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Chapter 4 Probability: Overview; Basic Concepts of Probability; Addition Rule;

Multiplication Rule: Basics; Multiplication Rule: Complements and Conditional Probability; Probabilities Through Simulation; Counting

Objective: Develop an understanding of probability values which will be used in statistics

and the basic skills necessary to determine probability values in different circumstances.

Chapter Problem: Are polygraph instruments effective as “lie detectors”? Question: Did

the subject actually lie? Possible results are: incorrect results are false positive, false negative; correct results are true positive, and true negative

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4-1 Overview of Probability; 4-2.1 Key Concept

4-1 Overview

 Probability considerations are prevalent in our everyday life like statistics; e.g. chance of rain, or win a state lottery

 Probability is the fundamentals of inferential statistics - methods of drawing conclusions about a population based on a sample of population

 Statisticians would reject an explanation based on very low probabilities

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

4-2.1 Key Concept

 This section introduces the basic concept of the probability of an event. Three different methods for finding probability values will be presented.

 The most important objective of this section is to learn how to interpret

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4-2.2 Basic Concepts of Probability

An event is any collection of results or outcomes of a procedure

 Example 1: Procedure: single birth; Event (result or outcome): female

 Example 2: Procedure: 3 births; Event (results or outcomes): 2 females and 1 male

 Example 3: Procedure: tossing a coin 5 times; Event (results or outcomes): 4 heads and 1 tail

A simple event is an outcome or an event that cannot be further broken

down into simpler components

 Examples: Female (F) for single birth procedure; 2 females and one male is not a simple event; but each of {FFM, FMF, MFF} is a simple event from 2F and 1M

The sample space for a procedure consists of all possible

simple events

;

i.e. the sample space consists of all outcomes that cannot be broken down

any further

Examples:

{

female, male} for single birth procedure

 {FFF, FFM, FMF, FMM, MFF, MFM, MMF, MMM} for 3 births procedure

 {Head, Tail} when tossing a coin

 {1, 2, 3, 4, 5, 6} when tossing a dice

 {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} is sample space containing the sum of tossing two dices

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4-2.3 Probabilities Notations and Rules

Notation: use P to denote a probability, A, B, and C denote specific events,

and P(A) denote the probability of event A occurring

 Examples (1) If A is single birth for female (F), then P(F) is the probability of female birth; (2) P(head) is the probability of tossing a coin with head as outcome

Rule 1 (Relative frequency ): approximation of probability P(A) for

probability of event A occurring is estimated as (based on the actual results):

 Example: toss a coin with the outcome head or tail; toss 10 times, 20 times, 30 times,…

Rule 2 (Classical Approach to Probability): Requires equally likely

outcomes, i.e. need to verify that the outcomes are equally likely

 Assume that a given procedure has n different simple events, and that each of those simple events has an equal chance of occurring. If event A can occur in s of these n ways, then  P(A) = (number of ways A can occur)/(number of different simple events) = s/n

 Examples: Determine the face 2 of a balanced and fair dice; Determine the probability of winning a lottery by selecting 6 numbers between 1 and 60. The probability of winning is 0.0000000200 repeated was trial time of number occured A times of number ) A ( P 

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4-2.4 Rules for Probabilities and Law of Large Numbers

Rule 3 (Subjective Probabilities): The probability P(A) of event A is

estimated by using knowledge of the relevant circumstances

Example:

 possibility of rain for tomorrow (weather forecast)

 Probability of an astronaut surviving a mission in a space shuttle (0.99 for now) relies on the technology and conditions

Rule 1 – use relative frequency approach, approximation only

Rule 2 – classical, need the outcomes to be equally likely

Rule 3 – subjective, educated guess

Law of Large Numbers

 As a procedure is repeated again and again, the relative frequency probability (rule 1) of an event tends to approach the actual probability

 This law tells us the relative frequency approximations tend to get better with more observation (e.g. toss a coin 100 times vs 10 times)

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4-2.5 Examples

Probability and outcomes that are not equally likely ; i.e. if there are only two

possible outcomes, it does NOT imply that the probability is ½

 Examples: (1) Presidential election from two major parties, or pass a course; (2) Car crash (6,511,100 cars crashed among 135,670,000 cars) Thus P(crash) =

6,511,100/135,670,000 = 0.048; classical approach is not suitable, since the two outcomes are not equal likely.

