11.2 – Probability – Events Involving “Not” and “Or”
Properties of Probability
1. The probability of an event is between 0 and 1, inclusive.
2. The probability of an impossible event is 0.
3. The probability of a certain event is 1.
Examples:
Rolling a fair die, what is the probability of each event?
a) The number 3 is rolled.
b) A number not 3 is rolled.
P(3) = 1/6
P(not 3) = 5/6 c) The number 9 is rolled.
d) A number < 7 is rolled.
P(9) = 0
P(< 7) = 1

11.2 – Probability – Events Involving “Not” and “Or”
Events Involving “Not”
Probability of a Complement
The probability that an event E will not occur is equal to one minus the probability that it will occur.
P(not E) = 1 – P(E) P(E

) = 1 – P(E)
Other forms of the equation:
P(E) + P(E

) = 1
P(E) = 1 – P(E

)
What is the probability of not drawing an ace from a standard deck of 52 cards?
P(not an ace) = 1 – P(ace)
P(not an ace) = 1 – (4/52)
P(not an ace) = 48/52 = 12/13

11.2 – Probability – Events Involving “Not” and “Or”
Events Involving “Or”
The probability that one event or another event will occur usually involves the union and addition .
Mutually Exclusive Events
Two events, A and B , are mutually exclusive events if they have no outcomes in common.
Mutually exclusive events can not occur simultaneously
Addition Rule of Probability (for A or B )
If A and B are any two events, then
P(A or B) = P(A) + P(B) – P(A and B)
If A and B are mutually exclusive, then
P(A or B) = P(A) + P(B)

11.2 – Probability – Events Involving “Not” and “Or”
Examples:
1. What is the probability of drawing a king or a diamond from a standard deck of 52 cards?
P(king or diamond) = P(K) + P(D) – P(K and D)
= 4/52 + 13/52 –
= 16/52 = 4/13
1/52

11.2 – Probability – Events Involving “Not” and “Or”
Examples:
2. What is the probability of a 2 or an odd number being rolled on a fair die?
Mutually exclusive events
P(2 or odd) = P(2) + P(odd)
= 1/6 + 3/6
= 4/6 = 2/3

11.3 – Conditional Probability – Events Involving “And”
Conditional Probability
The probability of an event based on the fact that some other event has occurred, will occur, or is occurring.
The probability of event B occurring given that event A has occurred is usually stated as “ the conditional probability of B, given A ; P(B/A)
P(B/A) =
P ( A

B )

P ( A )
P ( A and B )
P ( A )

11.3 – Conditional Probability – Events Involving “And”
Conditional Probability
Example:
A number from the sample space S = {2, 3, 4, 5, 6, 7, 8, 9} is randomly selected. Given the defined events A and B,
A: selected number is odd, and
B: selected number is a multiple of 3 find the following probabilities.
a) P(B) b) P(A and B) a) B = {3, 6, 9} b) P(A and B) =
P(B) = 3/8
P({3, 5, 7, 9}

{3, 6, 9})
= P({3, 9}) = 2/8 = 1/4 c) P(B/A) c) Probability of B given A has occurred:
P(A and B)
P(B/A) = =
1/4
P(A) 4/8
= 1/2

11.3 – Conditional Probability – Events Involving “And”
Conditional Probability
Example:
Given a family with two children, find the probability that both are boys, given that at least one is a boy.
Conditional Probability P(B/A) =
P ( A and
P ( A )
S= {gg, gb, bg, bb}
B )
A = at least one boy A = {gb, bg, bb}
B = both are boys B = {bb}
P(A and B) = P({gb, bg, bb}

{bb}) = P({bb}) = 1/4
P(A) = P({gb, bg, bb}) = 3/4
P ( A and B )
P ( A )
=
1/4
3/4
= 1/3

11.3 – Conditional Probability – Events Involving “And”
Independent Events
Two events are Independent if the occurrence of one of them has no effect on the probability of the other.
P(B/A) = P(B) or
P(A/B) = P(A)

