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4.1 - 1Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Chapter 4Probability
4-1 Review and Preview
4-2 Basic Concepts of Probability
4-3 Addition Rule
4-4 Multiplication Rule: Basics
4-5 Multiplication Rule: Complements and Conditional Probability
4-6 Counting
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Section 4-1Review and Preview
4.1 - 3Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Review
Necessity of sound sampling methods.
Common measures of characteristics of data
Mean
Standard deviation
4.1 - 4Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Preview
Rare Event Rule for Inferential Statistics:
If, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct.
Statisticians use the rare event rule for inferential statistics.
4.1 - 5Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Section 4-2 Basic Concepts of
Probability
4.1 - 6Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Part 1
Basics of Probability
4.1 - 7Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Events and Sample Space
Event
any collection of results or outcomes of a procedure
Simple Event
an outcome or an event that cannot be further broken down into simpler components
Sample Space
for a procedure consists of all possible simple events; that is, the sample space consists of all outcomes that cannot be broken down any further
4.1 - 8Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example A pair of dice are rolled. The sample space has
36 simple events:
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
where the pairs represent the numbers rolled on each dice.
Which elements of the sample space correspond to the event that the sum of each dice is 4?
4.1 - 9Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example Which elements of the sample space correspond
to the event that the sum of each dice is 4?
ANSWER:
3,1 2,2 1,3
4.1 - 10Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Notation for Probabilities
P - denotes a probability.
A, B, and C - denote specific events.
P(A) - denotes the probability of event A occurring.
4.1 - 11Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Basic Rules for Computing Probability
Rule 1: Relative Frequency Approximation of Probability
Conduct (or observe) a procedure, and count the number of times event A actually occurs. Based on these actual results, P(A) is approximated as follows:
P(A) = # of times A occurred
# of times procedure was repeated
4.1 - 12Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 20 on page 149
F = event of a false negative on polygraph test
Thus this is not considered unusual since it is more than 0.001 (see page 146). The test is not highly accurate.
0918.098
9)( FP
4.1 - 13Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Rounding Off Probabilities
When expressing the value of a probability, either give the exact fraction or decimal or round off final decimal results to three significant digits. All digits are significant except for the zeros that are included for proper placement of the decimal point.
Example:
0.1254 has four significant digits
0.0013 has two significant digits
4.1 - 14Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 21 on page 149
F = event of a selecting a female senator
NOTE: total number of senators=100
Thus this does not agree with the claim that men and women have an equal (50%) chance of being selected as a senator.
160.0100
16
1684
16)(
FP
4.1 - 15Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 28 on page 150
A = event that Delta airlines passenger is involuntarily bumped from a flight
Thus this is considered unusual since it is less than 0.05 (see directions on page 149). Since probability is very low, getting bumped from a flight on Delta is not a serious problem.
000195.015378
3)( AP
4.1 - 16Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Basic Rules for Computing Probability
Rule 2: Classical Approach to Probability (Requires Equally Likely Outcomes)
Assume that for a given procedure each simple event has an equal chance of occurring.
P(A) = number of ways A can occur
number of different simple events in the sample space
4.1 - 17Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleWhat is the probability of rolling two die and getting a sum of 4?
A = event that sum of the dice is 4
Assume each number is equally likely to be rolled on the die. Rolling a sum of 4 can happen in one of three ways (see previous slide) with 36 simple events so:
0833.036
3)( AP
4.1 - 18Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleWhat is the probability of getting no heads when a “fair” coin is tossed three times? (A fair coin has an equal probability of showing heads or tails when tossed.)
A = event that no heads occurs in three tosses
Sample Space (in order of toss):
TTTTTHTHTTHHHTTHTHHHTHHH ,,,,,,,
4.1 - 19Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleSample space has 8 simple events.
