2. Overview 5.1 Introducing Probability 5.2 Combining Events
5.3 Conditional Probability
5.4 Counting Methods

3. 5.1 Introducing Probability
Objectives:
By the end of this section, I will be
able to…
Understand the meaning of an experiment, an outcome, an event, and a sample space.
Describe the classical method of assigning probability.
Explain the Law of Large Numbers and the relative frequency method of assigning probability.

4. Probability
Defined as the long-term proportion of times the outcome occurs
Building Blocks of Probability
Experiment - any activity for which the outcome is uncertain
Outcome - the result of a single performance of an experiment
Sample space (S) - collection of all possible outcomes
Event - collection of outcomes from the sample space

5. Rules of Probability
The probability P(E) for any event E is always between 0 and 1, inclusive.
That is, 0 ≤ P(E ) ≤ 1.
Law of Total Probability:
For any experiment, the sum of all the
outcome probabilities in the sample space must equal 1.

6. Classical Method of Assigning Probabilities
Let N(E) and N(S) denote the number of outcomes in event E and the sample space S, respectively.
If the experiment has equally likely outcomes, then the probability of event E is then

7. Example 5.1 - Probability of drawing an ace
Find the probability of drawing an ace when drawing a single card at random from a deck of cards.

8. Example 5.1 continued Solution
The sample space for the experiment where a subject chooses a single card at random from a deck of cards is given in Figure 5.1.
FIGURE 5.1 Sample space for drawing a card at random from a deck of cards.

9. Example 5.1 continued
If a card is chosen at random, then each card has the same chance of being drawn.
Since each card is equally likely to be drawn, we can use the classical method to assign probabilities.
There are 52 outcomes in this sample space, so N(S) = 52.

10. Example 5.1 continued Let E be the event that an ace is drawn.
Event E consists of the four aces
{A♥, A♦, A♣, A♠}, so N(E ) = 4.
Therefore, the probability of drawing
an ace is

11. Tree Diagram
Device used to count the number of outcomes of an experiment
Graphical display to visualize a multistage experiment
Helps us to construct the sample space for
a multistage experiment

12. FIGURE 5.2 Tree diagram for the experiment of tossing
a fair coin twice.

13. Law of Large Numbers
As the number of times that an experiment is repeated increases, the relative frequency (proportion) of a particular outcome tends to approach the probability of the outcome.
For quantitative data, as the number of times that an experiment is repeated increases, the mean of the outcomes tends to approach the population mean.
For categorical (qualitative) data, as the number of times that an experiment is repeated increases, the proportion of times a particular outcome occurs tends to approach the population proportion.

14. Relative Frequency Method
The probability of event E is approximately equal to the relative frequency of event E.
Also known as the empirical method

15. Example 5.7 - Teen bloggers
A recent study found that 35% of all online teen girls are bloggers, compared to 20% of online teen boys. Suppose that the 35% came from a random sample of 100 teen girls who use the Internet, 35 of whom are bloggers. If we choose one teen girl at random, find the probability that she is a blogger.

16. Example 5.7 continued Solution Define the event.
B: The online girl is a blogger.
We use the relative frequency method to find the probability of event B:

17. Subjective Method
Should be used when the event is not (even theoretically) repeatable
The assignment of a probability value to an outcome based on personal judgment

18. Summary
Section 5.1 introduces the building blocks of probability, including the concepts of probability, outcome, experiment, and sample space.
Probabilities always take values between and 1, where 0 means that the outcome cannot occur and 1 means that the outcome is certain.

19. Summary
The classical method of assigning probability is used if all outcomes are equally likely.
The classical method states that the probability of an event A equals the number of outcomes in A divided by the number of outcomes in the sample space.

20. Summary
The Law of Large Numbers states that, as an experiment is repeated many times, the relative frequency (proportion) of a particular outcome tends to approach the probability of the outcome.
The relative frequency method of assigning probability uses prior knowledge about the relative frequency of an outcome.
The subjective method of assigning probability is used when the other methods are not applicable.

21. 5.2 Combining Events Objectives: By the end of this section, I will be
able to…
Understand how to combine events using complement, union, and intersection.
Apply the Addition Rule to events in general and to mutually exclusive events in particular.