Probability and outcomes that are equally likely

 Genotypes AA, Aa, aA, aa, what’s the probability that you select the genotype Aa?

 Sample space {AA, Aa, aA, aa} includes equally likely outcomes. Thus we can use Rule 2; P(Aa) = ¼

Crushing meteorites: what is the probability that your car will be hit by a

meteorite this year?

 Can you use Rule 1? Or Rule 2? or Rule 3?

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4-2.6 More Examples

 Finding the probability that NBA basketball player Reggie Miller makes a free throw after being fouled. At one point, he made 5915 free throws in 6679 attempts

 Which rule shall we apply?

 What is the simple event? Make a free throw or not

 Sample space is {make a free throw, not make a free throw} are not equally likely; Can’t use rule 2; Can only use rule 1 approach

 P(Miller makes a throw) = 5915/6679= 0.886

 Should cloning of humans be allowed?

 Gallup poll randomly selected adults  91 No, 902 Yes, 20 No opinion

 P(not allowed) = 91/1012=0.0899

 Gender of children, what the probability that when a couple has 3 children, they will have two boys.

 Sample space {mmm, mmf, mfm, mff, fmm, fmf, ffm, fff} – 8 different ways  Outcomes are equally likely; Use rule 2

 P(two boys, 1 girl) = 3/8=0.375

 Example 5: What is the probability of Thanksgiving day will be on (a) Wednesday or (b) Thursday for any year selected randomly?

 Example 6: Probability of a President from Alaska?

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4-2.7 Example 8 Chapter Problem

Probability of a Positive Test Result

Table 4-1 in the Text: Did the Subject Actually Lie?

Assume that 98 test results summarized in the Table 4-1, find the

probability that it is a positive test result.

 Sample space – total number of test results

 P(positive test) = ?

America Online Survey: “Will KFC gain or lose business after eliminating

trans fats?”1941 said gain business, 1260 said “the same”, 204 said lose

business. What’s probability that a randomly selected response states that

KFC would gain business i.e. find P(response of a gain in business) ?

Interpretation? Remember this survey involve a voluntary response sample

because the AOL users themselves decided whether to respond, i.e. do not

necessary reflect the opinions of the general population.

No (Did Not Lie) Yes (Lied)

Positive Test result 15 (false positive) 42 (true positive) Negative Test Result 32 (true negative) 9 (false negative)

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4-2.7 Probability Values

 The probability of an impossible event is 0

 The probability of an event that is certain to occur is 1

 For any event A, the probability of A is between 0 and 1 inclusive 0 P(A)  1

 Definition: The complementary of event A, denoted by A, consists of all outcomes in which event A does not occur.

 Example: In reality, more boys are born than girls. 205 newborn babies, 105 of whom are boys. If one baby is randomly selected from the group, what is the probability that the baby is not a boy?

 P (not selecting a boy) = P (boy) = 100/205= 0.488

 Example: Guessing on an SAT Test, a typical question on the SAT test requires the test taker to select one of five possible choices: A, B, C, D, or E

 What is the probability if you make a random guess and not being correct (or being incorrect)?

 P(not guessing the correct answer) = P (correct) = P(incorrect) =?

0 0.5 1

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4-2.8 Expression of P-value and Odds

 Rounding Off Probabilities

When expressing the value of a probability, either give the exact fraction or decimal or round off final decimal result to three significant digits. If the fraction is a simple fraction, leave it like that.

