11.2 – Probability – Events Involving “Not” and “Or” Properties of Probability 1. The...
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Transcript of 11.2 – Probability – Events Involving “Not” and “Or” Properties of Probability 1. The...
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. P(3) = 1/6
b) A number not 3 is rolled.
c) The number 9 is rolled.
d) A number < 7 is rolled.
P(not 3) = 5/6
P(9) = 0
P(< 7) = 1
11.2 – Probability – Events Involving “Not” and “Or”
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)
Other forms of the equation:
What is the probability of not drawing an ace from a standard deck of 52 cards?
P(E) + P(E) = 1
P(E) = 1 – P(E)
P(not an ace) = 1 – P(ace)
P(not an ace) = 48/52 = 12/13
P(not an ace) = 1 – (4/52)
Events Involving “Not”
P(E) = 1 – P(E)
11.2 – Probability – Events Involving “Not” and “Or”
The probability that one event or another event will occur usually involves the union and addition.
Mutually Exclusive Events
Addition Rule of Probability (for A or B)
Mutually exclusive events can not occur simultaneously
If A and B are any two events, then
If A and B are mutually exclusive, then
Events Involving “Or”
Two events, A and B, are mutually exclusive events if they have no outcomes in common.
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = P(A) + P(B)
11.2 – Probability – Events Involving “Not” and “Or”
=
1. What is the probability of drawing a king or a diamond from a standard deck of 52 cards?
1/52
4/13
P(king or diamond) =
Examples:
P(K) + P(D) – P(K and D)
4/52 + 13/52 –
= 16/52 =
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
2/3
P(2 or odd) = P(2) + P(odd)
1/6 + 3/6
= 4/6 =
The probability of an event based on the fact that some other event has occurred, will occur, or is occurring.
P(B/A) =
Conditional Probability
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)
11.3 – Conditional Probability – Events Involving “And”
)(
)(
AP
BAP
)(
)(
AP
BandAP
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) B = {3, 6, 9}
Conditional Probability
b) P(A and B) =
a) P(B) b) P(A and B) c) P(B/A)
11.3 – Conditional Probability – Events Involving “And”
P(B) = 3/8
P({3, 5, 7, 9} {3, 6, 9})
= P({3, 9}) = 2/8 = 1/4
c) Probability of B given A has occurred:
P(B/A) = P(A)
P(A and B)
4/8
1/4 1/2 = =
Example:
Given a family with two children, find the probability that both are boys, given that at least one is a boy.
P({gb, bg, bb}) =
Conditional Probability
P(A and B) =
A = at least one boy
11.3 – Conditional Probability – Events Involving “And”
P(A) =
3/4
P({gb, bg, bb} {bb}) = P({bb}) =
= 1/3
1/4
Conditional Probability P(B/A) =
3/4
1/4
B = both are boys
S= {gg, gb, bg, bb}
A = {gb, bg, bb}
B = {bb}
)(
)(
AP
BandAP
)(
)(
AP
BandAP=
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)
Independent Events
or
11.3 – Conditional Probability – Events Involving “And”
P(A/B) = P(A)
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, B: the selected card is red,find the following probabilities.
a) P(B) =
Independent Events
b) P(A and B) =
a) P(B) b) P(A and B) c) P(B/A)
11.3 – Conditional Probability – Events Involving “And”
= 1/2
P({Ah, Ad, Ac, As} {all red}) = P({Ah, Ad}) = 2/52
Events A and B are independent as P(B) = P(B/A).
c) P(B/A) = P(A)
P(A and B)
4/52
2/52 1/2 = =
52
26
If A and B are any two events then
P(A and B) = P(A) P(B/A)
Multiplication Rule of Probability - Events Involving “And”
11.3 – Conditional Probability – Events Involving “And”
P(A and B) = P(A) P(B)
If A and B are independent events then
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
Example:
Multiplication Rule of Probability - Events Involving “And”
11.3 – Conditional Probability – Events Involving “And”
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
Example:
11.5 – Expected ValueThe Expected Value of x is the sum of the products of the values of x and their corresponding probabilities.
E(x) = x1 P(x1) + x2 P(x2) + x3 P(x3) + … + xn P(xn)
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 ValueExample:
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) Probability P(x)
0 0.3
1 0.4
2 0.2
3 0.1
Calculate the Expected Value of the quantity of time that a third grader spends each day playing electronic games.
11.5 – Expected ValueE(x) = x1 P(x1) + x2 P(x2) + x3 P(x3) + … + xn P(xn)
# of Hours (x) Probability P(x)
0 0.3
1 0.4
2 0.2
3 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 ValueExample:
Find the expected number of boys for a three-child family. Assume girls and boys are equally likely.
