1

Chapter 5: Probability

5.1 Probability Rules5.2 The Addition Rule and Complements5.3 Independence and the Multiplication Rule5.4 Conditional Probability and the General Multiplication Rule5.5 Counting Techniques

5.1 Probability Rules5.2 The Addition Rule and Complements5.3 Independence and the Multiplication Rule5.4 Conditional Probability and the General Multiplication Rule5.5 Counting Techniques

1November 4, 2008

2

Example

Question: When you submit your 2007 IRS tax return in 2008, what are your chances of having it audited by the IRS if your income is less than $25,000? That is, what is the probability of being audited? That is, what is the chance that you will be audited by the government.

Information: In 1997, 1.5% were audited.

Income Filed Examined< $25K 59,211,700 1,076,945

$25K-$50K 27,263,000 259,794

$50K-$100K 17,019,200 196,582

> $100K 4,540,800 129,320

$0 −25,000 : 107694559211700

=0.018 =1.8%

2007 Tax Auditing

3

How can Probability Quantify Randomness?

Question: What does the word probability mean?

Possible Answer: Probability is a branch of mathematics that deals with calculating the likelihood of a given event's occurrence, which is expressed as a number between 1 and 0. An event with a probability of 1 can be considered a certainty: for example, the probability of a coin toss resulting in either "heads" or "tails" is 1, because there are no other options, assuming the coin lands flat. An event with a probability of .5 can be considered to have equal odds of occurring or not occurring: for example, the probability of a coin toss resulting in "heads" is .5, because the toss is equally as likely to result in "tails." An event with a probability of 0 can be considered an impossibility: for example, the probability that the coin will land (flat) without either side facing up is 0, because either "heads" or "tails" must be facing up. A little paradoxical, probability theory applies precise calculations to quantify uncertain measures of random events.

http://whatis.techtarget.com/definition/0,,sid9_gci549076,00.html

4

Another Definition

Probability is the branch of mathematics that studies the possible outcomes of given events together with the outcomes' relative likelihoods and distributions. In common usage, the word "probability" is used to mean the chance that a particular event (or set of events) will occur expressed on a linear scale from 0 (impossibility) to 1 (certainty), also expressed as a percentage between 0 and 100%. The analysis of events governed by probability is called statistics.

http://mathworld.wolfram.com/Probability.html

5

Randomness

• Randomness is often observed in the outcomes of a response variable in either an observational or experimental study.

• All the possible outcomes are known, but it is uncertain which outcome will occur for any given observation.

• Randomness is the opposite of deterministic where a given input doesn’t always produces the same result.

6

Creating Random Events

• A machine or procedure that produces random events is called a randomizer.

• Examples of Randomizers:– Rolling dice– Wheel of Fortune– Flipping a coin– Drawing a card from a shuffled deck

7

Applets

8

Terminology

The process of rolling the die several times with varying results is called a probability experiment. Each roll of the die or dice is called a trial or outcome or event. The number of times that a certain event (outcome) occurs divided by the total number so trials is called the cumulative proportion or relative frequency of the probability experiment. Suppose that in rolling one die 200 times, the number 2 occurs 45 times. The cumulative proportion of this event is 45/200 = 0.225.

9

Events and Sample Space

Definition: A simple event is an outcome from a probability experiment that is observed on a single repetition of the experiment. The sample space of a probability experiment is the set of all possible simple events from the probability experiment. An event is a collection of simple events; in other words, it is a subset of the sample space. An event that consists of more than one outcome is called a compound event.

10

Examples

Probability Experiment Sample Space

Coin Toss {H,T}

Roll a single Die {1,2,3,4,5,6}

True/False Quiz Question {T,F}

Roll two Dices {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),…,(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

11

Simple and Compound Events

Probability Experiment Sample Space Simple Event Compound Event

Roll two Dice {(1,1),(1,2),…,(6,6)} Die1 = 1

Die2= 4

{(1,4)}

Sum of Dice is 5{(1,4),(4,1),(2,3),(3,2)}

Choose a single Card* {(A,H),(2,H),(3,H),…,(A,S)(2,S),…,(K,S)}

Ace of Hearts{(A,H)}

Queen{(Q,H),(Q,D),(Q,C),(Q,S)}

* H = heart, D = diamond, C = club, S = spade

12

Examples

Probability Experiment Event Sample Space

Birth of a single child:Male(m) or female(f)

male {m} is a simple event.

S = {m,f}

Three births 2 males and 1 female is not a simple event.

{mmf,mfm,fmm} are each simple events.

S = {mmm,fff,ffm,mmf,mff,mfm,fmf,fmm}

Note: The event, 2 males and 1 female, is not simple, because it can happen as mmf, mfm or fmm. We would call this a compound event.

13

Probability

Definition : Suppose that a probability experiment has n equally likely outcomes. Let E be

an event for this probability experiment and suppose that this event can occur m times.

Then the probability of the event E, denoted P(E), is defined as P(E) =mn.  In other words,

if N(E) denotes the number of outcomes in E and N(S) is the number of outcomes in the sample

space, then P(E) =N(E)N(S)

.

Note:   Since 0 ≤m≤n, we have 0 ≤P(E) ≤1.

Remark : The probability of an event E can be approximated by the relative frequency of E

occuring in a probability experiment. That is, we perform k trials of the experiment and count

how many times that E occurs. Then relative  frequency of  E = fE =frequency of E

number of trials≈P(E).

14

Remark

If the relative frequencies of events are known in a population, then the probability of these events are exactly the relative frequencies. For example, if we know that a bag contains 2 red balls, 3 green balls, 4 white balls and 1 blue ball, then the probability of randomly selecting a red ball from the bag is the relative frequency of red balls i.e., 2/10 = 1/5 or 0.2. If the relative frequency of an event in a population is unknown, then it can be approximated by simulating the event.

