Transcript 6.2 Probability Models

Daniel S. Yates
The Practice of Statistics
Third Edition
Chapter 6:
Probability and Simulation:
The Study of Randomness
6.2 Probability Models
Copyright © 2008 by W. H. Freeman & Company
Essential Questions
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What is meant by a random phenomenon?
How do you define probability in term of relative frequency?
What is a sample space?
How do you determine the number of outcomes in a sample space?
What are the five rules for assignment of probability?
What is meant by {A U B} and {A ∩ B}?
How do you compute the probability of an event given the
probabilities of outcomes that make up the event?
How do you compute the probability of an event of equally likely
outcomes?
What is meant by two events are independent?
How do you determine if two events are independent?
How do you use the multiplication rule for independent events?
Sample Space
• The sample space S of random
phenomenon is the set of all possible
outcomes.
Event
• An event is any outcome or a set of
outcomes of a random phenomenon.
• That is, an event is a subset of the sample
space
Sample Space: Flipping a Coin and
Rolling a Die
Sample
Space
Sample Space for two die
Sample Space: Rolling 2 Dice
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
Multiplication Principle
• If you can do one task n1 number of ways
and a second task in n2 number of ways,
then both tasks can be done in n1 x n2
number of ways.
Describe the Sample Space
Problem 6.30
• In each of the following situations, describe a sample
space S for the random phenomenon. In some cases
you have some freedom in specifying S, especially in
setting the largest and the smallest value of S.
a). Choose a student in your class at random. Ask how much time
that student spent studying during the past 24 hours.
Answer: S = { All numbers between 0 and 24}
b). The Physicians’ Health Study asked 11,000 physicians to take
an aspirin every other day and observed how many of them had
heart attack in a five year period.
Answer: S = { All integers from 0 to 11,000}
c). In a test of a new package design, you drop a carton of a dozen
eggs from a height of 1 foot and count the number of broken
eggs.
Answer: S = { Integers from 0 to 12}
Probability Model
• A probability model is a mathematical
description of a random phenomenon
consisting of two parts: a sample space S
and a way of assigning probabilities to
events.
Assigning Probability
The probability of any event A is
count of outcomes in A
P( A) 
count of outcomes in S
Assigning Probability to Events
Suppose we roll two die and we are interested in assigning a probability to
total number of dots showing face up on the die.
1
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
Assigning Probability to Events in S
A
2
3
4
5
P(A)
1/36 1/18 1/12 1/9
6
7
5/36 1/6
8
9
5/36 1/9
10 11 12
1/12 1/18 1/36
Additional Problems
Problem 6.36
• Suppose you select a card from a standard deck of 52
playing cards. In how many ways can the selected card
be –
a). A red card?
Ans: 26 P(red card) = 26/52 = 0.5
b). A heart?
Ans: 13 P(A heart) = 13/52 = 0.25
c). A queen and a heart?
Ans: 1 P(A queen and a heart) = 1/52 = 0.019
d). A queen or a heart?
Ans: 16 P(A queen or a heart) = 16/52 = 0.308
e). A queen that is not a heart?
Ans: 3 P(A queen that is not a heart) = 3/52 = 0.058
Probability Models Summary
Part 2
Probability Rules
•
Facts that must be true for any
assignment of probabilities.
• Facts follow the idea of probability as “the
long-run proportion of repetitions on which
an event occurs
Probability Rules: Rule 1
The probability P( A)
of any event A satisfies
0  P( A)  1
Any probability is a number between 0 and 1.
Probability Rules: Rule 2
If S is the sample space
in a probability model,
then P ( S )  1
The sum of the probabilities of all possible
outcomes must equal 1.
Probability Rules: Rule 3
Two events A and B are disjoint
(also called mutally exclusive)
if they have no outcomes in common
and so can never occur simultaneously.
If A and B are disjoint,
P ( A or B)  P( A)  P( B)
This is the addition rule for disjoint events.
If two events have no outcomes in common, the
probability that one or the other occurs is the sum of their
individual probabilities.
Probability Rules: Rule 3
If  A  B  =,
P( A  B)  P( A)  P( B)
This is the addition rule for disjoint events.
Venn Diagram for Two Disjointed
Events
Probability Rules: Rule 4
The complement of any event A is
the event that A does not occur,
c
written as A .