# of Boys Probability Product
x P(x) x P(x)
0 1/8
1 3/8
2 3/8
3 1/8
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
0
3/8
6/8
3/8
bbb
bbg
bgb
bgg
gbb
gbg
ggb
ggg
11.5 – Expected ValueExample:
Finding Expected Winnings
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 Payoff Net P(x) x P(x)
1, 2, 3, 4, 5
6
A player pays $3 to play the following game:
11.5 – Expected ValueExample:
A player pays $3 to play the following game:
Expected value, E(x) = – $10/6 + $4/6
Expected value, E(x) = – $6/6
Expected value, E(x) = – $1
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 Payoff Net P(x) x P(x)
1, 2, 3, 4, 5 $1 - $2 5/6 - $10/6
6 $7 $4 1/6 $4/6
Finding Expected Winnings
11.5 – Expected Value
Previous Example:
Finding Expected WinningsA player pays $3 to play the following game:
A fair game is one in which the net winnings are zero.
Win $7 by rolling a 6 on a single die,Win $1 by rolling any other number.
Is this a fair game?
An unfair game against the player has negative expected winnings.
An unfair game in favor of the player has positive expected winnings.
It is an unfair game against the player as it has negative expected winnings (–$1) .
NO
11.5 – Expected ValueExample:
Expected Investment Profits
E(x) = -400 (0.2) + 800 (0.5) + 1300 (.3)
Mark intends to invest $6,000 in one of two companies. His research is presented in the tables below:
What are the expected profits (or loses) for each company?
Company ABC
Profit/Loss (x) Probability P(x)
- $400 0.2
$800 0.5
$1300 0.3
Company PDQ
Profit/Loss (x) Probability P(x)
$600 0.8
$1000 0.2
E(x) = -80 + 400 + 390
E(x) = $710
E(x) = 600 (0.8) + 1000 (0.2)
E(x) = 480 + 200
E(x) = $680Profit 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.
Information that has been collected but not yet organized or processed is called raw data.
12.1 – Visual Displays of 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
Frequency Distributions
Frequency Distribution is a method to organize data that includes many repeated items.
12.1 – Visual Displays of Data
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.
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.
12.1 – Visual Displays of Data
Number x Frequency f Relative Frequency f /n
1
2
3
4
1
3
4
2
1/10 = 0.1
2/10 = 0.2
4/10 = 0.4
3/10 = 0.3
Sum = 10 Sum of f/n = 1
Grouped Frequency Distributions
A Grouped Frequency Distribution is used when data sets contain a large numbers of items.
12.1 – Visual Displays of Data
All data items are assigned to their appropriate classes, and displayed in a table.
The data are arranged into groups, or classes.
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.
Guidelines for the Classes of a Grouped Frequency Distribution
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.
Tabulate a grouped frequency distribution and a relative frequency distribution.
12.1 – Visual Displays of Data
15 58 37 42 20 27 36 57 2942 51 28 46 29 58 55 43 40
56 36
The data contains values in the tens, twenties, thirties, forties and fifties.
Five classes: 10-19, 20-29, 30-39, 40-49, 50-59
Hours Frequency f Relative Frequency f /n
12.1 – Visual Displays of Data15 58 37 42 20 27 36 57 2942 51 28 46 29 58 55 43 40
56 36
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%
Histogram
The data from the frequency distribution or a grouped frequency distribution can be graphically display using a histogram.
0
1
2
3
4
5
1 2 3 4
Siblings
Fre
quen
cy
12.1 – Visual Displays of Data
A series of rectangles, whose lengths represent the frequencies, are placed next to each other.
Number x
Frequency f
Relative Frequency f /n
1
2
3
4
1
3
4
2
1/10 = 0.1
2/10 = 0.2
4/10 = 0.43/10 = 0.3
Sum = 10 Sum of f/n = 1
Example: Histogram of a Grouped Frequency Distribution
Fre
quen
cy
Hours
0
1
2
3
4
5
6
7
10-19 20-29 30-39 40-49 50-59
12.1 – Visual Displays of Data
Hours Frequency f
10 – 19
20 – 29
30 – 39
40 – 49
50 – 59
1
5
3
5
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.
Frequency Polygon
Data can also be displayed by a frequency polygon.
0
1
2
3
4
5
1 2 3 4
Fre
quen
cy
Siblings
12.1 – Visual Displays of Data
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.
Line Graph
A line graph is used when it is important to show how data changes with respect to another variable, such as time.
12.1 – Visual Displays of Data
Example: Line Graph
The line graph below shows the stock price of company PCWP over a 6-month span.
Pri
ce in
dol
lars
0
1
2
3
4
5
6
7
8
9
Jan Feb Mar Apr May June
Month
Line Graph
12.1 – Visual Displays of Data
Stem-and-Leaf Displays
The digits to the left of the vertical line (blue region), are the “stems,”
12.1 – Visual Displays of Data
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 2942 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 corresponding ones digits (green region) are the “leaves.”
Bar Graphs
A frequency distribution of non-numerical data can be presented in the form of a bar graph.
12.1 – Visual Displays of Data
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
0
1
2
3
4
5
6
7
8
A bar graph is given for the occurrence of vowels in this sentence.
Fre
quen
cy
A E I O U
Vowel
Circle Graphs or Pie ChartA 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.
12.1 – Visual Displays of Data
Example: Expenses
A general estimate of Amy’s monthly expenses are illustrated in the circle graph below.
Other 35%Rent25%
Food 30%
Clothing 10%