Ch5 Probability
Probability is a measure of the likelihood of a random phenomenon or chance behavior. Probability describes
the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty.
If we flip a coin 100 times and compute the proportion of heads observed after each toss of the coin, what will
the proportion approach?
The Law of Large Numbers
As the number of repetitions of a probability experiment increases, the proportion with which a certain
outcome is observed gets closer to the probability of the outcome.
Ch 5.1 Probability Rule
Objective A: Sample Spaces and Events
Experiment – any activity that leads to well-defined results called outcomes.
Outcome – the result of a single trial of a probability experiment.
Sample space, S – the set of all possible outcomes of a probability experiment.
Event, E – a subset of the sample space
Simple event, ei – an event with one outcome is called a simple event.
Compound event – consists of two or more outcomes.
Example 1: A die is tossed one time.
(a) List the elements of the sample space S.
S = {1, 2, 3, 4, 5, 6}
(b) List the elements of the event consisting of a number that is greater than 4.
E = {5, 6}
Example 2: A coin is tossed twice. List the elements of the sample space S, and list the elements of the event
consisting of at least one head.
HH
H
H
T
S = {HH, HT, TH, TT}
TT
T
H
T
E = {HT, TH, HH}
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Objective B: Requirements for Probabilities
1. Each probability must lie on between 0 and 1. (0 ≤ P (E) ≤ 1)
2. The sum of the probabilities for all simple events in S equals 1. (∑ 𝑃(𝑒𝑖 ) = 1)
If an event is impossible, the probability of the event is 0.
If an event is a certainty, the probability of the event is 1.
An unusual event is an event that has a low probability of occurring. Typically, an event with a
probability less than 0.05 is considered as unusual.
Probabilities should be expressed as reduced fractions or rounded to three decimal places.
Example 1: A probability experiment is conducted. Which of these can be considered a probability of an
outcome?
(a) 2/5
(b) -0.28
(c) 1.09
Yes
No
No
(0 ≤ P (E) ≤ 1)
(0 ≤ P (E) ≤ 1)
Example 2: Why is the following not a probability model?
Color
Red
Green
Yellow
Probability
0.28
0.56
0.37
Condition 1: (0 ≤ P (E) ≤ 1)
Condition 2: (∑ 𝑃(𝑒𝑖 ) = 1)
Check: 0.28 + 0.56 + 0.37 = 1.21 ≠ 1
Condition 2 was not met.
Example 3: Given: 𝑆 = {𝑒1 , 𝑒2 , 𝑒3 , 𝑒4 }
𝑃(𝑒1 ) = 𝑃(𝑒2 ) = 0.2 and 𝑃(𝑒3 ) = 0.5
Find: 𝑃(𝑒4 )
Condition 1: (0 ≤ P (E) ≤ 1)
Condition 2: (∑ 𝑃(𝑒𝑖 ) = 1)
P(𝑒1 ) + P(𝑒2 ) + P(𝑒31 )+ P(𝑒4 ) = 1
0.2 + 0.2+ 0.5 + P(𝑒4 ) = 1
0.9 + P(𝑒4 ) = 1
P(𝑒4 ) = 1- 0.9
P(𝑒4 ) = 0.1
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Objective C:Calculating Probabilities P( E ) 
N (E)
N (S )
Example 1: The age distribution of employees for this college is shown below:
Age
Under 20
20 – 29
30 – 39
40 – 49
50 and over
Number of Employees
25
48
32
15
10
∑ 𝑓 = 130
If an employee is selected at random, find the probability that he or she is in the following
age groups
(a) Between 30 and 39 years of age
32
16
= 65or 0.246 (3 decimal places)
130
(b) Under 20 or over 49 years of age
25+10
130
35
7
= 130 = 26 or 0.269 (3 decimal places)
Example 2: Let the sample space be S  1, 2,3, 4,5,6,7,8,9,10  . Suppose the outcomes are equally likely.
(a) Compute the probability of the event F  5,9 .