 Examples: P( ) = 0.021491, round it to 0.0215, 1/3 can leave it or 0.333, ½ can be 0.5 not 0.50, 432/7842 is exact but hard to see, so express it as 0.0551

 Remember 0  P-value  1

 Unusual Event? If the probability is less than 0.001 ( < 0.1%) is considered as an unusual event. We use this as rare event rule for inferential statistics as mentioned earlier. Example 12 in Text for the clinical experiment of the Salk vaccine for polio. 115 cases out of 200,745 with placebo, and 33 out of 201,229 with vaccine. What are the possible interpretations?

The actual odds against event A occurring are the ratios P(not A)/P(A), usually expressed in the form of a:b (or “a to b”), where a and a are integers having no common factor.

The actual odds in favor of event A are the reciprocal of the actual odds against that event, i.e. P(A)/P(not A). If the odds against A are a:b, then the odds in favor of A are b:a

The payoff odds against event A represent the ratio of the net profit (if you win) to the amount bet ; Payoff odds against event A = (net profit): (amount bet)

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4-2.9 Example of Odds, Summary, and Homework #11 (4-2)

If you bet $5 on the number 13 in roulette, your probability of winning is

1/38 and the payoff odds are given by the casino as 35:1 (means (net profit)

: (amount bet) = 35 : 1)

 Find the actual odds against the outcome 13

 How much net profit would you make it you win by betting on 13? (P (not)/P(yes))

 If the casino were operating just for the fun of it, and the payoff odds were changed to match the actual odds against 13, how much would you win if the outcome were 13?

Summary:

Rare event rule for inferential statisticsProbability rule

 Law of large numbers  Complementary event s  Rounding of probabilities  Odds

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4-3.1 Addition Rule for Probabilities

 Objective: learn the addition rule to find the probabilities P(A or B) that either event A occurs or event B occurs (or they both occur) as the single outcome of a procedure.

A compound event is any event combining two or more simple events.

 Notation: P(A or B) = P(event A occurs or event B occurs or they both occur) in a single trial.

 General rule for a compound event –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 the total in such a way that no outcome is

counted more than once.

 Example of addition rule: Chapter Problem, 98 people tested by Polygraph instruments

If a subject is randomly selected, what is the probability of selecting a subject who had a positive test result or lied?

No (Did Not Lie) Yes (Lied)

Positive Test result 15 (false positive) 42 (true positive) Negative Test Result 32 (true negative) 9 (false negative)

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4-3.2 Addition Rules for 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 (inclusive)

Intuitive addition rule

 To find P(A or B)

P(A or B) = (number of ways event A occur + number of ways event B occur(but not in

both A and B)/ The total number of outcomes in the sample space.

That is every outcome is counted only once (in either A or B, if the outcome also in both

A and B, only counted once)

Event A and B are disjoint (or mutually exclusive) if they cannot occur at

the same time.

Example: chapter problem, polygraph testing, determine whether the two

events are disjoint

A: getting a subject with a negative test result B: getting a subject who did not lie

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4-3.3 Venn Diagram for probability

Venn diagram for events that are not disjoint

P(AUB) = P(A) + P(B) –P(A

B)

 Venn diagram for disjoint event

P(A or B) =P (A) + P(B)

Rule of complementary events

 P(A)+P(not A) = 1  P(not A) = 1-p(A)  P(A) = 1-P(not A)

P(A) and P(not A) are disjoint, i.e. it is impossible for an event and its

complement to occur at the same time; correct notation is pronounced as

P(A bar)

Example: Women have a 0.25% rate of red/green color blindness. If a

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4-3.4 Summary and Homework #12 for 4-3

Summary: we have discussed:

Compound events

Formal addition rule

Intuitive addition rule

Disjoint events

Complementary events

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4-4.1 Multiplication Rule: Basics

 Objective: learn the basic multiplication rule for finding the probability that event A occurs in a first trial and even B occurs in a second trial

 P(A or B) is the probability for a single trial that an outcome of A or B or both; In section 4-3, we used P(A and B) to denote that events A and B both occur in the same trial, but in this section 4-4, we use P(A and B) differently; it means:

P(A and B)= P(event A occurs in a first trial and event B occurs in a second trial)

 Example 1: Find the probability for picking A for the first trial, and A for the second trial from same deck of card? (by putting the card A back or not putting

A back; two cases)

 Example 2: Find the probability for answering one true/false choice problem and one multiple choice problem (one out of 5 choices), and you get both right?