15

Rules of Probabilities

Let E denote an event and P(E) the probability that the event will occur. Then

1. 0 ≤P(E) ≤1

2.  Let  E1,E2 ,...,En{ }  denote all of the possible events in a probabiltiy experiment i.e.,

the sample space of the eperiment is  E1,E2 ,...,En{ } .  Then 

P(Eii=1

n

∑ ) =P(E1) + P(E2 ) + ...+ P(En) =1.

16

Probability Model

Definition: A probability model is a table that lists all possible outcomes of a probability experiment and their probabilities. The form of this table is shown below.

Outcome Probability

17

Example (probability model)

The following table contains the probabilities of choosing different color balls out of a bag. Could this be a probability model for the experiment?

Ball Color Probability

red 0.25

yellow 0.35

white 0.20

blue 0.25

E ={red,yellow,white,blue} = E1,E2 ,E3,E4{ }

P Ei( )i=1

4

∑ =P(E1) + P(E2 ) + P(E3) + P(E4 ) =0.25 + 0.35 + 0.20 + 0.25 =1.05 ≠1.0

18

Example (probability model)

Suppose that we have a bag of jelly beans in six colors (green, orange, blue, red, yellow and brown). We have determined by performing a number of trials that the probability of picking a particular color is:

Color Probability

green 0.16

orange 0.20

blue 0.24

red 0.13

yellow 0.14

brown 0.13€ 

S = green,orange,blue,blue,red,yellow,brown{ }

P green( ) = 0.16

P red( ) = 0.13

P green( ) + P orange( ) + P blue( ) + P red( ) + P yellow( ) + P brown( )

= 0.16 + 0.12 + 0.24 + 0.13+ 0.14 + 0.13 =1.0

Note: This last equation is the probability of picking a jelly bean that is of the colors green or orange or blue or … .

19

Example

Probability Experiment: Gender of children when a couple has three children.

Event: Exactly 2 boys among the three children. E = 2 boys in 3 births.

Sample Space: S = {fff, ffm, fmf, fmm, mff, mfm, mmf, mmm}. n = 8.

Event for 2 boys: {fmm, mfm, mmf}. m = 3.

Probability: P(E) = m/n = 3/8 = 0.375.

20

Law of Large Numbers

As the number of trials of a random event, which we denote by N , increases, the

relative frequency of the event E, which we denote by fE , approaches the probability

of the event, P(E). In mathematical language, limN→ ∞

fE =P(E).

21

Two Types of Probability

1. Classical Method: Using mathematics to compute the exact probability of an event.

2. Empirical Method: Approximate the probability of an event by using a probability experiment and using the relative frequency of the event to estimate the exact probability

22

Example (Empirical)

Find the long term number for heads in the probability experiment of flipping a coin.

S = {h,t}.

N = 10. Trial = {t, h, t, h, h, t, t, t, h, t}. fh = 4/10 = 0.4

N = 50. Trial = {h, h, h, h, t, t, h, t, h, t, h, t, t, h, t, t, t, t, h, h, t, t, h, t, t, h, h, h, t, t, h, h, h, h, t, t, t, t, h, t, t, h, h, h, t, h, h, h, t, t}. fh = 25/50 = 0.5

N = 100. Trial = {h, h, h, t, h, t, t, t, h, t, t, h, h, h, h, h, t, t, t, t, h, t, t, t, t, t, h, h, t, t, h, h, t, h, h, t, h, t, t, h, t, h, h, h, t, h, t, h, t, h, t, h, t, h, h, t, t, h, t, t, h, h, t, h, h, t, t, t, t, h, t, h, t, t, t, t, t,

fh= 48/100 = 0.48

As N >> 1, fh approaches 1/2.

h, h, t, t, t, h, h, t, h, h, t, t, t, h, h, t, h, h, h, t, t, h, h}.

23

Caution

Over the “short run”, we do not expect the proportion of a particular event to the same as the proportion in the “long run.” For example, if I flip a coin 6 times, then I do not necessary expect that “heads” will occur exactly 3 times.

24

Examples (classical)

Find the probability of select a heart out from a deck of cards each of which have equal chance of being selected.

E = {heart}

P(E) = 13/52 = 1/4

Find the probability of selecting a red card from a deck of cards each of which have equal chance of being selected.

E = {red card}

P(E) = 26/52 = 1/2

Find the probability of selecting a face card from a deck of cards each of which have equal chance of being selected.

E = {face card}

P(E) = 12/52 = 3/13

25

Example (classical)

The West Meade Golf Shop sells used golf balls. This past Saturday I went to the shop and was told that the bag of used golf balls coned 35 Titlists, 25 Maxflis, and 20 Top-Flites. I was told that I could reach into the bag and select a ball. What is the probability that I would select a Titleist?

Solution: Let E = {Titleist}. P(E) = 35/(35+25+20) = 35/80 = 0.4375.

26

Example (classical)

Suppose a probability model has the following sample space:

S = {1,2,3,4,5,6,7,8,9,10}

i.e., there are ten possible outcomes in the probability experiment.

(a) Compute the probability of the event of selecting three numbers with the outcome of {3,4,7}

(b) Compute the probability of the event of selecting one number with the outcome of it being an even integer.

Answers: (a) P(E) = 3/10; (b) P(E) = 1/2.

27

Example

Example: Three question quiz.

Problem: Find the possible outcomes for the student taking this three question quiz.

Sample Space: {CCC,CCI,CIC,CII,ICC,ICI,IIC,III}

Question: Chances of getting all three questions correct? Note that is a simple event.

Answer: 1/8 = 0.125

28

Question: What are the chances of getting one out of the three questions correct?

Answer: If A = {CII,ICI,IIC}, then P(A) = 3/8 = 0.375. Note that A is a compound event.

29

Probability Estimates from Survey Data

200 Vanderbilt students were surveyed about their main recreational habits: {listening to music, watch television, playing cards, exercising, other}. The following table summarizes the frequencies and relative frequencies in the survey.