Event A corresponds to TTT only so that:
125.08
1)( AP
4.1 - 20Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 36 on page 151
Let:
S = event that son inherits disease (xY or Yx)
D = event that daughter inherits disease (xx)
4.1 - 21Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 36 on page 151
(a) Father: xY Mother: XX
Sample space for a son:
YX YX
Sample space has no simple
events that represent a son that has the disease so:
02
0)( SP
4.1 - 22Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 36 on page 151
(b) Father: xY Mother: XX
Sample space for a daughter:
xX xX
Sample space has no simple
events that represent a daughter that has the disease so:
02
0)( DP
4.1 - 23Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 36 on page 151
(c) Father: XY Mother: xX
Sample space for a son:
Yx YX
Sample space has one simple
event that represents a son that has the disease so:
5.02
1)( SP
4.1 - 24Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 36 on page 151
(d) Father: XY Mother: xX
Sample space for a daughter:
Xx XX
Sample space has no simple
event that represents a daughter that has the disease so:
02
0)( DP
4.1 - 25Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 18 on page 149
Table 4-1 on page 137 (polygraph data)
Example
Did Not LieDid Not Lie Did LieDid Lie
Positive Test ResultPositive Test Result 1515
(false positive)(false positive)
4242
(true positive)(true positive)
NegativeTest ResultNegativeTest Result 3232
(true negative)(true negative)
99
(false negative)(false negative)
4.1 - 26Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
It is helpful to first total the data in the table:
(a) How many responses were lies:
ANSWER: 51
Example
Did Not LieDid Not Lie Did LieDid Lie TOTALSTOTALS
Positive Test ResultPositive Test Result 1515
(false positive)(false positive)
4242
(true positive)(true positive)
5757
NegativeTest ResultNegativeTest Result 3232
(true negative)(true negative)
99
(false negative)(false negative)
4141
TOTALSTOTALS 4747 5151 9898
4.1 - 27Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
(b) If one response is randomly selected, what is the probability it is a lie?
L = event of selecting one of the lie responses
(c)
Example
98
51)( LP
520.098
51
4.1 - 28Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Basic Rules for Computing Probability - continued
Rule 3: Subjective Probabilities
P(A), the probability of event A, is
estimated by using knowledge of the
relevant circumstances.
4.1 - 29Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 4 on page 148
Probability should be high based on experience (it is rare to be delayed because of an accident).
Guess: P=99/100 (99 out of 100 times you will not be delayed because of an accident)
ANSWERS WILL VARY
4.1 - 30Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Classical probability predicts the probability of flipping a (non-biased) coin and it coming up heads is ½=0.5
Ten coin flips will sometimes result in exactly 5 heads and a frequency probability of heads 5/10=0.5; but often you will not get exactly 5 heads in ten flips.
4.1 - 31Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Law of Large Numbers
As a procedure is repeated again and again, the relative frequency probability of an event tends to approach the actual probability.
Example: If we flip a coin 1 million times the frequency probability should be approximately 0.5
4.1 - 32Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Probability Limits
The probability of an event that is certain to occur is 1.
The probability of an impossible event is 0.
For any event A, the probability of A is between 0 and 1 inclusive. That is:
0 P(A) 1
Always express a probability as a fraction or decimal number between 0 and 1.
4.1 - 33Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Possible Values for Probabilities
4.1 - 34Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Complementary Events
The complement of event A, denoted by
A, consists of all outcomes in which the
event A does not occur.
4.1 - 35Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleIf a fair coin is tossed three times and
A = event that exactly one heads occurs
Find the complement of A.
4.1 - 36Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleSample space:
Event A corresponds to HTT, THT, TTH
Therefore, the complement of A are the simple events:
HHH, HHT, HTH, THH, TTT
TTTTTHTHTTHHHTTHTHHHTHHH ,,,,,,,
4.1 - 37Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Part 2
Beyond theBasics of Probability: Odds
4.1 - 38Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Odds
The actual odds in favor of event A occurring are the ratio P(A)/ P(A), usually expressed in the form of a:b (or “a to b”), where a and b are integers having no common factors.
The actual odds against event A occurring are the ratio P(A)/P(A), which is the reciprocal of the actual odds in favor of the event. If the odds in favor of A are a:b, then the odds against A are b:a.