N (F) = 2
N (S) = 10
𝑁 (𝐹)
2
1
P (F) = 𝑁 (𝑆) = 10 = 5 or 0.2
(b) Compute the probability of the event E = "an odd number."
E = {1, 3, 5, 7, 9} N (E) = 5
𝑁 (𝐸)
5
1
P (E) = 𝑁 (𝑆) = 10 = 2 or 0.5
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Example 3: Two dice are tossed. Find the probability that the sum of two dice is greater than 8?
N (S) = 36
N (E) = 10
E = {(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)}
𝑁 (𝐸)
10
P(E) = 𝑁 (𝑆) = 36 =
5
18
𝑜𝑟 ≈ 0.278
Example 3: If one card is drawn from a deck, find the probability of getting
(a) a club
P (a club) =
𝑁 (𝑎 𝑐𝑙𝑢𝑏)
𝑁 (𝑆)
13
1
= 52 = 4 or 0.25
4
(b) a 4 and a club
P (a 4 and a club) =
𝑁 (𝑎 4 𝑎𝑛𝑑 𝑎 𝑐𝑙𝑢𝑏)
𝑁 (𝑆)
1
= 52 ≈ 0.019
Example 4:Three equally qualified runners, Mark, Bill, and Alan, run a 100-meter sprint, and the order of finish
is recorded.
(a) Give a sample space S.
B
A
A
B
M
A
A
M
M
B
B
M
M
S = {MBA, MAB, BMA, BAM, AMB, ABM}
B
A
(b) What is the probability that Mark will finish last?
E = {BAM, ABM}
𝑁 (𝐸)
2
1
P (E) = 𝑁 (𝑆) = 6 = 3 or ≈ 0.333
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Ch5.2 The Addition Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive) Events
Event A and B are disjoint (mutually exclusive) if they have no outcomes in common.
Addition Rule for Disjoint Events
If E and F are disjoint events, then P( E or F )  P( E )  P( F ) .
Example 1:
A standard deck of cards contains 52 cards. One card is randomly selected
from the deck. Compute the probability of randomly selecting a two or three
from a deck of cards.
4
4
8
P (2 or 3) = P (2) + P (3) = 52 + 52 = 52 = 13 or ≈ 0.154
Objective B: General Addition Rule
The General Addition Rule
For any two events E and F, P( E or F )  P ( E )  P ( F )  P ( E and F ) .
Example 1:
A standard deck of cards contains 52 cards. One card is randomly selected from
the deck. Compute the probability of randomly selecting a two or club from a deck
of cards.
6
P (2 or club) = P (2) + P(club) – P(2 and club)
4
13
1
16
4
= 52 + 52 - 52 = 52 = 13 or ≈ 0.308
Objective C: Complement Rule
Complement Rule
If E represents any event and E c represents the complement of E , then P( E C )  1  P( E ) .
Example 1: The chance of raining tomorrow is 70%. What is the probability that it will not rain tomorrow?
P (not raining) = 1 – P (raining)
= 1- 0.7 = 0.3
Example 2: In a large department store, there are 2 managers, 4 department heads, 16 clerks,
and 4 stock persons. If a person is selected at random,
(a) find the probability that the person is a clerk or a manager;
P (clerk or manager) = P (clerk) +P (manager) –P (clerk and manager)
16
2
0
18
9
= 26 + 26 - 26 = 26 = 13
(b) find the probability that the person is not a clerk.
7
P (not a clerk)= 1-P (clerk)
16 26 16
10
5
= 1- 26 = 26 - 26 = 26 = 13
Example 3: A probability experiment is conducted in which the sample space of the experiment
is S  1, 2,3, 4,5,6,7,8,9,10,11,12 .
Let event E  2,3,5,6,7 , event F  5,6,7,8 , and event G  9,11
(a) List the outcome in E and F . Are E and F mutually exclusive?
E = {2, 3, 5, 6, 7}
F = {5, 6, 7, 8}
Since there are common elements of 5, 6, and 7 between E and F,
E and F are not mutually exclusive.
(b) Are F and G mutually exclusive? Explain.
Yes, there is no common element between F and G.