Recreation Frequency Relative Frequency

music 75 3/8

TV 50 1/4

card 15 3/40

exercise 35 7/40

other 25 1/8

The probability that a Vanderbilt student watches TV for his or her main recreation is approximately 0.25.

30

Tree Diagram

• An nice way of visualizing a sample space with a small number of outcomes.

• As the number of possible outcomes for each trial increases, the tree diagram becomes impractical.

31

Tree Example

S = mmm,mmf ,mfm,mff , fmm, fmf , ffm, fff{ }Simple Event: mmfNon-simple Event: Two girls and one boy {mff , fmf , ffm}

Probability of two girls and one boy: P(E) =38=0.375

32

Example

Background: An experimental study by the University of Wisconsin to determine if Echinacea is an effective treatment for the common cold.

Medical Experiment Multi-center randomized experiment Half of the volunteers are randomly chosen to receive the

herbal remedy and the other half will receive the placebo Clinic in Madison, Wisconsin has four volunteers

• Two men: Jamal and Ken

• Two women: Linda and Mary

• Probability Experiment Randomly pairing the four volunteers

Sample Space to receive the herbal remedy:

{(Jamal, Ken), (Jamal, Linda), (Jamal, Mary), (Ken, Linda), (Ken, Mary), (Linda, Mary)}

33

Assumption: The six possible outcomes in this sample space for receiving the Echinacea are equally likely. Hence, the probability that any simple event in the sample space will occur is 1/6.

Hence, if S = {(Jamal, Ken), (Jamal, Linda), (Jamal, Mary), (Ken, Linda), (Ken, Mary), (Linda, Mary)}, then the probability of picking Ken and Linda is 1/6.

The probability of picking one man and one women in a simple event is the probability of picking (Jamal, Linda) or (Jamal, Mary) or (Ken, Linda) or (Ken, Mary) i.e., 4(1/6) = 2/3.

The probability of picking a simple event containing only women is (Linda, Mary) which has the probability 1/6.

34

Example

Problem: Suppose we roll two dice (#1 & #2) once. What is the probability that the sum of the numbers on the dice is 7?

Sample Space: Let (x,y) denote the ordered pair where x is the number form die #1 and y is the number from die #2. Then

S = {(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)}

and N = 36.

Event: E = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}

Probability: P(E) = 6/36 = 1/6.

35

Problem: What is the probability of rolling “box cars?”

Event: E = (6,6) (simple event)

Probability: P(E) = 1/36

36

Example

Problem: What are the chances of a taxpayer being audited by the IRS in 2003?

Solution: The problem can be solved with a contingency table for the audits according to income level. We can compute the relative frequencies of being audited for each income level.

37

Sample Space: We define the sample space to be ordered pairs (x,y) where x is the income range and y is yes (audited) or no (not audited). For x we introduce some notation:

< $25K x = 1

$25K-$49.999K x = 2

$50K-$99.999K x = 3

$100K < x = 4.

S = {(1,yes),(1,no),(2,yes),(2,no),(3,yes),(3,no),(4,yes),(4,no)}

38

• Probability of being audited (any income): 310/80200 = 0.004 or 0.4%

• Probability of being audited and making an income over $100K: 80/10700 = 0.007 or 0.7%

• Probability of being audited and making an income less than $25K: 90/14100 = 0.006 or 0.6%.

• Probability of not being audited and making an income less than $25K: 14010/14100 = 0.994 or 99.6%.

Remark: In this example we did not use the sample space, but rather a contingency table of incomes and audits.

Income/Audited Yes No Total

< $25K (1) 90 14,011 14,100

$25K-$49.999K (2) 71 30,629 30,700

$50K-$99.999K (3) 69 24,631 24,700

≥ $100K (4) 80 10,620 10,700

Total 310 79,890 80,200

39

Subjective Probability

The probability of an event that is obtained on the basis of personal judgment is called subjective probability. This type of calculation is the opposite of objective probability (for example, empirical probability calculations).

Example: What was the probability of landing a man on the moon in the 1960’s? Any estimate of this probability would be subjective since we have no prior history of the event.

40

The Addition Rule and Complements

Section 5.2

Definition: Two events in a probability experiment are said to be disjoint if they have no common outcomes. Another term for the same concept is mutually exclusive i.e., both events cannot happen simultaneously.

Suppose that A and B are events. If they are disjoint, then the probability and A and B happening is zero i.e., P(A and B) = 0.

41

Intersection and Union of Sets

We need a little information about working with sets. In particular, we define the notions of

the intersection and union of two sets. Let A and B be two sets: A = a1,a2 ,K ,aN{ }  and

B= b1,b2 ,K ,bN{ } .  The union of A and B is the set:A∪ B= a1,a2 ,K ,aN ,b1,b2 ,K ,bN{ } . The

intersection of A and B is the set: A∩ B= α :α ∈A, α ∈B{ }  i.e., it is the set of objects that belong to both sets.  Two sets are said to be disjoint if they have no common elements

i.e., A∩ B=∅ (the so-called empty set).

€ 

Example : A = {a,1,b,2} B = {a,3,c}⇒ A∪B = {a,b,c,1,2,3} A ∩ B = {a}

€ 

Example : A = {a,b,c} B = {d,e}⇒ A∪B = {a,b,c,d,e} A ∩ B =∅ . Hence, A and B

are disjoint sets.

42

Venn Diagrams

http://kt2.exp.sis.pitt.edu:8080/venn/andor.jsp

There is a graphical way of looking at the intersection and union of sets. They are called Venn Diagrams.

43

Picture of Intersection/Union

44

Complement of a Set

45

Sets and EventsSuppose that A is a set of events (possibly a single event in the sample space) and B is another set of events (again, possibly a single event) in a common sample space. We consider A and B to be subsets of the sample space. We can perform the set operations intersection and union. For example, the union of A and B is set of events that arise in A or B. The intersection of A and B is the set of events that is common to both events. Similarly, we can talk about the complement of Ac i.e., the set of events that are not in A, but are in the sample space.