The payoff odds against event A occurring are the ratio of the net profit (if you win) to the amount bet.
payoff odds against event A = (net profit) : (amount bet)
4.1 - 39Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example
Problem 38, page 149
W = simple event that you win due to an odd number
Sample Space00, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36
(a) There are 18 odd numbers so that
P(W) = 18/38
4.1 - 40Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example
Problem 38, page 149
There are 20 events that correspond to winning from a number that is not odd (i.e. you do not win due to an odd number) so:
(b) Odds against winning are
9:10or 9/1018
20
38/18
38/20
)(
)(
WP
WP
38
20)( WP
4.1 - 41Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example
Problem 38, page 149
(c) Payoff odds against winning are 1:1 That is, $1 net profit for every $1 bet
Thus, if you bet $18 and win, your net profit is $18 which can be found by solving the proportion:
The casino returns $18+$18=$36 to you.
1
1
18
profitnet
4.1 - 42Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example
Problem 38, page 149
(d) Actual odds against winning are 10:9 That is, $10 net profit for every $9 bet
Thus, if you bet $18 and win, your net profit is $20 which can be found by solving the proportion:
The casino returns $18+$20=$38 to you.
9
10
18
profitnet
4.1 - 43Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Recap
In this section we have discussed:
Rare event rule for inferential statistics.
Probability rules.
Law of large numbers.
Complementary events.
Rounding off probabilities.
Odds.
4.1 - 44Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Section 4-3 Addition Rule
4.1 - 45Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Key Concept
This section presents the addition rule as a device for finding probabilities that can be expressed as P(A or B), the probability that either event A occurs or event B occurs (or they both occur) as the single outcome of the procedure.
The key word in this section is “or.” It is the inclusive or, which means either one or the other or both.
4.1 - 46Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Compound Event any event combining 2 or more simple events
Compound Event
Notation
P(A or B) = P (in a single trial, event A occurs or event B occurs or they both occur)
4.1 - 47Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find that total in such a way that no outcome is counted more than once.
General Rule for a Compound Event
4.1 - 48Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
A random survey of members of the class of 2005 finds the following:
What is the probability the student did not graduate or was a man?
Example
Number of Number of students who students who graduatedgraduated
Number of students Number of students who did not graduatewho did not graduate
TOTALSTOTALS
WomenWomen 672672 2222 694694
MenMen 582582 1919 601601
TOTALSTOTALS 12541254 4141 12951295
4.1 - 49Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Method 1: directly add up those who did not graduate and those who are men (without counting men twice):
22 + 19 + 582 = 623
Method 2: add up the total number who did not graduate and the total number of men, then subtract the double count of men:
41 + 601 - 19 = 623
Example
4.1 - 50Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
The probability the student did not graduate or was a man :
623/1295 = 0.481
Example
4.1 - 51Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Compound Event
Formal Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)
where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial of a procedure.
4.1 - 52Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
N = event that student did not graduate
M = event that student was a man
P(N) = 41/1295 = 0.0317
P(M) = 601/1295 = 0.464
P(N and M) = 19/1295 = 0.0147
P(N or M) = P(N) + P(M) - P(N and M)
= 0.0317 + 0.464 - 0.0147
= 0.481
Previous Example
4.1 - 53Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 20 on page 157
What is the probability the subject had a negative test result or lied?
Example
Did Not LieDid Not Lie Did LieDid Lie TOTALSTOTALS
Positive Test ResultPositive Test Result 1515
(false positive)(false positive)
4242
(true positive)(true positive)
5757
NegativeTest ResultNegativeTest Result 3232
(true negative)(true negative)
99
(false negative)(false negative)
4141
TOTALSTOTALS 4747 5151 9898
4.1 - 54Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
N = event that there is a negative test result
L = event that the subject lied
P(N) = 41/98 = 0.418
P(L) = 51/98 = 0.520
P(N and L) = 9/98 = 0.0918
P(N or L) = P(N) + P(L) - P(N and L)
= 0.418 + 0.520 - 0.0918
= 0.846
Example (cont.)
4.1 - 55Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 30 on page 158N=event that person refuses to respond
F=event that person’s age is 60 or older
Note: total number of people in the study is 1205
total number who refused to respond is 156 so:
P(N)=156/1205
total number who are 60 or older is 251 so:
P(F)=251/1205
total number who refuse to respond and 60 or older is 49 so:
P(N and F)=49/1205
P(N OR F) = 156/1205 + 251/1205 – 49/1205 = 358/1205 = 0.297
Example
4.1 - 56Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Disjoint or Mutually ExclusiveEvents A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap.)