(c) List the outcome in E or F . Find P( E or F ) by counting the
number of outcomes in E or F .
E or F = {2, 3, 5, 6, 7, 8}
N (E or F) = 6
N (S)= 12
6
1
P (E or F) = 12 = 2 or 0.5
(d) Determine P( E or F ) using the General Addition Rule.
P( E or F) = P (E) + P (F) – P (E and F)
5
4
3
6
1
= 12 + 12 - 12 = 12 = 2 or 0.5
(e) List the outcomes in E C . Find P( E C ) by counting the number of outcomes in E C .
𝐸 𝐶 = {1, 4, 8, 9, 10, 11, 12}
P (𝐸 𝐶 )=
𝑁(𝐸 𝐶 )
𝑁 (𝑆)
7
= 12
(f) Determine P( E C ) using the Complement Rule.
P (𝐸 𝐶 )= 1- P (E)
8
5
12
5
7
= 1 - 12 = 12 - 12 = 12
Objective D: Contingency Table
A contingency table relates two categories of data. It is also called a two-way table which consists of
a row variable and a column variable. Each box inside the table is called a cell.
Example 1: In a certain geographic region, newspapers are classified as being published daily morning,
daily evening, and weekly. Some have a comics section and other do not.
The distribution is shown here.
Have comics
(M)
Section
Morning
(E)
Evening
(W)
Weekly
(CY)
Yes
2
3
1
6
(CN)
No
3
4
2
9
5
7
3
15
If a newspaper is selected at random, find these probabilities.
(a) The newspaper is a weekly publication.
P(W) =
𝑁(𝑊)
𝑁(𝑆)
=
3
15
=
1
5
(b) The newspaper is a daily morning publication or has comics.
P (M or CY) = P (M) + P (CY) – P (M and CY)
5
6
2
9
3
= 15 + 15 - 15 = 15 = 5
Ch 5.3
Independence and the Multiplication Rule
Objective A: Independent Events
Two events are independent if the occurrence of event E does not affect the probability of event F .
Two events are dependent if the occurrence of event E affects the probability of event F .
Example 1: Determine whether the events E and F are independent or dependent. Justify
your answer.
(a)
E : The battery in your cell phone is dead.
F : The battery in your calculator is dead.
Independent
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(b)
E : You are late to class.
F :Your car runs out of gas.
Dependent
Objective B: Multiplication Rule for Independent Events
If E and F are independent events, then P( E and F )  P( E )  P( F )
Example 1: If 36% of college students are underweight, find the probability that if
three college students are selected at random, all will be underweight.
Independent case
P (1st underweight and 2nd underweight and 3rd underweight)
P (1st underweight) * P (2nd underweight) * P (3rd underweight)
= 0.36*0.36*0.36 ≈ 0.047
Example 2: If 25% of U.S. federal prison inmates are not U.S. citizens, find the probability that
two randomly selected federal prison inmates will be U.S. citizens.
P (1st U. S. citizen and 2nd U. S. citizen)
= P (1st U. S. citizen) * P (2nd U. S. citizen)
= 0.75 * 0.75 ≈ 0.563
Objective C: At-Least Probabilities
Probabilities involving the phrase “at least” typically use the Complement Rule. The phrase at
least means “greater than or equal to.” For example, a person must be at least 17 years old to
see an R-rated movie.
Example 1: If you make random guesses for four multiple-choice test questions
(each with five possible answers), what is the probability of getting at least one correct?
Direct method:
P (at least one correct)
= P (one correct) + P (two correct) + P (three correct) + P (four correct)
Indirect method:
10
What is the opposite of at least one correct? None correct.
4
4
4
4
4
5
5
5
5
5
P (none correct) = * * * = ( )4
P (at least one correct) = 1- P (none correct)
4
= 1- (5)4 ≈ 0.590
Example 2: According to the Department of Health and Human Services, 30% of 18- to 25-year-olds have
some form of mental illness.
(a) What is the probability two randomly selected 18- to 25-year-olds have some form of
mental illness?