46

Union/Intersection of Events

• The union of two events A and B is the new event consisting of events that are either in A or B.

• The intersection of two events A and B is the new event consisting of events that are in A and B.

47

Sets, Events and Probabilities

Suppose that we want to calculate the probability that a particular event in the same space will occur. For example in our three question pop quiz illustration, what is the probability that a student will have two and only two correct answers to the three questions. As we already know, the sample space can be viewed as a set and events are subsets of the sample space. That is, we have the event: A = {CCI,CIC,ICC}. Notice that the event B = {CII,ICI,IIC} is a disjoint event from the event A. In fact, A is disjoint from its complement:

Ac = {CCC,CII,CIC,IIC,III}.

The event, C, of getting one or two correct answers on the quiz is given by the set C which is the union of A and B i.e.,

C = {CII,ICI,IIC,CCI,CIC,ICC}.

We would like to determine the probability of the event C by using information about the probabilities of events A and B.

48

Notation

Let A and B be events (sets in the sample space). We introduce the notation:

A ∪ B↔ A or B  and A∩ B↔ A and B.That is we associate the mathematical symbol ∪  with the word or and the symbol ∪with the word and.  In particular, P(A∪ B)↔ P(A or B) and P(A∩ B)↔ P(A and B).

49

Disjoint Events

Let A and B be two disjoint events i.e., A ∩ B=∅={}.  Then P(A and B) =0.That is, the probability of two disjoint events is zero which makes senses since

they do not share any simple events. Disjoint events are also called  mutually exclusiveevents.

50

Probability of A or B for Disjoint Events

Theorem : Let A and B be disjoint (i.e., mutally exclusive) events. Then

P(A or  B) =P(A) + P(B).  Furthermore, if A1,A2 ,...,Ak are disjoint events,

then P(A1 or A2 or ... or Ak) = P(Ai ).i=1

k

∑

51

Example

Consider a deck of 52 cards (spades, hearts, clubs and diamonds; 2,3,…,10,J,Q,K,A). Consider the problem of drawing one card from this deck. The sample space has 52 simple events (drawing one of the cards). We characterize the sample space as ordered pairs: (count,suit) e.g., 10 of

diamonds is (10,D).

Question: What is the probability of drawing the ace of diamonds?

Answer: P(A,D) = 1/52 = 0.019 .

Question: What is the probability of a king?

Answer: P(king) = P(K,H) + P(K,D) + P(K,C) + P(K,S) = 4/52 = 1/13 = 0.077

Note that (K,H), (K,D), (K,C) and (K,S) are mutually exclusive events.

52

Question: What is the probability of drawing a heart?

Answer: P(heart) = P(A,H) + P(2,H) +…+ P(K,H) = 13/52 = 1/4 = 0.25

Note that (A,H),(2,H),…,(K,H) are mutually exclusive events.

Question: What is the probability of drawing a king or queen?

Answer: P(king or queen) = P(king) + P(queen)

P(king or queen) = 4/52 + 4/52 = 8/52 = 2/13 = 0.154

53

Probability of A or B for any two Events

Theorem : Let A and B be any events. Then P(A or  B) =P(A) + P(B)−P(A and B).  Furthermore, if A1,A2 ,...,Ak are any events, then 

P(A1 or A2 or ... or Ak) = P(Ai )−P(A1 and A2 and ... and Ak).i=1

k

∑

54

Example

Consider a deck of 52 cards (spades, hearts, clubs and diamonds; 2,3,…,10,J,Q,K,A). Consider the problem of drawing one card from this deck. The

sample space has 52 simple events (drawing one of the cards).

Question: What is the probability of drawing the ace or a diamond?

Answer: Let A = ace and B = diamond. These are not disjoint events.

P(A) = 4/52 = 1/13, P(B) = 13/52

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

P(A and B) = P(ace and diamond) = 1/52

Hence, P(ace or diamond) = 1/13 + 13/52 - 1/52 = (4+13-1)/52 = 16/52 = 4/13

55

Example

Two hundred and fifty Vanderbilt students were analyzed for the IQ and their ability to do a certain mathematical puzzle.

The results are summarized in the following contingency table.

Puzzle/IQ Average(90-120)

High(>120 )

Couldn’t do Puzzle

75 30

Could do Puzzle

20 125

56

Let A = high IQ. Then P(A) = (30+125)/250 = 155/250 = 0.62

Let B = could. Then P(B) = (20+125)/250 = 145/250 = 0.58

Question: What is the probability that a student has a high IQ or could do the puzzle?

Answer: P(A or B) = P(A) + P(B) - P(A and B) = 0.62 + 0.58 - ?

From the table, P(A and B) = 125/250 = 0.50

Hence, P(A or B) = 0.62 + 0.58 - 0.50 = 0.70

Puzzle/IQ Average(90-120)

High(>120 )

Total

Couldn’t do Puzzle

75 30 105

Could do Puzzle

20 125 145

Total 95 155 250

57

Another Approach

Let us set the situation as ordered pairs:

S = {(couldn’t,average),(couldn’t,high),(could,average),(could high)}.

We would like to find the probable of choosing the ordered pair: (could, high). There are 250 order pairs. We want to calculate the probability of choosing ordered pairs {(could,---),(---,high)}. Then

P[(could,---)] = (20+125)/250 = 145/250 = 0.58

P[(---,high)] = (30+125)/250 = 155/250 = 0.62

P[(could,high)] = 125/145 = 0.50

Then

P[(could,---) or (---,high)] = 0.58 + 0.62 - 0.50 = 0.70

58

Example (gender/marriage)

Consider the following contingency table for present marital status and gender of people in the U.S. over the age of 18 in 2003.

Males Females Total

Never Married 28.6 23.3 51.9

Married 62.1 62.8 124.9

Widowed 2.7 11.3 14.0

Divorced 9.0 12.7 21.7

Total 102.4 110.1 212.5

We want to calculate the probability that a person in the U.S. over the age of 18 has some particular characteristic. We use the relative frequencies from this table to compute the probabilities.