A tree diagram is a picture of the possible outcomes of a procedure, shown as line segment emanating from one starting point; example on the board

 Example 3: Use the chapter problem – polygraph test to find the probability that the 1st selected had a positive test result and the 2nd had negative test result, if randomly select 2 subjects from this test without replacement.

 First selection P(positive test result)=?

 Second selection: P(negative test result)=?

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17

4-4.2 Multiplication Rule: Basics

What did we learn from those examples?

 We need to adjust the probability of the second event to reflect the outcome of the first event (the total # of samples)

 Conditional Probability: P(B|A) represents the probability of event B occurring after event A has already occurred. (read it as B given A)

 Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other.

 If the events A and B are not independent, then we say A and B are dependent  Example: outcome of lottery of NY and California, dependent or independent?  Formal Multiplication Rule

(1) P(A and B) = P(A)  P(B|A)

(2) If A and B are independent events, P(B|A) is the same as P(B); i.e. P(A and B) = P(A)  P(B)

 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 of the previous occurrence of event A.

 Use Intuitive Multiplication Rule is recommended and note multiplication rule can be easily extended to several events, e.g. tossing a coin 3 times and getting all heads.  Example– a biologist experiments with a sample of two vascular plants (V) and four

nonvascular plants (N). She wants to randomly select two of the plants for further experimentation. Find the probability that the 1st selected is N and the second is also an N without replacement.

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4-4.3 Examples of multiplication rule

 Example 1: Quality Control in Manufacturing Pacemakers quality: 250,000 implanted in U.S. failure rate is 0.0014 per year.

Case 1: Consider small sample of 5 pacemaker: 3 good ones, 2 defective

A medical researcher wants to randomly select two pacemakers for experimentation. Find the probability that the first selected pacemaker is good and the second one is also good; Assume that two random selections are made with (1) replacement (statistical interest) and (2) without replacement (practical case in medical research)

5% Guideline for small sample from a large population: If a sample size is  5% of population, and if the calculations are cumbersome, then the treat it as being

“independent” (whether or not with replacement); Pollsters use this guideline when they survey small # from a population of millions.

Case 2: Quality Control in Manufacturing; application of 5% guideline

 A batch of 100,000 pacemakers, 99,950 are good, 50 are defective.

(1) If two pacemakers are randomly selected without replacement, find the probability that they both are good. (2) If twenty pacemakers are randomly selected without replacement, find the probability that they are all good.

Solution: 5% of 100,000 is 5000; for (1), sample size is 2 < 5000, but calculation is simple and

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4-4.4 More Examples

Example 2: Assume that two people are randomly selected and also assume

that birthdays occur on the days of the week with equal frequencies.

a. Find the probability that the two people are born on the same day of the week b. Find the probability that the two people are born on Monday

Example 3: Effectiveness of Gender Selection: A geneticist developed a

procedure for increasing likelihood of female babies. In an initial test, 20

couples use the method and the result consist of 20 females among 20 babies.

Assume that the gender-selection procedure has no effect, find the probability

of getting 20 females among 20 babies by chance. Does the resulting

probabilities provide strong evidence to support the geneticist’s claim that the

procedure is effective in increasing the likelihood that babies will be females?

Example 4: Redundancy for increased reliability: to increase reliability is to

use redundancy system like aircraft engines. Assume that the probability of an

electrical system failure is 0.001. If the engine in an aircraft has one electrical

system, what is the probability that it will work? If the engine in an aircraft has

two independent electrical systems, what is the probability that the engine can

function with a working electrical system? Interpretation?