Previous Example:
G = students who graduated
N = students who did not graduate
N and G are disjoint because no student who did not graduate also graduated
4.1 - 57Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Disjoint or Mutually ExclusiveEvents A and B are not disjoint if they overlap.
Previous Example:
M = male students
G = students who graduated
M and G are not disjoint because some students who graduated are also male
4.1 - 58Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 10 and 12 on page 157
10.These are disjoint since a subject treated with Lipitor cannot be a subject given no medication.
12. These are not disjoint since it is possible for a homeless person to be a college graduate.
4.1 - 59Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Venn DiagramA Venn diagram is a way to picture how sets overlap.
Venn Diagram for Events That Are Not Disjoint and overlap.
Venn Diagram for Disjoint Events which do not overlap.
4.1 - 60Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Complementary Events
It is impossible for an event and its complement to occur at the same time.
That is, for any event A,
are disjoint
AA and
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Probability Rule of Complementary Events
P(A) + P(A) = 1
= 1 – P(A)
P(A) = 1 – P(A)
P(A)
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Venn Diagram for the Complement of Event A
4.1 - 63Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
ExampleProblem 16 on page 157
is the probability that a screened driver is not intoxicated
)(IP
991.000888.01)(1)( IPIP
4.1 - 64Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Recap
In this section we have discussed:
Compound events.
Formal addition rule.
Intuitive addition rule.
Disjoint events.
Complementary events.
4.1 - 65Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Section 4-4 Multiplication Rule:
Basics
4.1 - 66Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Tree Diagrams
A tree diagram is a picture of the possible outcomes of a procedure, shown as line segments emanating from one starting point. These diagrams are sometimes helpful in determining the number of possible outcomes in a sample space, if the number of possibilities is not too large.
4.1 - 67Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Tree Diagrams
This figure summarizes the possible outcomes for a true/false question followed by a multiple choice question.
Note that there are 10 possible combinations.
4.1 - 68Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Tree Diagrams
A bag contains three different colored marbles: red, blue, and green. Suppose two marbles are drawn from the bag and after the first marble is drawn, it is put back into the bag before the second marble is drawn. Construct a tree diagram that depicts all possible outcomes.
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Example: Tree Diagrams
4.1 - 70Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
Use the previous example of drawing two marbles with replacement to compute the probability of drawing a red marble on the first draw and a blue marble on the second draw.
4.1 - 71Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
P(Y) = number of ways Y can occur
number of different simple events in the sample space
9
1)( YP
Y = event of drawing red first then blue
4.1 - 72Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
Consider each event separately:
3
1)( RP
R = event of drawing red on first draw
B = event of drawing blue on second draw
3
1)( BP
4.1 - 73Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Notation
P(A and B) =
P(event A occurs in a first trial and
event B occurs in a second trial)
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Key Concept
The basic multiplication rule is used for finding P(A and B), the probability that event A occurs in a first trial and event B occurs in a second trial.
4.1 - 75Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
The probability of drawing red on first draw and drawing blue on second draw is the product of the individual probabilities:
9
1
3
1
3
1) and ( BRP
4.1 - 76Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Key Concept
NEXT EXAMPLE SHOWS:
If the outcome of the first event A somehow affects the probability of the second event B, it is important to adjust the probability of B to reflect the occurrence of event A.
This is Conditional Probability
4.1 - 77Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Tree Diagrams
A bag contains three different colored marbles: red, blue, and green. Suppose two marbles are drawn from the bag without replacing the first marble after it is drawn. Construct a tree diagram that depicts all possible outcomes.
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Example: Tree Diagrams
4.1 - 79Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
Use the previous example to compute the probability of drawing a red marble on the first draw and a blue marble on the second draw.