59

Question: What is the probability that a person in this census is a female?

Answer: P(female) = 110.1/212.5 = 0.518

Question: What is the probability that a person in this census is widowed?

Answer: P(widowed) = 14.0/212.5 = 0.066

Question: What is the probability that a person in this census is a widowed or divorced?

Answer: P(widowed or divorced) = P(widowed) + P(divorced) - P(widowed and divorced)

P(widowed or divorced) = 14.0/212.5 + 21.7/212.5 - 0 = 0.168

Males Females Total

Never Married 28.6 23.3 51.9

Married 62.1 62.8 124.9

Widowed 2.7 11.3 14.0

Divorced 9.0 12.7 21.7

Total 102.4 110.1 212.5

60

Question: What is the probability that a person in this census is a married or female?

Answer: P(married or female) = P(married) + P(female) - P(married and female)

P(married or female) = 124.9/212.5 + 110.1/212.5 - 62.8/215.5 = 172.2/212.5 = 0.799

Question: What is the probability that a person in this census is a male or divorced?

Answer: P(male or divorced) = P(male) + P(divorced) - P(male and divorced)

P(male or divorced) = 102.4/212.5 + 21.7/212.5 - 9.0/212.5 = 115.1/212.5 = 0.542

Males Females Total

Never Married 28.6 23.3 51.9

Married 62.1 62.8 124.9

Widowed 2.7 11.3 14.0

Divorced 9.0 12.7 21.7

Total 102.4 110.1 212.5

61

Complement of an Event

Definition : Suppose E is an event in a sample space S. The complement of

the event E, denoted E c , is the set of all events in S that do not include E.

Example: Three question quiz.

S = {CCC,CCI,CIC,CII,ICC,ICI,IIC,III}

Let E be the event of having one correct answer:

E = {CII,ICI,IIC}.

Then

Ec = {CCC,CCI,CIC,ICC,III}.

62

Probability of the Complement of an Event

Theorem : If P(E) is the probability of an event E and P(E c ) is the

probability of the complement of E, then P(E c ) =1−P(E).

Example: Three question quiz. Find the probability of not having one, and only one, correct answer on the quiz.

Let E be the event of having one correct answer, E = {CII,ICI,IIC}, then Ec = {CCC,CCI,CIC,ICC,III}. If each event in the sample space is equally likely, then P(E) = 3/8 and P(Ec) = 5/8 = 1 - 3/8.

63

Example

The following table shows the relative frequencies of the size of farms in the U.S.

Size in Acres (x) Relative Frequency

x < 10 0.084

10 ≤ x < 50 0.265

50 ≤ x < 100 0.161

100 ≤ x < 180 0.149

180 ≤ x < 260 0.077

260 ≤ x < 500 0.106

500 ≤ x < 1000 0.076

1000 ≤ x < 2000 0.047

2000 ≤ x 0.035

(1) What is the probability that a farm in the U.S. will be between 100 and 500 acres?

(2) What is the probability that a farm will be greater than or equal to 10 acres?

(1) P(100 ≤ x <500) = P(100 ≤ x <180) + P(100 ≤ x <180) + P(180 ≤ x <260) + P(260 ≤ x <500)

= 0.149 + 0.077 + 0.106 = 0.332

(2) P(x ≥ 10) = 1 - P(x < 10) = 1 - 0.084 = 0.916

64

Example

Roulette consists of a wheel with 38 slots, numbered 0,1,2,…,36,00 i.e., 38 slots. The odd-numbered slots are red and the even-numbered slots are black. The slots, 0 & 00, are colored green. The wheel is spun and a metal ball circles the wheel until it lands in a numbered slot.

(a) What is the probability that the ball lands on a green or red slot?

(b) What is the probability that it does not land in a green slot?

Let A = event of landing on red slot, B = event of landing on green slot. Then P(A) = 18/38 = 9/19 and P(B) = 2/38 = 1/19.

• Then P(A or B) = P(A) + P(B) - P(A and B) = 9/19 + 1/19 - 0 = 10/19

(b) P(Bc) = 1 - P(B) = 1 - 1/19 = 18/19.

65

Independence and the Multiplication Rule

Definition: Two events A and B are called independent events if the fact that A occurs does not affect the probability of B occurring and vice-versa. When the probability of A effects the probability of B, then we say that the events are dependent.

Examples: Independent

(a) Rolling a die and getting a 4 and then rolling it a second time and getting a 2.

(b) Drawing a card from a deck and get a 10 of diamonds and then replacing the card, shuffling and then drawing a 10 of diamonds again.

(c) Suppose a patient at the Vanderbilt Hospital is selected at random from a group of patients. Let A be the event that the patient has atherosclerosis. Let B be the event that the patient is a smoker. A and B are not independent i.e., they are dependent since smoker has a high incidence of heart disease.

Section 5.3

66

Dependent Events

Examples: Dependent Events

(a) Being a lifeguard and getting a suntan.

(b) Parking in a no-parking zone and getting a parking ticket.

(c) Being a Vanderbilt student and getting a good education.

67

Remark

Disjoint events and independent events are different concepts. Two events are disjoint if they one event occurs, then the other one will not happen. Two events are independent if the probability of one event does not event the probability of the other event. Hence, if two events are disjoint, then they cannot be independent

68

Probability of A and B for Independent Events

Let A and B be two independent events in a sample space. Then

P(A and B) = P(A)P(B).

Note: If A and B are disjoint events, then P(A and B) = 0.

69

Example

Suppose we flip a coin twice in a row. What is the probability that we will see two heads?

Sample Space: S = {h,t}.

Note that the events are independent.

Let A = {h} and B = {h}. Then P(A and B) = P(A)P(B) = (0.5)(0.5) = 0.25 .

Question: Is this the same as asking the probability of the event: given the first flip is head, then the second flip is also a head?

