Transcript 6.3 Probability Models

Section 6.3
Probability Models
Statistics AP
Mrs. 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
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?
AP Statistics, Section 6.3, Part 1
3
GENERAL MULTIPLICATION RULE
The 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
Venn Diagrams: Disjoint Events
S
A
B
5
Venn Diagrams: Disjoint Events
Rule #3 (addition
rule for disjoint
events!)
P(A or B) = P(A) + P(B)
S
A
B
6
Venn Diagrams:
Non-disjoint Events
P(A or B) = P(A) + P(B) – P(A and B)
S
B
A
A and B
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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
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Example

What’s the probability of either
Claire or Alex getting promoted
P(C or A)?

P(C) = 0.7
 P(A) = 0.5
 P(C and A) = 0.3
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Example




What’s the probability of either
Claire or Alex getting promoted
P(C or A)?
P(C) = 0.7
P(A) = 0.5
P(C and A) = 0.3
A
A and C
C
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Example

What’s the probability
of either Claire or
Alex getting promoted
P(C or A)?
A
A and C
0.2
0.3
C
0.4
0.1
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Example

P(C and Ac)?

P(A and Cc)?

P(Ac and Cc)?
A
A and C
0.2
0.3
C
0.4
0.1
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Example

Are the events, Alex gets
a promotion and Claire
gets a promotion
independent? Why?
A
A and C
0.2
0.3
C
0.4
0.1
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Charts!

Woohoo!

Charts are
AWESOME!
AP Statistics, Section 6.3, Part 1
14
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
22,512
62,689
18,669
103,870
Total
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Age
18-29

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
22,512
62,689
18,669
103,870
Total

30-64
A=is young (between 18 and 29)
P(A)=
16
Age
18-29

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
22,512
62,689
18,669
103,870
Total

30-64
B=married
P(B)=
17
Age
18-29
Married

Total
43,808
8,270
59,920
13,930
7,184
751
21,865
Widowed
36
2,523
8,385
10,944
Divorced
704
9,174
1,263
11,141
22,512
62,689
18,669
103,870
Total

65 and over
7,842
Never Married

30-64
A=is young (between 18 and 29)
B=married
P(A and B)=
18
Age
18-29
Married
30-64
65 and over
Total
7,842
43,808
8,270
59,920
13,930
7,184
751
21,865
Widowed
36
2,523
8,385
10,944
Divorced
704
9,174
1,263
11,141
22,512
62,689
18,669
103,870
Never Married
Total



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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Age
18-29
Married
30-64
65 and over
Total
7,842
43,808
8,270
59,920
13,930
7,184
751
21,865
Widowed
36
2,523
8,385
10,944
Divorced
704
9,174
1,263
11,141
22,512
62,689
18,669
103,870
Never Married
Total



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”
20
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…
P(A  B)
P(A | B) 
P(B)

It is usually easier to use a tree diagram,
venn diagram, or table to solve these
problems!!!
AP Statistics, Section 6.3, Part 1
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Age
18-29
Married
30-64
65 and over
Total
7,842
43,808
8,270
59,920
13,930
7,184
751
21,865
Widowed
36
2,523
8,385
10,944
Divorced
704
9,174
1,263
11,141
22,512
62,689
18,669
103,870
Never Married
Total

A=is young (between 18 and 29)
B=married
P(A | B)=

P(B | A) =


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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
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Conditional Probabilities and
Venn Diagrams

What is the probability
of Claire being
promoted given that
Alex got promoted?
A
A and C
0.2
0.3
C
0.4
0.1
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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?
0.97
G
0.03
D
F1 Good
0.679
F1
0.7
0.3
0.98
F1 Defective 0.021
G
F2 Good 0.294
D
F2 Defective 0.006
F2
0.02
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
Given that a randomly chosen part is defective, what is the
probability that it came from factory F1?
0.97
G
0.03
D
F1 Good
0.679
F1
0.7
0.3
0.98
F1 Defective 0.021
G
F2 Good 0.294
D
F2 Defective 0.006
F2
0.02
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Summary…You should be able to:

Construct a Venn Diagram and use it to
calculate probabilities
 Particularly

Fill in a table of probabilities and use it to
calculate probabilities
 Especially

useful for nondisjoint events
useful for conditional probabilities
Construct a Tree Diagram and use it to
calculate probabilities
AP Statistics, Section 6.3, Part 1
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Summary…Formulas

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)
AP Statistics, Section 6.3, Part 1
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