4.1 - 80Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
P(Y) = number of ways Y can occur
number of different simple events in the sample space
6
1)( YP
Y = event of drawing red first then blue
4.1 - 81Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
Multiplication Method
3
1)( RP
R = event of drawing red on first draw
B = event of drawing blue on second draw (given that there are now only two marbles in the bag)
2
1)( BP
4.1 - 82Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example: Computing Probability withTree Diagrams
The probability of drawing red on first draw and drawing blue on second draw is the product of the individual probabilities:
6
1
2
1
3
1) and ( BRP
4.1 - 83Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Conditional ProbabilityKey Point
Without replacement, we must adjust the probability of the second event to reflect the outcome of the first event.
4.1 - 84Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Conditional ProbabilityImportant Principle
The probability for the second event B should take into account the fact that the first event A has already occurred.
4.1 - 85Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Notation for Conditional Probability
P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B|A as “B given A.”)
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Multiplication Rule and Conditional Probability
P(A and B) = P(A) • P(B|A)
4.1 - 87Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Intuitive Multiplication Rule
When finding the probability that event A occurs in one trial and event B occurs in the next trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B takes into account the previous occurrence of event A.
4.1 - 88Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 14 on page 168
If three are selected without replacement, what is probability they all had false positive test results?
Example
Did Not LieDid Not Lie Did LieDid Lie TOTALSTOTALS
Positive Test ResultPositive Test Result 1515
(false positive)(false positive)
4242
(true positive)(true positive)
5757
NegativeTest ResultNegativeTest Result 3232
(true negative)(true negative)
99
(false negative)(false negative)
4141
TOTALSTOTALS 4747 5151 9898
4.1 - 89Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 14 on page 168
On first selection there are 98 subjects, 15 of which are false positive:
Example
98
15)( 1 FP
selectionth on positive false a is thereevent that nFn
4.1 - 90Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 14 on page 168
On second selection there are 97 subjects (without replacement)
If the first selection was false positive, there are 14 false positives left:
Example
97
14)|( 12 FFP
4.1 - 91Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 14 on page 168
On third selection there are 96 subjects (without replacement)
If the first and second selections were false positive, there are 13 false positives left:
Example
96
13) and |( 213 FFFP
4.1 - 92Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 14 on page 168
ANSWER:
This is considered unusual since probability is less than 0.05
Example
) and |()|()() and and ( 213121321 FFFPFFPFPFFFP
00299.096
13
97
14
98
15
4.1 - 93Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Dependent and Independent
Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are similarly independent if the occurrence of any does not affect the probabilities of the occurrence of the others.) If A and B are not independent, they are said to be dependent.
4.1 - 94Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Problem 10 on page 168
Finding that your calculator works
Finding that your computer works
Assuming your calculator and your computer are not both running off the same source of power, these are independent events.
Examples
4.1 - 95Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Previous example of drawing two marbles with replacement
Since the probability of drawing the second marble is not affected by drawing the first marble, these are independent events.
Examples
4.1 - 96Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Dependent Events
Two events are dependent if the occurrence of one of them affects the probability of the occurrence of the other, but this does not necessarily mean that one of the events is a cause of the other.
4.1 - 97Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Previous example of drawing two marbles without replacement
Since the probability of drawing the second marble is affected by drawing the first marble, these are dependent events.
Examples
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Multiplication Rule for Independent Events
Note that if A and B are independent events, then P(B|A)=P(B) and the multiplication rule is then:
P(A and B) = P(A) • P(B)
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Applying the Multiplication Rule
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Applying the Multiplication Rule
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Caution
When applying the multiplication rule, always consider whether the events are independent or dependent, and adjust the calculations accordingly.
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Multiplication Rule for Several Events
In general, the probability of any sequence of independent events is simply the product of their corresponding probabilities.
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Example
Page 169, problem 26
These events are independent since the probability of getting a girl on any try is not affected by the occurence of getting a girl on a previous try.
th tryon birth girl a is thereevent that nGn
2
1)( nGP
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Example
Page 169, problem 26
ten factors of 1/2
10
2
1
2
1...
2
1
2
1
)(...)()(
) and ... and and (
1021
1021
GPGPGP
GGGP
4.1 - 105
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ExamplePage 169, problem 26
Calculator: use “hat key” to evaluate powers: 000977.05.0
2
1 1010
10 ^ 5.0)5.0( 10
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ExamplePage 169, problem 26
Since:
We see that getting 10 girls by chance alone is unusual and conclude that the gender selection method is effective.
