1. Elementary Statistics
Probability 3
Psychology
Business
Weather forecast
Elementary Statistics
Medicine
Sports
Games

2. Basic Concepts of Probability
Section 3.1
Basic Concepts of Probability

3. Probability experiment:
Important Terms
Probability experiment:
Roll a die
An action through which counts, measurements or responses are obtained
Sample space:
{ }
The set of all possible outcomes
Event:
{ Die is even }={ }
Basic Definitions are given first. Each concept is then illustrated.
A subset of the sample space.
Outcome:
{4}
The result of a single trial

4.                  Another Experiment
Probability Experiment: An action through which counts, measurements, or responses are obtained
Sample Space: The set of all possible outcomes
Event: A subset of the sample space.
Outcome: The result of a single trial
Choose a car from production line
          
A quality control example utilizing the newly defined terms
     

5. Classical (equally probable outcomes)
Types of Probability
Classical (equally probable outcomes)
Empirical
Probability blood pressure will decrease after medication
The three types of probability and an example for each. Formulas for classical and empirical probabilities are given. In classical probability all simple outcomes are equally likely.
Intuition
Probability the line will be busy

6. Tree Diagrams Start 36 outcomes Two dice are rolled. Describe the
sample space.
Start
1st roll 1 2 3 4 5 6
Illustration of a tree diagram using the roll of two dice.
2nd roll
36 outcomes

7. Sample Space 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
The sum is 4 can be successful in 3 ways. P(4) = 3/36 = 1/12 = 0.083
The sum is 11 can be successful in 2 ways. P11) = 2/36 = 118 =0.056
Find the probability the sum is 4
3/36 = 1/12 = 0.083
Find the probability the sum is 11
2/36 = 1/18 = 0.056
Find the probability the sum is 4 or 11
5/36 = 0.139

8.  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.
Sometimes it is easier to calculate the probability that an event won’t happen. Then subtract from 1 to find the probability that it will
Solution:
P(defective) = 5/12
P(not defective) = 1 - 5/12 = 7/12 = 0.583 

9. Conditional Probability and the
Section 3.2
Conditional Probability and the
Multiplication Rule

10. 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 cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective?
           
2 illustrations for conditional probability, one with dependent events the other with independent events.
          
Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) = 4/11

11. Independent Events the second die is a 4, given the first was a 4.
 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}
2 illustrations for conditional probability, one with dependent events the other with independent events. Note that knowing what happens on the first die, does not affect the probability of rolling a 4 on the second.
The conditional probability, P(B|A) = 1/6

12. 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.
Intuitive examples for pairs of independent events and pairs of dependent events
A= taking an aspirin each day
B= having a heart attack
A= being a female
B= being under 64” tall

13. Independent Events
If events A and B are independent, then P(B|A) = P(B)
Conditional Probability
Probability
 12 cars are on a production line where 5 are defective and 2 cars are selected at random.
A= first car is defective
B= second car is defective.
The probability of getting a defective car for the second car depends on whether the first was defective. The events are dependent.
Distinguish between 2 events that are independent and 2 events that 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.

14. Contingency Table
The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:
Omaha
Seattle
Miami
Total
Yes
100
150
400 No
125
130 95
350
Undecided 75
170 5
250
300
450
1000
One of the responses is selected at random. Find:
1. P(Yes)
2. P(Seattle)
3. P(Miami)
4. P(No, given Miami)
A 3 by 3 contingency table

15. Solutions 100 150 125 130 95 350 75 170 5 250 Omaha Seattle Miami
Total
Yes No
Und
300
450
400
1000
= 400 / 1000 = 0.4
= 450 / 1000 = 0.45
1. P(Yes)
2. P(Seattle)
3. P(Miami)
4. P(No, given Miami)
Notice how the sample space changes when an event is “given”.
=250 / 1000 = 0.25
= 95 / 250 = 0.38
Answers: 1) ) ) ) 0.38

16.                      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 cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective.
          