22. Complement of A Symbolized by AC Collection of outcomes not in event A
Complement comes from the word “to complete”
Any event and its complement together make up the complete sample space

23. Example 5.10 - Finding the probability of the complement of an event
If A is the event “observing a sum of 4 when the two fair dice are rolled,” then your roommate is interested in the probability of AC, the event that a 4 is not rolled. Find the probability that your roommate does not roll a 4.

24. Example 5.10 continued Solution Which outcomes belong to AC?
By the definition, AC is all the outcomes in the sample space that do not belong in A.
There are the following outcomes in A:
{(3,1)(2,2)(1,3)}.

25. Example 5.10 continued
Figure 5.10 shows all the outcomes except the outcomes from A in the two-dice sample space.

26. Example 5.10 continued
There are 33 outcomes in AC and 36 outcomes in the sample space.
The classical probability method then gives the probability of not rolling a 4 to be
The probability is high that, on this roll at least, your roommate will not land on Boardwalk.

27. Probabilities for Complements
For any event A and its complement AC,
P(A) + P(AC) = 1.
P(A) = 1 - P(AC)
P(AC ) = 1 - P(A)

28. Union of Events
The event representing all the outcomes that belong to A or B or both
Denoted as A B
Associated with “or”
FIGURE 5.11 Union of events A, B.

29. Intersection of Events
The event representing all the outcomes that belong to both A and B
Denoted as A B
Associated with “and”
FIGURE 5.12 Intersection of Events A, B.

30. Example 5.11 - Union and intersection
Let our experiment be to draw a single card at random from a deck of cards. Define the following events: A: The card drawn is an ace. H: The card drawn is a heart. a. Find A H. b. Find A H.

31. Example 5.11 continued Solution
a. A H is event containing all outcomes are either aces or hearts or both (the ace of hearts).
A H is all cards shown in Figure 5.13.
Figure 5.13

32. Example 5.11 continued Solution
b. The intersection of A and H is the event containing the outcomes that are common to both A and H.
There is only one such outcome: the ace of hearts (see Figure 5.13).

33. Addition Rule
The probability that either one event or another event may occur
Count the probabilities of the outcomes in A
Add the probabilities of the outcomes in B
Subtract the probability of the intersection (overlap)

34. Example 5.12 - Addition Rule applied to a deck of cards
Suppose you pay $1 to play the following game. You choose one card at random from a deck of 52 cards, and you will win $3 if the card is either an ace or a heart. Find the probability of winning this game.

35. Example 5.12 continued Solution
Using the same events defined in Example 5.11, we find P(A or H) = P(A H).
By the Addition Rule, we know that
P(A H) = P(A) + P(H ) - P(A H)
There are 4 aces in a deck of 52 cards, so by the classical method (equally likely outcomes), P(A) = 4/52.

36. Example 5.12 continued
There are 13 hearts in a deck of 52 cards, so P(H ) = 13/52.
From Example 5.11, we know that A H represents the ace of hearts.
Since each card is equally likely to be drawn, then P(ace of hearts) = P(A H) = 1/52.
P(A H) = P(A) + P(H ) - P(A H)

37. Mutually Exclusive Events
Also known as disjoint events
Events having no outcomes in common
Addition Rule for Mutually Exclusive Events
If A and B are mutually exclusive events, P(A B) = P(A) + P(B).
FIGURE 5.14
Mutually exclusive
events.

38. Summary
Combinations of events may be formed using the concepts of complement, union, and intersection.
The Addition Rule provides the probability of Event A or Event B to be the sum of their two probabilities minus the probability of their intersection.
Mutually exclusive events have no outcomes in common.

39. 5.3 Conditional Probability
Objectives:
By the end of this section, I will be
able to…
Calculate conditional probabilities.
Explain independent and dependent events.
Solve problems using the Multiplication Rule.
Recognize the difference between sampling with replacement and sampling without replacement.

40. Conditional probability
For two related events A and B, the probability of B given A
Denoted P(B|A)
FIGURE 5.16 How conditional probability works.

41. Calculating Conditional Probability
The conditional probability that B will occur, given that event A has already taken place, equals

42. Independent Events
If the occurrence of an event does not affect the probability of a second event, then the two events are independent
Events A and B are independent if
P(A| B) = P(A) or if P(B| A) = P(B)
Otherwise the events are said to be dependent.