11.3 – Conditional Probability – Events Involving “And”
Independent Events
Example:
A single card is randomly selected from a standard 52-card deck. Given the defined events A and B,
A: the selected card is an ace, find the following probabilities.
B: the selected card is red, a) P(B) b) P(A and B) c) P(B/A)
26 a) P(B) = = 1/2
52 b) P(A and B) = P({Ah, Ad, Ac, As}

{all red}) = P({Ah, Ad}) = 2/52 c) P(B/A) =
P(A and B)
P(A)
=
2/52
4/52
= 1/2
Events A and B are independent as P(B) = P(B/A).

11.3 – Conditional Probability – Events Involving “And”
Multiplication Rule of Probability - Events Involving “And”
If A and B are any two events then
P(A and B) = P(A)

P(B/A)
If A and B are independent events then
P(A and B) = P(A)

P(B)
Example:
A jar contains 4 red marbles, 3 blue marbles, and 2 yellow marbles.
What is the probability that a red marble is selected and then a blue one without replacement?
P(Red and Blue) = P(Red)

P(Blue/Red)
= 4/9

3/8
= 12/72
= 1/8 = 0.1667

11.3 – Conditional Probability – Events Involving “And”
Multiplication Rule of Probability - Events Involving “And”
Example:
A jar contains 4 red marbles, 3 blue marbles, and 2 yellow marbles.
What is the probability that a red marble is selected and then a blue one with replacement?
P(Red and Blue) = P(Red)

P(Blue)
= 4/9

3/9
= 12/81
= 4/27 = 0.148

11.5 – Expected Value
The Expected Value of x is the sum of the products of the values of x and their corresponding probabilities.
E(x) = x
1
P(x
1
) + x
2
P(x
2
) + x
3
P(x
3
) + … + x n
P(x n
)
The expected value is a calculation that serves as the best prediction of a value. It is the probability-weighted average of all possible outcomes.
The expected value of a possible future event assists in making mathematically sound decisions. It is often used when making investments, determining a price for numerous services, prioritizing events, and in calculating Return on Investment.

11.5 – Expected Value
Example:
A third grade class was surveyed regarding the number of hours that they played electronic games each day. The probability distribution is given in the table below:
# of Hours (x)
0
1
2
3
Probability P(x)
0.3
0.4
0.2
0.1
Calculate the Expected Value of the quantity of time that a third grader spends each day playing electronic games.

11.5 – Expected Value
E(x) = x
1
P(x
1
) + x
2
P(x
2
) + x
3
P(x
3
) + … + x n
P(x n
)
# of Hours (x)
0
1
2
3
Probability P(x)
0.3
0.4
0.2
0.1
Expected value, E(x) = 0 (0.3) + 1 (0.4) + 2 (0.2) + 3 (0.1)
Expected value, E(x) = 0 + 0.4 + 0.4 + 0.3
Expected value, E(x) = 1.1 hours

11.5 – Expected Value
Example:
Find the expected number of boys for a three-child family. Assume girls and boys are equally likely.
bbb bbg bgb bgg gbb gbg ggb ggg
# of Boys x
0
1
2
3
Probability
P(x)
1/8
3/8
3/8
1/8
Product x P(x)
0
3/8
6/8
3/8
Expected value, E(x) = 0 + 3/8 + 6/8 + 3/8
Expected value, E(x) = 12/8
Expected value, E(x) = 3/2
Expected value, E(x) = 1.5 boys

11.5 – Expected Value
Example:
Finding Expected Winnings
A player pays $3 to play the following game:
Win $7 by rolling a 6 on a single die,
Win $1 by rolling any other number.
What are the expected net winnings for the game?
Number
1, 2, 3, 4, 5
6
Payoff Net P(x) x P(x)