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4-4.5 Summary and HW #13 for 4-4

probability rules

 In addition rule, P(A or B) is addition, the word “or” suggests “addition”. Make sure that every outcome only counted once (if there are common members in event A and B, count only once) : P(A or B)=P(A)+P(B)P(A ∩ B)

 In multiplication rule, P(A and B) is multiplication, the word “and” suggest

“Multiplication”. Make sure the probability of event B take into account the previous occurrence of event A: P(A and B) = P(A)P(B|A) or P(A and B) = P(A) P(B)

Summary

 Notation for P(A and B)  Tree diagrams

 Notation for conditional probability  Independent events

 Formal and intuitive multiplication rules

HW #13, page 167-170, # 5, 7, 13 – 21 odd, 27, 29

(note that problem # 17

Start

P(A and B) Multiplication rule

A and B

Independent? P(A and B)=P(A)P(B) P(A and B)=P(A)P(B|A)

YES NO

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4-5.1 Multiplication rule: complements and conditional

probability

Objective:

1. Find the probability of getting at least one of some specified event;

2. Study the concept of conditional probability which is the probability of an event

given the additional information that some other event has already occurred

Complements:

 The complement of getting at least one (equivalent to one or more) item of a particular type is that you get no items of that type.

 To find the probability of at least one of something, calculate the probability of none, then subtract that result from1. i.e. P(at least one) = 1 – P(none).

E.g.

getting at least 1 girl among 3 children, means getting 1, 2 or 3 girls. The complement is getting no girls, i.e. all 3 children are boys

 Example 1: Find the probability of a couple having at least 1 girl among 3 children.

Assume that boys and girls are equally likely and the gender of a child is independent of any brothers or sisters and its interpretation.

 A= at least one is a girl; Not A = no girls, all boys

 P(not A)=P(boy, boy, boy), and P(A) = 1 – P(not A)

 Interpretation of the example: There is 7/8 probability that if a couple has 3 children, at least one of them is a girl.

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4-5.2 Multiplication rule: Conditional Probability

Conditional Probability P(B|A) of an event B is a probability obtained with

additional information that some other event A has already occurred.

P(A and B) = P(A)

P(B|A), thus we have P(B|A)=P(A and B)/P(A)

Intuitive Approach to conditional probability: the conditional probability of

B given A can be found by assuming that event A has occurred and working

under that assumption, calculating the probability the event B will occur

Example 2: If 1 of the 98 test subjects is randomly selected, find the probability

that the person tested positive is actually lied, i.e. find P(positive | lied) (1) use

conditional probability and (2) use intuitive approach .

If 1 of the 98 test subjects is randomly selected, find the probability that the

person actually lied, and tested positive P(lied | positive)=? Use two approaches

 P(positive test | lied) = 0.824, i.e. subject who lied has 82.4% probability will be tested positive  P(subject lied | positive test) = 0.737, subject who tested positive actually has 73.7% probability

that the subject lied

P(positive | lied)

P(lied | positive), i.e. P(A|B)

P(B|A); demonstrate the fact

called “confusion of the inverse”

P(cancer | positive test)

P(positive test | cancer); but 95% of physicians

estimated P(cancer | positive test) 10 times too high; what the interpretation

here?

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4-5.3 Summary and HW #14 for 4-5

Example: A homeowner finds that there is a 0.1 probability

that a flashlight does not work when turned on.

If she has three flashlights, find the probability that at least one of them

works when there is a power failure.

Find the probability that the second flashlight works given that the first

flashlight works.

Summary: in this section we have discussed

Concept of “at least one”

Conditional probability

Intuitive approach to conditional probability

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4-6.1 Probabilities Through Simulation

Objective: learn a very different approach for finding probabilities that can

overcome much of the difficulty encountered with the formal methods

discussed in the preceding sections of this chapter.

Definition: A

simulation

of a procedure is a process that behaves the same

way as the procedure, so that similar results are produced.

Example 1: When testing techniques of gender selection, medical researchers

need to know probability values of different outcomes, such as the probability

of getting at least 60 girls among 100 children. Assuming that male and

female births are equally likely, describe a simulation that results in genders of

100 newborn babies.