43. Strategy for Determining Whether Two Events Are Independent
Find P(B)
Find P(B|A)
Compare the two probabilities.
If they are equal, then A and B are independent events.
Otherwise, A and B are dependent events.

44. Example 5.17 - Are successive coin tosses independent?
Let our experiment be tossing a fair coin twice. Determine whether the following events are independent. A: Heads is observed on the first toss. B: Heads is observed on the second toss.

45. Example 5.17 continued Solution Tree diagram for this experiment
Figure 5.17

46. Example 5.17 continued Step 1:
To find P(B), the probability of observing heads on the second toss, we concentrate on the second flip of the experiment.
At Stage 2, there are four branches, two of which are labeled “Heads.”
Therefore, since each outcome is equally likely, the probability of observing heads on the second toss is P(B) = 2/4 = 1/2.

47. Example 5.17 continued Step 2: Next we find P(B | A).
Again consider Stage 2, but this time restrict your attention to the two “upper branches.”
One upper branch is labeled “Heads.”
Therefore, since each outcome is equally likely, P(B|A) = 1/2.

48. Example 5.17 continued Step 3:
Finally, we have P(B) = P(B|A), and we conclude that events A and B are independent events.
That is, successive coin tosses are independent.

49. Multiplication Rule
Used to find probabilities of intersections of events
P(A B) = P(A) P(B|A )
or equivalently
P(A B) = P(B) P(A|B)

50. Multiplication Rule for Two Independent Events
If A and B are any two independent events, P(A B) = P(A) P(B).

51. Example 5.20 - Multiplication Rule for Independent Events
Metropolitan Washington, D.C., has the highest proportion of female top-level Executives in the United States: 27%. We take a random sample of two top-level Executives and find the probability that both are female.

52. Example 5.20 continued Solution Define the following events:
A: First top-level executive is female.
B: Second top-level executive is female.

53. Example 5.20 continued
Because we are choosing the executives at random, it makes sense to assume that the events are independent.
Outcome of choosing the first executive will not affect the outcome of choosing the second executive.

54. Example 5.20 continued Using the Multiplication Rule for Independent
Events
P(A B) = P(A) P(B)
= (0.27)(0.27) =

55. Alternative Method for Determining Independence
If P(A) P(B) = P(A B),
then events A and B are independent.
If P(A) P(B) ≠ P(A B),
then events A and B are dependent.

56. Multiplication Rule for n Independent Events
If A, B, C, . . .
are independent events, then
P(A B C ) = P(A) P(B) P(C ) . . .

57. Sampling Sampling with replacement
the randomly selected unit is returned to
the population after being selected
It is possible for the same unit to be sampled more than once
Successive draws can be considered independent
Sampling without replacement
the randomly selected unit is not
returned to the population after being selected
It is not possible for the same unit to be sampled more than once
Successive draws should be considered dependent

58. Summary
Section 5.3 discusses conditional probability P(B | A), the probability of an event B given that an event A has occurred.
We can compare P(B | A) to P(B) to determine whether the events A and B are independent.
Events are independent if the occurrence of one event does not affect the probability that the other event will occur.

59. Summary
The Multiplication Rule for Independent Events is the product of the individual probabilities.
Sampling with replacement is associated with independence, while sampling without replacement means that the events are not independent.

60. 5.4 Counting Methods Objectives: By the end of this section, I will be
able to…
Apply the Multiplication Rule for Counting to solve certain counting problems.
Use permutations and combinations to solve certain counting problems.
Compute probabilities using combinations.

61. Multiplication Rule for Counting
Suppose an activity consists of a series of events in which there are a possible outcomes for the first event, b possible outcomes for the second event, c possible outcomes for the third event, and so on.
Then the total number of different possible outcomes for the series of events is: a · b · c · …

62. Example 5.28 - Counting with repetition: Famous initials
Some Americans in history are uniquely identified by their initials. For example, “JFK” stands for John Fitzgerald Kennedy, and “FDR” stands for Franklin Delano Roosevelt. How many different possible sets of initials are there for people with a first, middle, and last name?

63. Example 5.28 continued Solution
Selecting three initials is an activity consisting of three events.
Note that a particular letter may be repeated, as in “AAM” for A. A. Milne, author of Winnie the Pooh.