         
A = first car is defective B = second car is defective.
P(A) = 5/12
P(B|A) = 4/11
P(A and B) = 5/12 × 4/11 = 5/33=

17. When two events A and B are independent, then
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.

18. Section 3.3
The Addition Rule

19. Compare “A and B” to “A or B”
The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B).
The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B).
A and B A B A B
A or B

20. Mutually Exclusive Events
Two events, A and B are mutually exclusive, if they cannot occur in the same trial.
A = A person is under 21years old
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
Mutually exclusive
P(A and B) = 0 B
When event A occurs it excludes event B in the same trial.

21. Non-Mutually Exclusive Events
If two events can occur in the same trial, they are non-mutually exclusive.
A = A person is under 25
B = A person is a lawyer
A = A person was born in Philadelphia
B = A person watches West Wing on TV.
Non-mutually exclusive
P(A and B) ¹ 0
A and B A B

22. The Addition Rule A= the card is a king B = the card is red.
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 - 2/52
= 28/52 = 0.538

23. When events are mutually exclusive,
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)

24. Contingency Table One of the responses is selected at random. Find:
The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:
Omaha
Seattle
Miami
Total
Yes
100
150
400 No
125
130 95
350
Undecided 75
170 5
250
300
450
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
2. P(Miami and Seattle)
3. P(Miami or Yes)
4. P(Miami or Seattle)

25. Contingency Table One of the responses is selected at random. Find:
Omaha
Seattle
Miami
Total
Yes
100
150
400 No
125
130 95
350
Undecided 75
170 5
250
300
450
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
2. P(Miami and Seattle)
= 250/1000 * 150/250 = 150/1000 = 0.15
= 0

26. Contingency Table 3 P(Miami or Yes) 4. P(Miami or Seattle) Omaha
Total
Yes
100
150
400 No
125
130 95
350
Undecided 75
170 5
250
300
450
1000
250/ / /1000
=500/1000 = 0.5
3 P(Miami or Yes)
4. P(Miami or Seattle)
250/ / /1000
=700/1000 = 0.7

27. P(A and B) = P(A) *P(B|A) P(A or B) = P(A) + P(B) - P(A and B)
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.

28. Section 3.4
Counting Principles

29. 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? 2 *
Dessert 2
Soup * 3
Main
The probability of ordering a particular soup and particular main dish and a particular desert is 1/12 or 0.083
Start
= 12 meals

30. Factorials This is called n factorial and written as n!
Suppose you want to arrange n objects in order.
There are n choices for 1st place
Leaving n-1 choices for second, then n –2 choices for third place and so on until there is one choice of last place.
Using the Fundamental Counting Principle, the number of ways of arranging n objects is:
This is called n factorial and written as n!

31. The number of permutations for n objects is n!
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 (where r  n) is:
 You are required to read 5 books from a list of 8. In how many
different orders can you do so?
Emphasize that in counting permutations, order is important.
There are permutations of 8 books reading 5.

32. Combinations
A combination is an selection of r objects from a group of n objects.
The number of combinations of n objects taken r at a time is
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.
In counting combinations, order does not matter. An object is either selected or it is not.
There are 56 combinations of 8 objects taking 5.

33. 2 4 1 3 2 1 3 1 4 1 2 3 3 4 2 4 Combinations of 4 objects choosing 2
The next two slides show the difference between permutations and combinations. With 4 objects, there are 6 ways to chose 2 without regard to order. 3 4 2 4
Each of the 6 groups represents a combination

34. 2 4 1 3
Permutations of 4 objects choosing 2 1 2 1 2 1 3 1 3 1 4 1 4 2 3 2 3
If order is important, there are 12 ways of arranging 4 objects, choosing 2. If there are r members in each subset, then each subset can be arranged in r! ways. In this case there are 2 members in each subset so each subset (combination) can be arranged in 2! = 2*1 = 2 ways 3 4 3 4 2 2 4
Each of the 12 groups represents a permutation.