The complement rule states that
P( A )  1  P( A)
c
The probability that an event does not occur is 1
minus the probability that the event does occur.
Venn Diagram for the Complement
Ac of Event A
Summary of Probability Rules
Example
Marital
status:
Never
married
Married
Widowed
Divorced
Probability:
.298
.622
.005
.075
• What is P(Married)?
– P(Married)=.622
Example
Marital
status:
Never
married
Married
Widowed
Divorced
Probability:
.298
.622
.005
.075
• What is P(Never married or Divorced)?
– Since “Never married and Divorced are disjoint,
P(Never married or Divorced)= .298+.075=.373
(Addition Rule for disjoint events)
Example
Marital
status:
Never
married
Married
Widowed
Divorced
Probability:
.298
.622
.005
.075
• What is P(not Married)?
– P(not Married)= 1-.622=.378 (Complement Rule)
Benford’s Law
Example 6.15
• Benford’s Law is the distribution of first digits in
tax records, payment records, invoices, etc.
• The next slide gives the distribution for legitimate
records.
• Since crooks avoid using too many round
number and fake data by using random digits,
the illegitimate records will end up with too many
first digits 6 or greater and too few 1s and 2s.
• This distribution is handy in spotting illegitimate
records.
Example
First Digit
Probability:
1
2
3
4
5
6
7
8
9
.301 .176 .125 .097 .079 .067 .058 .051 .046
A  {first digit is 1}
B  {first digit is 6 or greater}
P ( A)  .301
P ( B )  .067  .058  .051  .046  .222
Example
First Digit
Probability:
1
2
3
4
5
6
7
8
9
.301 .176 .125 .097 .079 .067 .058 .051 .046
P( A)  .301
P( B)  .067  .058  .051  .046  .222
P( A or B)  .301  .222  .523 (Addition Rule)
Example
First Digit
Probability:
1
2
3
4
5
6
7
8
9
.301 .176 .125 .097 .079 .067 .058 .051 .046
C  {first digit is odd}
P(C )  P(1)  P(3)  P(5)  P(7)  P(9)
 .301  .125  .079  .058  .046  .609
P( B or C)  .609  .222 (Sets are not disjoint)
=P(1)  P(3)  P(5)  P(6)  P(7)  P(8)  P(9)
 .301  .125  .079  .067  .058  .051  .046  .727
Special Case – Equally Likely
Outcomes
Some random phenomenon have balance which
produces equally likely outcome, such as flipping coins
and drawing playing cards.
Most random phenomenon do not have equally likely
outcomes, so the general rule is more important.
Venn Diagrams: Disjoint Events
P( A or B)  P( A)  P( B)
S
A
B
Example: Flipping one coin. Heads or Tails can not occur at the same
time for each event. “OR”
Venn Diagrams:
Non-disjoint Events
S
B
A
A and B
Example: Flipping two coins at the same time.
Two head or two tails can exist at the same time.
The Probability of Event A and
Event B P(A and B)
This rule applies only to independent events.
Are these events Independent?
• Event A: Randomly selected person is a
man.
• Event B: Randomly selected person is
pregnant.
Proof that independent events
cannot be disjoint events
• Theorem – If A and B are both non-empty, independent events, then
A and B can not be disjoint (i.e., they have to have outcomes in
common).
• Proof ( by contradiction)
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Assume that A and B are non-empty, independent events.
Since A and B are independent,
then P(A and B) = P(A)●P(B).
Suppose A and B are disjoint,
Then P(A and B) = P( Null) = 0
Then P(A) = 0 or P(B) = 0.
This means A and B are empty sets.
But this contradicts our assumption that A and B are non-empty sets.
Therefore, we conclude that A and B are not disjoint and do intersect.
One Last Problem
Choose a person aged 19 to 25 years at random and
ask, “In the past seven days, how many times did you go
to an exercise or fitness center or work out?” Based on
a large sample survey, here is a probability model for the
answer you will get:
Days
0
1
2
3
4
5
6
7
Probability
0.68
0.05
0.07
0.08
0.05
0.04
0.01
0.02
• Is this a legitimate probability model?
• What is the probability that the person you choose worked out
either 2 or 3 days in the past seven days?
• What is the probability that the person you choose worked out
at least one day in the past seven days?