70

Example

Suppose we have a two question pop quiz. We perform a probability experiment and find the following data about the outcome of taking the quiz.

Outcomes II IC CI CC

Probabilities 0.26 0.11 0.05 0.58

Suppose that we consider the events (answering the questions) are not independent. For example, if you answer the first question correct, the probability of answer the second question correct is not necessarily the same as if you had answered the first question incorrectly.

Let A = event that the first question is answered correctly, irregardless of the answer to the second question.

Let B = event that the second question is answered correctly, irregardless of the answer to the first question.

71

P(A) = P(CI or CC) = P(CI) + P(CC) = 0.05 + 0.58 = 0.63

P(B) = P(IC or CC) = P(IC) + P(CC) = 0.11 + 0.58 = 0.69

P(A and B) = P(CC) = 0.58

If A and B were independent, then P(A and B) = P(A) P(B) = (0.63)(0.69) = 0.43

Notice that we computed different values for P(A and B), depending on the whether we assume independence or not of the two events.

Outcomes II IC CI CC

Probabilities 0.26 0.11 0.05 0.58

72

Example

The E.P.T. Pregnancy Test states that the test is “99% accurate in detecting typical pregnancy hormone levels.” Suppose that we randomly select 12 pregnant women.

(a) What is the probability that all 12 of them will test positively?

(b) What is the probability that at least one will not test positively?

Let A = event that one will test positively. Hence, P(A) = 0.99

(a) P(all test positively) = P(A and A and … and A) = P(A)12 = (0.99)12 = 0.886385

(b) P(at least one test negatively) = 1 - P(all test positively) = 0.113615

73

Caution

Don’t assume that events are independent unless you have given this assumption careful thought and it seems plausible.

74

Summary

Summary of the Laws of Probability

Let A and B be events (simple or compound) in a sample space: S ={E1,E2 ,...,En}1.  0 ≤P(A) ≤1

2.   P(Eii=1

n

∑ ) =1

3.  If A and B are disjoint events, then P(A or B) =P(A) + P(B).4.  If A and B are any events, then P(A or B) =P(A) + P(B)−P(A and B).

5.  P(Ac) =1−P(A).6.  If A and B are independent events, then P(A and B) =P(A)P(B).7. If A and B are disjoint events, then P(A and B) =0.

75

Conditional Probability and the General Multiplication Rule

Section 5.4

Suppose that we have two events, A and B.

(a) P(A or B) = P(A) + P(B) - P(A and B)

(b) If they are independent events, then P(A and B) = P(A)P(B).

Recall that two events are independent if the the probability of A is independent of B and vice-versa.

Question: What happens if the events are not independent i.e., they are dependent. That is, the probability of one dependents on the probability of the other.

Definition: The symbol P(A|B) means the probability of an event A given that an event B has occurred. This probability is called a conditional probability.

76

Conditional Events

P(E|F): the probability of E given F

In other words, the probability of E, given the F has already happened.

77

Condition Probability Formula

€ 

Let A and B be events. The conditional probability of A, given B, is given by

P(A | B) =P(A and B)

P(B). Furthermore, P(B | A) =

P(A and B)

P(A).

78

Example

Twenty-five percent of Vanderbilt professors in the age range 50-60 years old have hypertension. In this same group, five percent also have diabetes. Given that an individual in this age group has hypertension, what is the probability that he or she will also have diabetes?

A =event of having hypertension

B=event of having diabetes

P(A) =0.25P(A and B) =0.05

⇒ P(B |A) =P(A and B)

P(A)=0.050.25

=15=0.20

79

Example

Income/Audited Yes No Total

< $25K 90 14,011 14,100

$25K-$49.999K 71 30,629 30,700

$50K-$100K 69 24,631 24,700

$100K < 80 10,620 10,700

Total 310 79,890 80,200

Contingency Table for IRS filers.

Probability Table for IRS filers.

Income/Audited Yes No Total

< $25K 0.0011221 0.1747007 0.1758104

$25K-$49.999K 0.0008852 0.3819077 0.3827930

$50K-$100K 0.0008603 0.3071119 0.3079800

$100K < 0.0009975 0.1324189 0.1334164

Total 0.0038653 0.9961346 1.000

Probabilities of audit for different income classes:

P(yes and <$25K) = 90/80200 = 0.0011221

P(no and <$25K) = 14011/80200 = 0.1747007

P(yes and $25K-50K) = 71/80200 = 0.0008862

P(yes) = 310/80200 = 0.0038653 (overall)

S = {(<$25K,yes),(<$25K,no),($25-50K,yes), ($25-50K,no),($50-100K,yes),($50-100K,no),($100K<,yes),($100K<,no)}.

P[(<$25K,yes)] = 0.0011 P[($100K<,no)] = 0.1324

The sum of events in the sample space: 0.0011+0.1747+0.0009+0.3819+0.0009+0.3071+0.001+0.1324 = 1.0

Here we consider two dependent events with the events being income level and the other being audited.

80

Question: Let A be the event {audit=yes} and let B be the event {income>$100K}. Find P(A|B) i.e., the probability that an individual will be audited, given that his or her income is greater than $100K.

Answer: We note that P(A and B) = P[(yes,$100K<)] = 0.0009975. Furthermore, the probability that taxpayer has an income greater than $100K is P($100K<) = 0.1344164. Therefore,

P(yes | $100K<) = P[(yes,$100K<)] /P($100K<) = 0.0009975/0.1334164 = 0.0074766

Income/Audited Yes No Total

< $25K 0.0011221 0.1747007 0.1758104

$25K-$49.999K 0.0008852 0.3819077 0.3827930

$50K-$100K 0.0008603 0.3071119 0.3079800

$100K < 0.0009975 0.1324189 0.1334164

Total 0.0038653 0.9961346 1.0

P(A | B) =P(A and B)

P(B)

Probability Table for IRS filers (from previous page)

81

Income/Audited Yes No Total

< $25K 0.0011221 0.1747007 0.1758104

$25K-$49.999K 0.0008852 0.3819077 0.3827930

$50K-$100K 0.0008603 0.3071119 0.3079800

$100K < 0.0009975 0.1324189 0.1334164

Total 0.0039 0.9961 1.0

With some rounding, we can form a conditional probability table.