64. Example 5.28 continued
Then there are a = 26 ways to choose the first initial, b = 26 ways to choose the second initial, and c = 26 ways to choose the third initial.
Thus, by the Multiplication Rule for Counting, the total number of different sets of initials is
26 · 26 · 26 = 17,576

65. Example 5.29 - Counting without repetition: Intramural singles tennis
A local college has an intramural singles tennis league with five players, Ryan, Megan, Nicole, Justin, and Kyle. The college presents a trophy to the top three players in the league. How many different possible sets of three trophy winners are there?

66. Example 5.29 continued Solution
The major difference between Examples and 5.29 is that in Example 5.29 there can be no repetition.
Ryan cannot finish in first place and second place.

67. Example 5.29 continued Solution
Five possible players could finish in first place, so a = 5.
Only four players left, so b = 4.
Only three players, giving c = 3.

68. Example 5.29 continued Solution
By the Multiplication Rule for Counting, the number of different possible sets of trophy winners is
5 · 4 · 3 = 60

69. Factorial symbol
For any integer n ≥ 0, the factorial symbol n! is defined as follows:
0! = 1
1! = 1
n! = n(n - 1)(n - 2) · · · 3 · 2 · 1

70. Combinations An arrangement of items in which
r items are chosen from n distinct items.
repetition of items is not allowed.
the order of the items is not important.
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula:

71. Example 5.37 - Permutations with nondistinct items
How many distinct strings of letters can we make by using all the letters in the word STATISTICS?

72. Example 5.37 continued Solution
Each string will be ten letters long and include 3 S’s, 3 T’s, 2 I’s, 1 A, and 1 C.
The ten positions shown here need to be filled.
___ ___ ___ ___ ___ ___ ___ ___ ___ ___

73. Example 5.37 continued The string-forming process is as follows:
Step 1 Choose the positions for the three S’s.
Step 2 Choose the positions for the three T’s.
Step 3 Choose the positions for the two I’s.
Step 4 Choose the position for the one A.
Step 5 Choose the position for the one C.

74. Example 5.37 continued 10C3 ways to place the three S’s in Step 1.
7C3 positions for the three T’s.
4C2 ways to place the two I’s.
2C1 ways to position the A.
1C1 way to place the C.

75. Example 5.37 continued Calculate the number of distinct letter strings
There are 50,400 distinct strings of letters that can be made using the letters in the word STATISTICS.

76. Example 5.41 - Florida Lotto
You can win the jackpot in the Florida Lotto by correctly choosing all 6 winning numbers out of the numbers 1–53. a. What is the number of ways of winning the jackpot by choosing all 6 winning numbers? b. What is the number of outcomes in this sample space? c. If you buy a single ticket for $1, what is your probability of winning the jackpot? d. If you mortgage your house and buy 500,000 tickets, what is your probability of winning the jackpot (assuming that all the tickets are different)?

77. Example 5.41 continued Solution a. Number of ways of winning the
jackpot by correctly choosing all 6 numbers is
N(Jackpot) = 6C6 · 47C0 = 1 · 1 = 1

78. Example 5.41 continued
Solution
b. The size of the sample space is

79. Example 5.41 continued Solution
c. If you buy a single ticket, your probability of winning the jackpot is given by

80. Example 5.41 continued Solution
d. If you buy 500,000 tickets and they are all unique,

81. Example 5.41 continued
Since unique tickets are mutually exclusive, use Addition Rule for Mutually Exclusive Events to add probabilities.
After mortgaging your $500,000 house and buying lottery tickets with the proceeds, there is a better than 97% probability that you will not win the lottery.

82. Summary
The Multiplication Rule for Counting provides the total number of different possible outcomes for a series of events.
A permutation nPr is an arrangement in which
• r items are chosen from n distinct items.
• repetition of items is not allowed.
• the order of the items is important.
In a permutation, order is important.

83. Summary
In a combination, order does not matter. A combination nCr is an arrangement in which • r items are chosen from n distinct items. • repetition of items is not allowed. • the order of the items is not important.

84. Summary
Step 1 : Confirm that the desired probability involves a combination. Step 2: Find N(E), the number of outcomes in event E. Step 3: Find N(S), the number of outcomes in the sample space. Step 4: Assuming that each possible combination is equally likely