Calculator: use “hat key” to evaluate powers:
05.0000977.0
4.1 - 107
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Treating Dependent Events as Independent
Some calculations are cumbersome, but they can be made manageable by using the common practice of treating events as independent when small samples are drawn from large populations. In such cases, it is rare to select the same item twice (sample with replacement).
4.1 - 108
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The 5% Guideline for Cumbersome Calculations
If a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent).
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Example
Page 169, problem 30
a) If we select without replacement, then randomly selecting an ignition system are not independent. But since 3/200 = 0.015 =1.5%, we could use the 5% guideline and regard these events as independent.
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Example
Page 169, problem 30
b) If these events are not independent (dependent) then:
) and |()|()() and and ( 213121321 GGGPGGPGPGGGP
926.0198
193
199
194
200
195
4.1 - 111
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Example
Page 169, problem 30
c) If these events are independent then:
)()()() and and ( 321321 GPGPGPGGGP 3
200
195
200
195
200
195
200
195
927.0975.0 3
4.1 - 112
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Example
Page 169, problem 30
d) The answer from part (b) is exact so it is better.
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Summary of Fundamentals
In the addition rule, the word “or” in P(A or B) suggests addition. Add P(A) and P(B), being careful to add in such a way that every outcome is counted only once.
In the multiplication rule, the word “and” in P(A and B) suggests multiplication. Multiply P(A) and P(B), but be sure that the probability of event B takes into account the previous occurrence of event A.
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Recap
In this section we have discussed:
Notation for P(A and B).
Notation for conditional probability.
Independent events.
Formal and intuitive multiplication rules.
Tree diagrams.
4.1 - 115
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Section 4-5 Multiplication Rule:Complements and
Conditional Probability
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Key Concepts
Probability of “at least one”:Find the probability that among several trials, we get at least one of some specified event.
Conditional probability:Find the probability of an event when we have additional information that some other event has already occurred.
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Complements: The Probability of “At Least One”
The complement of getting at least one item of a particular type is that you get no items of that type.
“At least one” is equivalent to “one or more.”
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Example
Page 175, problems 6
If not all 6 are free from defects, that means that at least one of them is defective.
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Example
Page 175, problems 8
If it is not true that at least one of the five accepts an invitation, then all five did not accept the invitation.
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Finding the Probability of “At Least One”
To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is,
P(at least one) = 1 – P(none).
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Example
Page 175, problem 10
P(at least one girl) = 1 – P(all boys)
)(...)()(1
) and ... and and (1)boys all(1
821
821
BPBPBP
BBBPP
th tryon birth boy a is thereevent that nBn
996.0
00391.012
11
8
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Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Example
Page 175, problem 10
The probability of having 8 children and none of them are girls (all boys) is 0.00391 which means this is a rare event.
4.1 - 123
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Example
Page 175, problem 12
P(at least one working calculator)
= 1 – P(all calculators fail)
Note: these are independent events and
fails calculatorth event that nFn
04.096.01)( nFP
P(a calculator fails) = 1 – P(a calculator does not fail)
4.1 - 124
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Example
Page 175, problem 12
998.0
0016.0104.01
)()(1
) and (1)fail scalculator all(1
2
21
21
FPFP
FFPP
4.1 - 125
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ExamplePage 175, problem 12
With one calculator,
P(working calculator) = 0.96
With two calculators,
P(at least one working calculator)
= 0.998
The increase in chance of a working calculator might be worth the effort.
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Conditional ProbabilityA conditional probability of an event is a probability obtained with the additional information that some other event has already occurred.
P(B|A) denotes the conditional probability of event B occurring, given that event A has already occurred.
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Intuitive Approach to Conditional Probability
The conditional probability of B given A can be found by assuming that event A has occurred, and then calculating the probability that event B will occur.