11.5 – Expected Value
Example:
Finding Expected Winnings
A player pays $3 to play the following game:
Win $7 by rolling a 6 on a single die,
Win $1 by rolling any other number.
What are the expected net winnings for the game?
Number
1, 2, 3, 4, 5
6
Payoff
$1
$7
Net
- $2
$4
P(x)
5/6
1/6 x P(x)
- $10/6
$4/6
Expected value, E(x) = – $10/6 + $4/6
Expected value, E(x) = – $6/6
Expected value, E(x) = – $1

11.5 – Expected Value
A fair game is one in which the net winnings are zero.
An unfair game against the player has negative expected winnings.
An unfair game in favor of the player has positive expected winnings.
Previous Example:
Finding Expected Winnings
A player pays $3 to play the following game:
Win $7 by rolling a 6 on a single die,
Win $1 by rolling any other number.
Is this a fair game ?
NO
It is an unfair game against the player as it has negative expected winnings (–$1) .

11.5 – Expected Value
Example:
Expected Investment Profits
Mark intends to invest $6,000 in one of two companies. His research is presented in the tables below:
Company ABC
Profit/Loss (x) Probability P(x)
- $400
$800
$1300
0.2
0.5
0.3
Company PDQ
Profit/Loss (x) Probability P(x)
$600
$1000
0.8
0.2
What are the expected profits (or loses) for each company?
E(x) = -400 (0.2) + 800 (0.5) + 1300 (.3) E(x) = 600 (0.8) + 1000 (0.2)
E(x) = -80 + 400 + 390
E(x) = $710 Profit
E(x) = 480 + 200
E(x) = $680 Profit

12.1 – Visual Displays of Data
In statistics:
A population includes all of the items of interest.
A sample includes some of the items in the population.
The study of statistics is usually divided into two main areas.
Descriptive statistics : has to do with collecting, organizing, summarizing, and presenting data (information).
Inferential statistics : has to do with drawing inferences or conclusions about populations based on information from samples.

12.1 – Visual Displays of Data
Information that has been collected but not yet organized or processed is called raw data .
Raw data are often quantitative (or numerical ), but can also be qualitative (or non-numerical ).
Quantitative data : The number of siblings in ten different families: 3, 1, 2, 1, 5, 4, 3, 3, 8, 2
Quantitative data can be sorted in mathematical order. The number siblings can appear as 1, 1, 2, 2, 3, 3, 3, 4, 5, 8
Qualitative data: The makes of five different automobiles:
Toyota, Ford, Nissan, Chevrolet, Honda

12.1 – Visual Displays of Data
Frequency Distributions
Frequency Distribution is a method to organize data that includes many repeated items.
It lists the distinct values ( x ) along with their frequencies ( f ).
The relative frequency of each distinct item is the fraction, or percentage, of the data set represented by each item.

12.1 – Visual Displays of Data
Example:
Ten students in a math class were polled as to the number of siblings in their families {3, 2, 2, 1, 3, 4, 3, 3, 4, 2}. Construct a frequency distribution and a relative frequency distribution for the data.
Number x
1
2
3
4
Frequency f Relative Frequency f /n
1 1/10 = 0.1
3
4
2
Sum = 10
3/10 = 0.3
4/10 = 0.4
2/10 = 0.2
Sum of f/n = 1

12.1 – Visual Displays of Data
Grouped Frequency Distributions
A Grouped Frequency Distribution is used when data sets contain a large numbers of items.
The data are arranged into groups, or classes .
All data items are assigned to their appropriate classes, and displayed in a table.
Guidelines for the Classes of a Grouped Frequency
Distribution
1.
Make sure each data item will fit into one and only one, class.
2.
Make all the classes the same width.
3.
Make sure that the classes do not overlap.
4.
Use from 5 to 12 classes.

