5 P(C and A) = 0.3

AC

A and C

11

Example What’s the probability

of either Claire or Alex getting promotedP(C or A)? A

0.2

C0.4A and C

0.3

0.1

12

Example P(C and Ac)?

P(A and Cc)?

P(Ac and Cc)?A0.2

C0.4A and C

0.3

0.1

13

Example Are the events, Alex gets

a promotion and Claire gets a promotion independent? Why? A

0.2

C0.4A and C

0.3

0.1

Charts!

Woohoo! Charts are AWESOME!

AP Statistics, Section 6.3, Part 1 14

15

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

16

Age18-29 30-64 65 and over Total

Married 7,842 43,808 8,270 59,920Never Married 13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944Divorced 704 9,174 1,263 11,141Total 22,512 62,689 18,669 103,870

A=is young (between 18 and 29) P(A)=

17

Age18-29 30-64 65 and over Total

Married 7,842 43,808 8,270 59,920Never Married 13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944Divorced 704 9,174 1,263 11,141Total 22,512 62,689 18,669 103,870

B=married P(B)=

18

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 10,944Divorced 704 9,174 1,263 11,141Total 22,512 62,689 18,669 103,870

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

19

Age18-29 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 10,944Divorced 704 9,174 1,263 11,141Total 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”)

The probability that a a person is young, given that they are married

This is known as a “conditional probability”

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Age18-29 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 10,944Divorced 704 9,174 1,263 11,141Total 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”)

The probability that a a person is young, given that they are married

This is known as a “conditional probability”

Conditional Probabilities Conditional probability measures the probability of an

event A occurring given that B has already occurred. There is a formula for this in your packets. Sometimes it can be confusing for solving real-life problems…

It is usually easier to use a tree diagram, venn diagram, or table to solve these problems!!!

AP Statistics, Section 6.3, Part 1 21

P(B)B)P(AB)|P(A

22

Age18-29 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 10,944Divorced 704 9,174 1,263 11,141Total 22,512 62,689 18,669 103,870

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

P(B | A) =

Conditional Probabilities and Tables Bag A contains 5 blue and 4 green marbles. Bag B contains 3

yellow, 4 blue, and 2 green marbles. Given you have a green marble, what is the probability it came from Bag A?

AP Statistics, Section 6.3, Part 1 23

24

Conditional Probabilities and Venn Diagrams

What is the probability of Claire being promoted given that Alex got promoted?

A0.2

C0.4A and C

0.3

0.1

Probabilities with Tree Diagrams Example: A videocassette recorder (VCR) manufacturer

receives 70% of his parts from factory F1 and the rest from factory F2. Suppose 3% of the output from F1 are defective, while only 2% of the output from F2 are defective. What is the probability the part is defective?

25

F1

F2

G

D

G

D

F1 Good

F1 Defective

F2 Good

F2 Defective

0.7

0.3

Example: A videocassette recorder (VCR) manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose 3% of the output from F1 are defective, while only 2% of the output from F2 are defective. What is the probability the part is defective?

Example: A videocassette recorder (VCR) manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose 3% of the output from F1 are defective, while only 2% of the output from F2 are defective. What is the probability the part is defective?

0.97

0.03

Example: A videocassette recorder (VCR) manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose 3% of the output from F1 are defective, while only 2% of the output from F2 are defective. What is the probability the part is defective?

0.98

0.02

0.679

0.021

0.294

0.006

Example: A videocassette recorder (VCR) manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose 3% of the output from F1 are defective, while only 2% of the output from F2 are defective. What is the probability the part is defective?

Given that a randomly chosen part is defective, what is the probability that it came from factory F1?

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F1

F2

G

D

G

D

F1 Good

F1 Defective

F2 Good

F2 Defective

0.7

0.3

0.97

0.03

0.98

0.02

0.679

0.021

0.294

0.006

Summary…You should be able to:

AP Statistics, Section 6.3, Part 1 27

Construct a Venn Diagram and use it to calculate probabilitiesParticularly useful for nondisjoint events

Fill in a table of probabilities and use it to calculate probabilitiesEspecially useful for conditional probabilities

Construct a Tree Diagram and use it to calculate probabilities

Summary…Formulas

AP Statistics, Section 6.3, Part 1 28

Probability of nondisjoint events

( ) ( ) ( ) ( )P A B P A P B P A B Multiplication Rule

P(A and B) = P(A)P(B|A) Conditional Probabilities

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