Transcript 1, 2, 3, 4, 5, 6

Chapter
3
Probability
Psychology
Weather forecast
Business
Elementary Statistics
Larson
Farber
Games
Sports
Medicine
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Important Terms
Probability experiment:
Roll a die
An action through which counts, measurements or
responses are obtained
Sample space:
{1 2 3 4 5 6}
The set of all possible outcomes
Event:
{ Die is even }={ 2 4 6 }
A subset of the sample space.
Outcome:
{4}
The result of a single trial
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Another Experiment
Probability Experiment: An action through which
counts, measurements, or responses are obtained
Choose a car from production line
Sample Space: The set of all possible outcomes
Q Q Q
Q Q Q   Q
Event: A subset of the sample space.
    
Outcome: The result of a single trial

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Types of Probability
Classical (equally probable outcomes)
Number of outcomes in event E
P(E) 
Number of outcomes in sample space
Empirical
Frequency of event E
P(E) 
Total Frequency
Probability blood pressure will decrease after medication
Subjective
Probability the line will be busy
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Two dice are rolled.
Describe the
sample space.
Tree Diagrams
Start
1st roll
1
2
3
4
5
6
1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
2nd roll
36 outcomes
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Sample Spaces and Probabilities
Two dice are rolled and the sum is noted.
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
Find the probability the sum is 4
Find the probability the sum is 11
Find the probability the sum is 4 or 11
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Complementary Events
The complement of event E is event E. E consists of all the
events in the sample space that are not in event E.
E
E
P(E´) = 1 - P(E)
 The day’s production consists of 12 cars, 5 of which are
defective. If one car is selected at random, find the
probability it is not defective.
Solution:
P(defective) = 5/12
P(not defective) = 1 - 5/12 = 7/12 = 0.583
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Conditional Probability
The probability an event B will occur, given (on the
condition) that another event A has occurred.
We write this as P(B|A) and say “probability of B, given A”.
Two planes are selected from a production line of 12
planes where 5 are defective. What is the probability the
2nd plane is defective, given the first plane was defective?
Q Q Q Q Q Q Q Q Q Q Q Q
Q Q Q Q Q Q Q Q Q
QQ
Given a defective plane has been selected, the conditional
sample space has 4 defective out of 11. So, P(B|A) = 4/11
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Conditional Probability
 Two dice are rolled, find the probability
the second die is a 4, given the first was a 4.
Original sample space: {1, 2, 3, 4, 5, 6 }
Given the first die was a 4, the conditional sample space is :
{1, 2, 3, 4, 5, 6}
The conditional probability, P(B|A) = 1/6
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Independent Events
Two events A and B are independent if the probability of
the occurrence of event B is not affected by the occurrence
(or non-occurrence) of event A.
A= Being female
B=Having type O blood
A= 1st child is a boy
B= 2nd child is a boy
Two events that are not independent are dependent.
A= taking an aspirin each day
B= having a heart attack
A= being a female
B= being under 64” tall
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Independent Events
If events A and B are independent, then P(B|A) = P(B)
Conditional Probability
Probability
12 planes are on a production line where 5 are defective and
2 planes are selected at random.
A= first plane is defective
B= second plane is defective.
The probability of getting a defective plane for the second plane
depends on whether the first was defective.
The events are dependent.
 Two dice are rolled. A= first is a 4 and B = second is a 4
P(B)= 1/6 and P(B|A) = 1/6. The events are independent.
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Contingency Table
The results of responses when a sample of adults in
3 cities were asked if they liked a new juice is:
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Yes)
5. P(Not Seattle)
2. P(Seattle)
6. P(Seattle, given yes)
3. P(Miami)
7. P(Yes, given Seattle)
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4. P(No, given Miami)
8. P(Miami, given Omaha)
Solutions
Yes
No
Und
Total
1. P(Yes)
2. P(Seattle)
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
= 400 / 1000 = 0.4
= 450 / 1000 = 0.45
3. P(Miami)
4. P(No, given Miami)
=250 / 1000 = 0.25
= 95 / 250 = 0.38
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Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38
Solutions
Omaha
Seattle
Miami
Total
Yes
100
150
150
400
No
125
130
95
350
75
170
5
250
300
450
250
1000
Undecided
Total
5. P(Not Seattle)
= 1 - 450 / 1000 = 0.55
6. P(Seattle, given yes)
= 150 / 400 = 0.375
=150 / 450 = 0.333
7. P(Yes, given Seattle)
8. P(Miami, given Omaha)
= 0 / 250 = 0
Answers: 5) 0.55 6) 0.375 7) 0.333 8) 0
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Solutions
Omaha
Seattle
Miami
Total
Yes
100
150
150
400
No
125
130
95
350
75
170
5
250
300
450
250
1000
Undecided
Total
Are events A= Seattle and B= Yes independent events?
If events are independent P(B|A) = P(B)
P(Yes) = 0.4
P(Yes |Seattle) = 150/450 = 0.333
Since 0.333  0.4 the events are NOT independent.
Are events A = Miami and B = Omaha independent events?
If events are independent P(B|A) = P(B)
P(Omaha|Miami) = 0
P(Omaha) = 0.3
Since 0  0.3 the events are NOT independent.
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Multiplication Rule
To find the probability that two events, A and B will occur in
sequence, multiply the probability A occurs by the conditional
probability B occurs, given A has occurred.
P( A and B) = P(A) × P(B|A)
 Two planes are selected from a production line of 12 where 5 are
defective. Find the probability both planes are defective.
Q Q Q Q Q Q Q Q Q Q Q Q
Q Q Q Q Q Q Q Q Q
QQ
A = first plane is defective B = second plane is defective.
P(A) = 5/12
P(B|A) = 4/11
P(A and B) = 5/12 × 4/11 = 5/33= 0.1515
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Multiplication Rule
Two dice are rolled. Find the probability
both are 4’s.
A= first die is a 4 and B= second die is a 4.
P(A) = 1/6
P(B|A) = 1/6
P(A and B) = 1/6 × 1/6 = 1/36 = 0.028
When two events A and B are independent, then
P (A and B) = P(A) × P(B)
Note for independent events P(B) and P(B|A) are the same.
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Mutually Exclusive Events
Two events, A and B are mutually exclusive, if they
cannot occur in the same trial.
A= A person is under 21
B= A person is running for the U.S. Senate
A = A person was born in Philadelphia
B = A person was born in Houston
A
B
Mutually exclusive
P(A and B) = 0
When event A occurs it excludes event B in the same trial.
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Non-Mutually Exclusive Events
If two events can occur in the same trial, they are nonmutually exclusive.
A = A person is under 25
B = A person is a lawyer
A = A person was born in Philadelphia
B = A person watches 20/20 on TV.
A and B
Non-mutually exclusive
P(A and B)  0
A
B
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The Addition Rule
The probability that one or the other of two events will
occur is:
P(A) + P(B) - P(A and B)
A card is drawn from a deck. Find the probability it