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Problem 22 on page 175
Table 4-1 on page 137 (polygraph data)
Example
Did Not LieDid Not Lie Did LieDid Lie
Positive Test ResultPositive Test Result 1515
(false positive)(false positive)
4242
(true positive)(true positive)
NegativeTest ResultNegativeTest Result 3232
(true negative)(true negative)
99
(false negative)(false negative)
4.1 - 129
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Example
Page 175, problem 22
(a)There are 47 subjects who did not lie, 32 of which had a negative test result.
681.047
32lie)not didsubject |result test negative( P
4.1 - 130
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Conditional Probability Formula
Conditional probability of event B occurring, given that event A has already occurred
P(B A) = P(A and B)
P(A)
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Example
Page 175, problem 22
(a)Using the formula
681.0 98/47
98/32
lie)not didsubject (
result) test negative had and lienot didsubject (
lie)not didsubject |result test negative(
P
P
P
4.1 - 132
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Confusion of the Inverse
To incorrectly believe that P(A|B) and P(B|A) are the same, or to incorrectly use one value for the other, is often called confusion of the inverse.
4.1 - 133
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Example
Page 175, problem 22
(b) Using the formula
780.0 98/41
98/32
result) test negative(
lie)not didsubject andresult test negative(
result) test negative|lienot didsubject (
P
P
P
4.1 - 134
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Example
Page 175, problem 22
(c) The results from parts (a) and (b) are not equal.
4.1 - 135
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Recap
In this section we have discussed:
Concept of “at least one.”
Conditional probability.
Intuitive approach to conditional probability.
4.1 - 136
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Section 4-6Counting
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Key Concept
In many probability problems, the big obstacle is finding the total number of outcomes, and this section presents several methods for finding such numbers without directly listing and counting the possibilities.
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Fundamental Counting Rule
For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways.
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Example
A coin is flipped and then a die is rolled. What are the total number of outcomes?
4.1 - 140
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Example
A coin is flipped and then a die is rolled. What are the total number of outcomes?
4.1 - 141
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Example
A coin is flipped and then a die is rolled. What are the total number of outcomes?
ANSWER: There are 2 outcomes for the coin flip and 6 outcomes for the die roll. Total number of outcomes:
1262
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Example
Page 185, Problem 32 (a)
If 20 newborn babies are randomly selected, how many different gender sequences are possible?
4.1 - 143
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Example
Page 185, Problem 32 (a)
ANSWER:
This is a sequence of 20 events each of which has two possible outcomes (boy or girl). By the fundamental counting rule, total number of gender sequences will be:
576,048,122222 20
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Example
Page 185, Problem 28
A safe combination consists of four numbers between 0 and 99. If four numbers are randomly selected, what is the probability of getting the correct combination on the first attempt?
4.1 - 145
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ExamplePage 185, Problem 28
ANSWER:
Total number of possible combinations is
Since there is only one correct combination:
It is not feasible to try opening the safe this way.
000,000,100100100100100100 4
00000001.0000,000,100
1n)combinatiocorrect ( P
4.1 - 146
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Notation
The factorial symbol ! denotes the product of decreasing positive whole numbers.
For example,
4! 4 3 2 1 24.
By special definition, 0! = 1.
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A collection of n different items can be arranged in order n! different ways. (This factorial rule reflects the fact that the first item may be selected in n different ways, the second item may be selected in n – 1 ways, and so on.)
Factorial Rule
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Example
Page 183, Problem 6
Find the number of different ways that the nine players on a baseball team can line up for the National Anthem by evaluating 9 factorial.
880,362123456789!9
4.1 - 149
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Factorial on Calculator
Calculator
9 MATH PRB 4:!
To get:
Then Enter gives the result
!9
4.1 - 150
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Permutations Rule(when items are all different)
(n - r)!n rP = n!
If the preceding requirements are satisfied, the number of permutations (or sequences) of r items selected from n available items (without replacement) is
Requirements:
1. There are n different items available. (This rule does not apply if some of the items are identical to others.)
2. We select r of the n items (without replacement).
3. We consider rearrangements of the same items to be different sequences. (The permutation of ABC is different from CBA and is counted separately.)
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Example
A bag contains 4 colored marbles: red, blue, green, yellow. If we select 3 of the four marbles from the bag without replacement, how many different color orders are there?