Income/Audited Yes No Total

< $25K 0.006 0.994 1.000

$25K-$49.999K 0.002 0.998 1.000

$50K-$100K 0.003 0.997 1.000

$100K < 0.007 0.993 1.000

Given income is less than $25K,

0.0011221/0.1758104 = 0.0062

0.1747007/0.1758104 = 0.9937

Given income is greater than $100K,

0.0009975/0.1334164 = 0.0074

0.1324189/0.1334164 = 0.9925

P(A | B) =P(A and B)

P(B)

P(A and  B)

A =audit

B=income

Probability Table for IRS filers (from previous page)

82

Remark

Income/Audited Yes No Total

< $25K 90 14,011 14,100

$25K-$49.999K 71 30,629 30,700

$50K-$100K 69 24,631 24,700

$100K < 80 10,620 10,700

Total 310 79,890 80,200

We cam use the contingency table directly i.e., without finding the conditional proportions.

P(yes | $100K <) =

P(yes and $100K <)P($100K <)

=80 / 80200

10700 / 80200=

8010700

≈0.0074766

83

Example

A survey asked 100 people their opinions about gender and combat in the military. Here is the results of this survey.

Gender/Combat Yes (y) No (n) Total

Male (M) 32 18 50

Female (F) 8 42 50

Total 40 60 100

Question: What is the probability the answer is yes, given this it is a female?

Answer: P(y|F) = P(y and F)/P(F) = (8/100)/(50/100) = 8/50 = 4/25 = 0.16

Question: What is the probability the subject is male, given the answer it is no?

Answer: P(M|n) = P(M and n)/P(n) = (18/100)/(60/100) = 18/60 = 3/10 = 0.30

P(y and F) = 8/100

P(n and M) =18/100

84

Example

Problem: Analyzing the Triple Blood Test for Down Syndrome. Take blood from a pregnant woman and perform the biochemical analysis (see if there is an extra copy of chromosome 21). The test can be either positive or negative. Unfortunately, it is not always accurate.

Results of Test:

1. True Positive: Test is positive and baby has extra chromosome.

2. False Positive: Test is positive and baby does not have extra chromosome.

3. True Negative: Test is negative and baby does not have extra chromosome.

4. False Negative: Test is negative and baby has extra chromosome.

85

Study: 5,282 women of age 35 or older took the Triple Blood Test and after they had their child, the accuracy of the test was analyzed. Here is a contingency table for the study.

Syndrome/Test Positive (p) Negative (n) Total

Yes (D) 48 6 54

No (Dc) 1307 3921 5228

Total 1355 3927 5282

P(p) = 1355/5282 = 0.2565 P(n) = 3927/5282 = 0.7434

P(D and p) = 48/5282 = 0.0091 P(D and n) = 6/5282 = 0.0011

P(D|p) = P(D and p)/P(p) = 0.0091/0.2565 = 0.035

P(D|n) = P(D and n)/P(n) = 0.0011/0.7434 = 0.0015

86

Conclusion: If the sample of 5,282 is representative of all women who take the test, then for women who test positively, only approximately 4% [P(D|p)] of the fetuses have Down Syndrome. However, there is 0.9% chance of the fetus having the disease and 0.1% chance of a negative test still having the syndrome [P(D|n)]. Hence, a negative test is a good indicator of the disease not be present.

Syndrome/Test Positive (p) Negative (n) Total

Yes (D) 0.009 0.001 0.010

No (Dc) 0.247 0.742 0.990

Total 0.257 0.743 1.0

Probability Table: P(A), P(B), P(A and B)

87

ExampleThe following table show the results of a study of 137,243 men in the U.S. The study investigated

the association between cigar smoking and death from cancer.

(a) What is the probability that a randomly select man form the study who died from cancer was a former cigar smoker?

(b) What is the probability that a randomly select man who was a former cigar smoker died from cancer?

Died from Cancer

Did not die from cancer

Total

Never smoked cigars

782 120,747 121,529

Former cigar

smoker91 7,757 7,848

Current cigar

smoker141 7,725 7,866

Total 1,014 136,229 137,243

Let A =died from cancer and B=former cigar smoker

(a) P(B |A) =P(A and B)

P(A)=

91 /1372431014 /137243

=91

1014=0.08974

(b) P(A |B) =P(A and B)

P(B)=

91 /1372437848 /137243

=91

7848=0.01159

88

Counting Techniques

In order to calculate probabilities using the classical method, one must be able to count how many times a particular event will occur. For sample spaces that are small, this is usually an easy task. However, for sample spaces that are large, we will need some “counting formulas.”

Section 5.5

89

Multiplication Rule of Counting

This rule is often useful in counting outcomes (events) that have a tree structure. For example, how many different ways can the births of three children occur?

Count of Final Events =2⋅2⋅2 =23 =8

90

Generalization

Suppose that we have 3 branch points with the first branch point 2 possibilities, then 3 possibilies, then 2 possibilities.

Count of Final Events =2⋅3⋅2 =12

Theorem : Suppose there are r1 possibilities at first step, then r2 possibilities at second

step, then r3 possibilities at third step, then r4 possibilities at fourth step, ..., then rk at the

kth step. The total number of outcomes is then r1 ⋅r2 ⋅r3L rk.

91

Example

Problem: A girl has five blouses and three skirts. How many different outfits can she put together?

Answer: (5)(3) = 15

Problem: A girl has five blouses and three skirts and 4 pairs of shoes. How many different outfits (including the shoes) can she put together?