is a king or it is red.
A= the card is a king B = the card is red.


P(A) = 4/52 and P(B) = 26/52 but P( A and B) = 2/52
P(A or B) = 4/52 + 26/52 - 2/52
= 28/52 = 0.538
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The Addition Rule
A card is drawn from a deck. Find the probability
the card is a king or a 10.
A = the card is a king and B = the card is a 10.

P(A) = 4/52 and P(B) = 4/52 and P( A and B) = 0/52
P(A or B) = 4/52 + 4/52 - 0/52 = 8/52 = 0.054
When events are mutually exclusive,
P(A or B) = P(A) +P(B)
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Contingency Table
The results of responses when a sample of adults in 3 cities
was asked if they liked a new juice is:
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
4. P(Miami or Yes)
2. P(Omaha and No)
5. P(Omaha or No)
3. P(Miami and Seattle)
6. P(Miami or Seattle)
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Contingency Table
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
= 250/1000 * 150/250 = 150/1000 = 0.15
2. P(Omaha and No)
= 300/1000 * 125/300 = 125/1000 = 0.125
3. P(Miami and Seattle)
=0
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Contingency Table
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
4. P(Miami or Yes)
250/1000 + 400/1000 - 150/1000
=500/1000 = 0.5
5. P(Omaha or No)
300/1000 + 350/1000 - 125/1000
=525/1000 = 0.525
6. P(Miami or Seattle)
250/1000 + 450/1000 - 0/1000
=700/1000 = 0.7
Answers: 4) 0.5 5) 0.525 6) 0.7
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Summary
For complementary events P(E') = 1 - P(E)
Subtract the probability of the event from one.
The probability both of two events occur
P(A and B) = P(A) *P(B|A)
Multiply the probability of the first event by the conditional
probability the second event occurs, given the first occurred.
Probability at least one of two events occur
P(A or B) = P(A) + P(B) - P(A and B)
Add the simple probabilities but to prevent double counting,
don’t forget to subtract the probability of both occurring
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Fundamental Counting Principle
If one event can occur m ways and a second event can occur n ways,
the number of ways the two events can occur in sequence is m*n.
This rule can be extended for any number of events occurring in
a sequence.
 If a meal consists of 2 choices of soup, 3 main dishes and 2 desserts, how many
different meals can be selected?
Soup
Main
Dessert
Start
2
*
3
*
2
= 12
meals
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Permutations
A permutation is an ordered arrangement
The number of permutations for n objects is n!
n! = n*(n - 1)*(n -2)…..3*2*1
The number of permutations of n objects taken r at a time is
n!
( n  r )!
 You are required to read 5 books from a list of 8. In how many
different orders can you do so?
n Pr 
8!
8  7  6  5  4  3  2 1

 6720
8 P
5 
(8  5)!
3  2 1
You have 6720 permutations of 8 books reading 5.
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Combinations
A combination is an selection or r objects from a group of
n objects.
The number of combinations of n objects taken r at a time is
n
Cr
n!

( n  r )! r!
 You are required to read 5 books from a list of 8. In how many
different ways can you choose the books if order does not matter.
8 C5 
8!
8  7  6  5  4  3  2 1

 56
(8  5)!5!
3  2 1  5  4  3  2 1
There are 56 combinations of 8 objects taking 5.
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Permutations of 4 objects taking 2
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Each of the 12 groups represents a permutation.
Combinations of 4 objects taking 2
Each of the 6 groups represents a combination
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Distinguishable Permutations
• What if when we are selecting items, and order
matters and some are the same?
n= the total number of items
n k= the number of each choice
n!
n1!n2 !n3!nk !
Ex- How many ways can we order the letters of illinois?
8!
 3360
3!2!1!1!1!
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