Solution (1) flip the coin 100 times, heads for female, tail for male (2) use a

calculator or computer

Example 2: Swinging Sammy Skor's batting process was simulated to get an

estimate of the probability that Sammy will get a hit. Let 1 = HIT and 0 =

OUT. The output from the simulation was as follows: 0 0 0 1 0 0 1 0 0 1 1 1 0

0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1; Estimate the probability

that he gets a hit.

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4-6.2 Summary, TI-83/84 Calculator and HW #15

Same Birthdays: the probability that in a randomly selected group of 25

people, at least 2 share the same birthday

Solution: begin by representing birthdays by integers from 1 through 365,

Jan 1 is 1, Jan 2

nd

is 2, ….Dec 31 is 365. Use the computer random number

generators to generate the number

Use TI-83/84 Calculator

 Press MATH key, select PRB, then choose randInt, enter the minimum of 1, the maximum of 365, and 25 for the number of values, all separated by

commas, then press STO to L1, press enter to get the set of number, then use STAT SORTA(L1), then press enter. The data now is sorted, use STAT EDIT to view the data in L1.

Summary

 The definition of a simulation

 How to create a simulation

 Ways to generate random numbers (read the book for other ways not mentioned in the lecture)

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4-7.1 Counting

Objective 4-7: Learn different methods for finding numbers of different

possible outcomes without directly listing and counting the possibilities

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.

Example 1: Identity theft: social security number

 If the theft claims that the number was generated randomly, what is the probability of getting your ss# when randomly generated nine digits?

Example 2: Chronological Order: In a history test, arrange the following event

in chronological order:

a)

Boston Tea Party

b)

Teapot Dome Scandal

c)

The civil war

If a student makes random guesses, Find the probability that this student

chooses the correct chronological order?

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4-7.2 Permutations Rule (when items are all different)

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.)

If the preceding requirements are satisfied, the number of permutations (or

sequences) of r items selected from n available items (without replacement) is

Factorial Rule: 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.).

Examples :

 Routing problems like Verizon to route telephone calls through shortest networks  Federal Express wants to find the shortest routes for its deliveries

 American Airlines wants to find the shortest route for returning crew member to their homes

)! ( ! r n n Pr n

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4-7.3 Permutations Rule (when some items are identical to others)

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.

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

 Example 1: In designing a test of a gender-selection method with 14 couples. How many ways can 11 girls and 3 boys be arranged in sequence? That is, find the

number of permutations of 11 girls and 3 boys. (A: 364 different ways)

 Example 2: How many different way to arrange the letters in the word: “arrange”?  Example 3: How many different way to arrange the books in a bookshelf with 3

algebra books, 5 English books, 7 Social Science books according to the subject? ! ! ! ! 2 1 n nk n n   

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4-7.4 Combinations Rule

 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.)

If the preceding requirements are satisfied, the number of combinations of r items selected from n different items is

 Permutations versus Combinations

 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.

 Clinical Trial: when testing a new drug on humans, a clinical test is normally done in 3 phases. Phase 1 is conducted with a relatively small number of healthy volunteers. Assume that we want to treat 8 healthy humans with a new drug and we have 10 suitable volunteers available.

1. If the subjects are selected and treated in sequence, so that the trail is discontinued if anyone

presents with a particular adverse reaction, how many different sequential arrangements are possible if 8 people are selected from the 10 that are available?

2. If 8 subjects are selected from 10 that are available, and the 8 selected subjects are all treated at

the same time, how many different treatment groups are possible?

! )! ( ! r r n n Cr n

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4-7.5 Examples, Summary, and HW #16

Florida Lottery – the Florida Lotto game is typical of state lotteries. You

must select six numbers between 1 and 53. You win the jackpot if the same

six numbers are drawn in any order. Find the probability of winning the

jackpot.

If you need to select 3 of Frank Sinatra’s top-10 songs to be played as a

tribute to Sinatra in a MTV music awards ceremony and the order of the

songs is important so that they fit together well, how many different

sequences are possible?

Summary:

 The fundamental counting rule.

 The factorial rule.

 The permutations rule (when items are all different).

 The permutations rule (when some items are identical to others).

 The combinations rule.

References

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