P (1st with some form of mental illness and 2nd with some form of mental illness)
= P (1st with some form of mental illness) * P (2nd with some form of mental illness)
= 0.30*0.30 = 0.09
(b) What is the probability six randomly selected 18- to 25-year-olds have some form of
mental illness?
P (1st with mental illness and 2nd with mental illness and 3rd with mental illness and 4th
with mental illness and 5th with mental illness and 6th with mental illness)
= P (1st with mental illness) * P (2nd with mental illness) * P (3rd with mental illness) *
P (4th with mental illness) * P (5th with mental illness) * P (6th with mental illness)
= 0.30 * 0.30 * 0.30 * 0.30 * 0.30 * 0.30 = 0.000729
(c) What is the probability at least one of six randomly selected 18- to 25-year-olds has some
form of mental illness?
30% of 18- to 25-year-olds have some form of mental illness therefore 1-30% do not have
some form of mental illness. 1-0.30 = 0.70
P (at least one of 6 has mental illness)
= 1 – P (none with mental illness)
= 1 - 0.706
= 1-0.117649
= 0.882351
≈ 0.882
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(d) Would it be unusual that among four randomly selected 18- to 25-year-olds, none has
some form of mental illness?
P (all four have no mental illness) = (0.70)4
= 0.2401
Since 0.2401 ≤ 0.05, it would not be unusual for none of the four to have mental illness.
Ch 5.4 Conditional Probability and the General Multiplication Rule
Objective A: Conditional Probability and the General Multiplication Rule
A1. Multiplication Rule for Dependent Events
If E and F are dependent events, then P( E and F )  P( E )  P( F | E ) .
The probability of E and F is the probability of event E occurring times the probability of event F
occurring, given the occurrence of event E .
Example 1: A box has 5 red balls and 2 white balls. If two balls are randomly selected (one after the
other), what is the probability that they both are red?
(a) With replacement
P (1st red and 2nd red)
= P (1st red) * P (2nd red)
5 5
25
= 7 * 7 = 49
(b) Without replacement
P (1st red and 2nd red)
= P (1st red) * P (2nd red│1st red)
5 4
20 10
= 7 * 6 = 42 = 21
Example 2: Three cards are drawn from a deck without replacement. Find the probability that all
are jacks.
Without replacement  Dependent case
P (1st jack and 2nd jack and 3rd jack)
4
3
2
1
14
1
1
= 52 * 51 * 50 = 13 * 17 * 25 = 5525 ≈ 0.00018
(Almost zero percent of a chance)
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A2. Conditional Probability
If E and F are any two events, then P( F E ) 
P( E and F ) N ( E and F )

.
P( E )
N (E)
The probability of event F occurring, given the occurrence of event E , is found by dividing the
probability of E and F by the probability of E .
Example 1:At a local Country Club, 65% of the members play bridge and swim, and 72% play bridge.
If a member is selected at random, find the probability that the member swims, given that
the member plays bridge.
Let B be the event of playing bridge.
Let S be the event of swimming.
Given: P (B and S) = 0.65
P (B) = 0.72
Find: P (S│ B)
𝑃 (𝑆 and 𝐵) 0.65
P (S │B) = 𝑃 (𝐵) = 0.72 ≈ 0.903
Objective B: Application
Example 1: Eighty students in a school cafeteria were asked if they favored a ban on smoking in the
cafeteria. The results of the survey are shown in the table.
(Fa)
(O)
(N)
Class
Favor
Oppose
No opinion
(F)
Freshman
15
27
8
50
(S)
Sophomore
23
5
2
30
38
32
10
80
If a student is selected at random, find these probabilities.
(a) The student is a freshman or favors the ban.
P(F or Fa) = P (F) + P (Fa) – P (F and Fa)
50
80
38
15
73
+ 80 - 80 = 80
(b) Given that the student favors the ban, the student is a sophomore.
P (S│Fa) =
23/80
𝑃 (𝑆 𝑎𝑛𝑑 𝐹𝑎)
𝑃 (𝐹𝑎)
23
= 38/80 = 38
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