Answer: (5)(3)(4) = 60

92

Factorials and Sequences

Suppose that we have n elements in a set: S = a1,a2 ,...,an{ } .  Suppose

that construct sequences using elements of S:  b1,b2 ,...,bn{ }  where the first element, b1,  

can be any of the elements of S, the second element, b2 ,  can be any of the elements of

S− b1{ } , the third element, b3, be any element of S− b1,b2{ }  and so forth.

How many possible sequences are there?

Answer :  Using the Multiplication Rule:  n⋅(n−1)⋅(n−2)L (3)(2)(1) =n!

1. How many sequences can be constructed in the above way from the set S = {a,b,c,d}.

Answer: 4! = (1)(2)(3)(4) = 24

2. A salesman must travel to five cities to promote his products. How many different trips are possible if any route between two cities is possible?

Answer: Sample trip: {city1, city2, city3, city4, city5}. Hence, 5! = 120.

93

PermutationsSuppose we have n distinct elements in a set: S = a1,a2 ,...,an{ } .  If we

choose any r distinct elements in this set, then how many different arrangements 

of the r elements can be constructed where order is important (e.g.,  A,B,C{ }  is different

from  C,A,B{ }  or  B,A,C{ }).

Answer :  nPr =n!

n−r( )!

Note:  nPn =n!

n−n( )!=

n!0!

=n!1

=n!

nPn−1 =n!

n−(n−1)( )!=

n!1!

=n!1

=n!

94

Example

How many different ways (permutations) can four letters, {a,b,c,d} be arranged, taking two a time, e.g., {a,b}, {a,d}, {c,d}, etc?

4 P2 =4!

(4 −2)!=242

=12

{a,b},{b,a},{a,c},{c,a},{a,d},{d,a},{b,c},{c,b},{c,d},{d,c},{b,d},{d,b} --- 12 arrangements

95

Permutations with Duplicates

Suppose we have n elements in a set: S = a1,a2 ,...,aN{ }  where some or all of the elements can

be repeated. Suppose that there are k distinct elements in S.  Let k1 be the number of repeats of a1,  k2 be the number of repeats of a2 ,..., and km be the number of repeats of am.  If we choose all n  elements in this set, then how many different arrangements can be constructed where order is 

important (e.g.,  A,B,C{ }  is different from  C,A,B{ }  or  B,A,C{ } ).  Note that k1 + k2 + ...+ km =N.

Answer :  %Pn =n!

k1!( ) k2 !( )L kn!( )

Example:   How many ways can 7 girls and 2 boys be arranged in a sequence e.g.,  B,G,B,G,G,G,G,G,G{ } ?

Answer :  %P9 =9!

7!( ) 2!( )=(7!)(8)(9)7!( ) 2!( )

=722

=36

96

Combinations

Suppose we have n distinct elements in a set: S = a1,a2 ,...,an{ } .  If we

choose any r distinct elements in this set, then how many different arrangements 

of the r elements can be constructed where order is not important (e.g.,  A,B,C{ }  is treated

the same as  C,A,B{ }  or  B,A,C{ } ).

Answer :  nCr =n!

n−r( )! r!

Note:  nCn =n!

n−n( )! n!=1

nCn−1 =n!

n−(n−1)( )! (n−1)!=

n!(1!)(n−1)!

=n

97

Examples

Example : Find all of the combinations of five distinct objects, a,b,c,d,e,

taken two at a time.

Answer : 5 C2 =5!

(5 −2)! 2!=

5!3! 2!

=4 ⋅52

=10

Example:  In the U.S. Senate, there are 21 members ofn the Committee on Banking, Housing and

Urban Affairs.  Nine of these 21 members are selected to be on the Subcommittee on Economic

Policy.  How many different committee structures are possible for this subcommittee?

Answer : 21C9 =21!

(21−9)! 9!=

21!12! 9!

=9⋅10 ⋅11L 211⋅2 ⋅3L 9

=293,930

98

Examples

Example : In the Pick 5 lottery game, one must pick 5 balls from a group of 39 balls that are numbered 1-39.

The order of the 5 balls is not important. What is the probability of winning this game?

Answer : P(winning) =1

combinations of 5 balls=

1

39C5

=139!

5!⋅(39 −5)!

=5!⋅34!39!

=5!

35 ⋅36 ⋅37 ⋅38 ⋅39=

1575,757

=1.73684 ×10−6

Example:  How many ways can two particular horses finish in a 10-horse race?

Answer :  Notice that order is important in counting the different ways.  Hence, 10P2 =10!

(10 −2)!=10!8!

=9⋅10 =90.

99

Example

Example : How many different 10-letter words (real or imaginery) can be formed from the letter in the word: STATISTICS?

Answer : Each word must have 3 S's, 3 T's, 2 I's, 1 A and 1 C. Each word will have 10 letters. If we neglected the counting of each

letter, then there are 10! arrangements of the ten letters. However, we have restrictions on the number of times each letter appears.

Therefore, we use muliplication rule:

(number of ways for S)(number of ways for T)(number of ways for I)(number of ways for A)(number of ways for C)

= 10 C3( ) 7C3( ) 4C2( ) 2C1( ) 1C1( ) =10!

(3!)(10 −3)!⎛⎝⎜

⎞⎠⎟

7!(3)!(7 −3)!

⎛⎝⎜

⎞⎠⎟

4!(2)!(4 −2)!

⎛⎝⎜

⎞⎠⎟

2!(1)!(2 −1)!

⎛⎝⎜

⎞⎠⎟

1!(1)!(1−1)!

⎛⎝⎜

⎞⎠⎟

=10!

(3!)(7)!⎛⎝⎜

⎞⎠⎟

7!(3)!(4)!

⎛⎝⎜

⎞⎠⎟

4!(2)!(2)!

⎛⎝⎜

⎞⎠⎟

2!(1)!(1)!

⎛⎝⎜

⎞⎠⎟

1!(1)!(0)!

⎛⎝⎜

⎞⎠⎟=

10!3!( ) 3!( ) 2!( ) 1!( ) 0!( )

=10!

3!( ) 3!( ) 2!( ) 1!( ) 1!( )=50,400