4.1 - 152
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Example
ANSWER:
241
1234
!1
!4
)!34(
!434
P
4.1 - 153
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Example
Page 183, Problem 12
In horse racing, a trifecta is a bet that the first three finishers in a race are selected, and they are selected in the correct order. Find the number of different possible trifecta bets in a race with ten horses by evaluating
310P
4.1 - 154
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Example
Page 183, Problem 12
ANSWER:
7201
8910
1234567
12345678910
!7
!10
)!310(
!10310
P
4.1 - 155
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Permutations on Calculator
Calculator
10 MATH PRB 2:nPr 3
To get:
Then Enter gives the result
310P
4.1 - 156
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Permutations Rule(when some items are identical to others)
n1! . n2! .. . . . . . . nk! n!
If the preceding requirements are satisfied, and if there are n1 alike, n2 alike, . . . nk alike, the number of permutations (or sequences) of all items selected without replacement is
Requirements:
1. There are n items available, and some items are identical to others.
2. We select all of the n items (without replacement).
3. We consider rearrangements of distinct items to be different sequences.
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Example
Page 185, Problem 26
Find the number of different ways the letters AGGYB can be arranged.
4.1 - 158
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Example
Page 185, Problem 26
ANSWER: there are two letters of the five that are alike. Then the total number of arrangements will be
6012
12345
!1!1!1!2
!5
4.1 - 159
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Example
Page 185, Problem 32 (b)
How many different ways can 10 girls and 10 boys be arranged in a sequence?
4.1 - 160
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Example
Page 185, Problem 32 (b)
ANSWER: there are twenty total children in the arrangement. There are two groups of 10 that are alike. This gives a total number of arrangements:
756,184!10!10
!20
4.1 - 161
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Example
Page 185, Problem 32 (c)
What is the probability of getting 10 girls and 10 boys when twenty babies are born?
4.1 - 162
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Example
Page 185, Problem 32 (c)
ANSWER: as found on previous slides
The total number of boy/girl arrangements of 20 newborns is:
The total number of ways 10 girls and 10 boys can be arranged in a sequence is
576,048,1220
756,184!10!10
!20
4.1 - 163
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Example
Page 185, Problem 32 (c)
ANSWER:
176.0576,048,1
756,184
born) are babies 20 when boys 10 and girls 10(
P
4.1 - 164
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(n - r )! r!n!
nCr =
Combinations Rule
If the preceding requirements are satisfied, the number of combinations of r items selected from n different items is
Requirements:
1. There are n different items available.
2. We select r of the n items (without replacement).
3. We consider rearrangements of the same items to be the same. (The combination of ABC is the same as CBA.)
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Example
Page 183, Problem 8
Find the number of different possible 5 card poker hands by evaluating
552C
4.1 - 166
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Example
Page 183, Problem 8
ANSWER:
960,598,2 !5
4849505152
!5!47
!52
!5)!552(
!52552
C
4.1 - 167
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Combinations on Calculator
Calculator
52 MATH PRB 3:nCr 5
To get:
Then Enter gives the result
552C
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Example
Page 184, Problem 16 (a)
What is the probability of winning a lottery with one ticket if you select five winning numbers from 1,2,…,31.
Note: numbers selected are different and order does not matter.
4.1 - 169
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Example
Page 184, Problem 16 (a)
ANSWER:
Total number of possible combinations is:
Since only one combination wins:
911,169!5!26
!31531
C
00000589.0911,169
1winning)( P
4.1 - 170
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Example
Page 184, Problem 16 (b)
What is the probability of winning a lottery with one ticket if you select five winning numbers from 1,2,…,31.
Note: assume now that you must select the numbers in the same order they were drawn so that order does matter.
4.1 - 171
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Example
Page 184, Problem 16 (b)
ANSWER:
Total number of possible combinations is:
Since only one permutation wins:
320,389,20!26
!31531 P
0000000490.0320,389,20
1winning)( P
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When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.
Permutations versus Combinations
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Recap
In this section we have discussed: The fundamental counting rule.
The permutations rule (when items are all different).
The permutations rule (when some items are identical to others).
The combinations rule.
The factorial rule.