12.1 – Visual Displays of Data
Example:
Twenty students, selected randomly were asked to estimate the number of hours that they had spent studying in the past week (in and out of class). The responses are below.
15 58 37 42 20 27 36 57 29
42 51 28 46 29 58 55 43 40
56 36
Tabulate a grouped frequency distribution and a relative frequency distribution.
The data contains values in the tens, twenties, thirties, forties and fifties.
Five classes: 10-19, 20-29, 30-39, 40-49, 50-59

12.1 – Visual Displays of Data
15 58 37 42 20 27 36 57 29
42 51 28 46 29 58 55 43 40
56 36
Hours Frequency f Relative Frequency f /n
10 – 19
20 – 29
30 – 39
40 – 49
50 – 59
1
5
3
5
6
1/20 = 0.05 = 5%
5/20 = 0.25 = 25%
3/20 = 0.15 = 15%
5/20 = 0.25 = 25%
6/20 = 0.30 = 30%
Sum = 20 Sum of f/n = 1 or 100%

12.1 – Visual Displays of Data
Histogram
The data from the frequency distribution or a grouped frequency distribution can be graphically display using a histogram .
A series of rectangles, whose lengths represent the frequencies, are placed next to each other.
Number x
Frequency f
Relative
Frequency f /n
3
4
1
2
1
3
1/10 = 0.1
3/10 = 0.3
4/10 = 0.4
4
2 2/10 = 0.2
Sum = 10 Sum of f/n = 1
5
4
3
2
1
0
1 2 3
Siblings
4

12.1 – Visual Displays of Data
Example: Histogram of a Grouped Frequency Distribution
Hours Frequency f
10 – 19
20 – 29
30 – 39
40 – 49
1
5
3
5
7
6
5
2
1
4
3
0
10-19 20-29 30-39 40-49 50-59
Hours
50 – 59 6
In the table, the numbers 10, 20, 30, 40, and 50 are called the lower class limits .
The numbers 19, 29, 39, 49, and 59 are called the upper class limits .
The class width for the distribution is the difference of any two successive lower (or upper) class limits. The class width is 10.

12.1 – Visual Displays of Data
Frequency Polygon
Data can also be displayed by a frequency polygon .
Plot a single point at the appropriate height for each frequency, connect the points with a series of connected line segments and complete the polygon with segments that trail down to the axis.
5
4
3
2
1
0
1 2 3
Siblings
4

12.1 – Visual Displays of Data
Line Graph
A line graph is used when it is important to show how data changes with respect to another variable , such as time.
Example: Line Graph
The line graph below shows the stock price of company
PCWP over a 6-month span.
6
5
4
3
2
1
0
9
8
7
Jan Feb Mar Apr May June
Month

12.1 – Visual Displays of Data
Line Graph

12.1 – Visual Displays of Data
Stem-and-Leaf Displays
The stem and leaf display is another method to present data.
It preserves the exact data values.
15 58 37 42 20 27 36 57 29
42 51 28 46 29 58 55 43 40
56 36
1 5
2 0 7 8 9 9
3 6 6 7
4 0 2 2 3 6
5 1 5 6 7 8 8
The digits to the left of the vertical line (blue region), are the “stems,”
The corresponding ones digits (green region) are the “leaves.”

12.1 – Visual Displays of Data
Bar Graphs
A frequency distribution of non-numerical data can be presented in the form of a bar graph .
It is similar to a histogram except that the rectangles (bars) usually are not touching each other and sometimes are arranged horizontally rather than vertically.
Example: Bar Graph
A bar graph is given for the occurrence of vowels in this sentence.
4
3
2
1
0
8
7
6
5
A E I O U
Vowel

12.1 – Visual Displays of Data
Circle Graphs or Pie Chart
A graphical alternative to the bar graph is the circle graph , or pie chart .
Each sector or wedge, shows the relative magnitude of the categories.
The entire circle measures 360°. The measure of each sector angle should correspond to the percentage of the data being represented.
Example: Expenses
A general estimate of Amy’s monthly expenses are illustrated in the circle graph below.
Clothing 10%
Other 35%
Rent25%
Food 30%