Probability Formal study of uncertainty The engine that drives statistics.

Probability

•Formal study of uncertainty•The engine that drives statistics

Introduction

• Nothing in life is certain• We gauge the chances of

successful outcomes in business, medicine, weather, and other everyday situations such as the lottery (recall the birthday problem)

History

• For most of human history, probability, the formal study of the laws of chance, has been used for only one thing: gambling

History (cont.)• Nobody knows exactly when

gambling began; goes back at least as far as ancient Egypt where 4-sided “astragali” (made from animal heelbones) were used

History (cont.)• The Roman emperor Claudius

(10BC-54AD) wrote the first known treatise on gambling.

• The book “How to Win at Gambling” was lost.

Rule 1: Let Caesar win IVout of V times

Approaches to Probability

• Relative frequencyevent probability = x/n, where x=# of occurrences of event of interest, n=total # of observations

• Coin, die tossing; nuclear power plants?

• Limitationsrepeated observations not practical

Approaches to Probability (cont.)

• Subjective probabilityindividual assigns prob. based on personal experience, anecdotal evidence, etc.

• Classical approachevery possible outcome has equal probability (more later)

Basic Definitions

• Experiment: act or process that leads to a single outcome that cannot be predicted with certainty

• Examples:1. Toss a coin2. Draw 1 card from a standard deck of

cards3. Arrival time of flight from Atlanta to

RDU

Basic Definitions (cont.)

• Sample space: all possible outcomes of an experiment. Denoted by S

• Event: any subset of the sample space S;typically denoted A, B, C, etc.Simple event: event with only 1 outcomeNull event: the empty set Certain event: S

Examples

1. Toss a coin onceS = {H, T}; A = {H}, B = {T} simple events

2. Toss a die once; count dots on upper faceS = {1, 2, 3, 4, 5, 6}A=even # of dots on upper face={2, 4, 6}B=3 or fewer dots on upper face={1, 2, 3}

Laws of Probability

1)(,0)(.2

event any for ,1)(0 1.

SPP

AAP

Laws of Probability (cont.)

3. P(A’ ) = 1 - P(A)For an event A, A’ is the complement of A; A’ is everything in S that is not in A.

AA'

S

Birthday Problem• What is the smallest number of

people you need in a group so that the probability of 2 or more people having the same birthday is greater than 1/2?

• Answer: 23No. of people 23 30 40 60Probability .507.706.891.994

Example: Birthday Problem

• A={at least 2 people in the group have a common birthday}

• A’ = {no one has common birthday}

502.498.1)'(1)(

498.365

343

365

363

365

364)'(

:23365

363

365

364)'(:3

APAPso

AP

people

APpeople

Unions and Intersections

S

A B

A

A

Mutually Exclusive Events

• Mutually exclusive events-no outcomes from S in common

S

AB

A =

Laws of Probability (cont.)

Addition Rule for Disjoint Events:

4. If A and B are disjoint events, then

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

• 5. For two independent events A and B

P(A B) = P(A) × P(B)

Laws of Probability (cont.)

General Addition Rule

6. For any two events A and B

P(A B) = P(A) + P(B) – P(A B)

P(AB)=P(A) + P(B) - P(A B)

S

A B

A

Example: toss a fair die once

• S = {1, 2, 3, 4, 5, 6}• A = even # appears = {2, 4, 6}• B = 3 or fewer = {1, 2, 3}• P(A B) = P(A) + P(B) - P(A B)

=P({2, 4, 6}) + P({1, 2, 3}) - P({2})

= 3/6 + 3/6 - 1/6 = 5/6

Laws of Probability: Summary

• 1. 0 P(A) 1 for any event A• 2. P() = 0, P(S) = 1• 3. P(A’) = 1 – P(A)• 4. If A and B are disjoint events, then

P(A B) = P(A) + P(B)• 5. If A and B are independent events,

thenP(A B) = P(A) × P(B)

• 6. For any two events A and B,P(A B) = P(A) + P(B) – P(A B)

Probability Models

The Equally Likely Approach(also called the Classical

Approach)

Assigning Probabilities

• If an experiment has N outcomes, then each outcome has probability 1/N of occurring

• If an event A1 has n1 outcomes, then

P(A1) = n1/N

We Need Efficient Methods for Counting Outcomes

Product Rule for Ordered Pairs

• A student wishes to commute to a junior college for 2 years and then commute to a state college for 2 years. Within commuting distance there are 4 junior colleges and 3 state colleges. How many junior college-state college pairs are available to her?

Product Rule for Ordered Pairs

• junior colleges: 1, 2, 3, 4• state colleges a, b, c• possible pairs:(1, a) (1, b) (1, c)(2, a) (2, b) (2, c)(3, a) (3, b) (3, c)(4, a) (4, b) (4, c)

Product Rule for Ordered Pairs

• junior colleges: 1, 2, 3, 4• state colleges a, b, c• possible pairs:(1, a) (1, b) (1, c)(2, a) (2, b) (2, c)(3, a) (3, b) (3, c)(4, a) (4, b) (4, c)

4 junior colleges3 state collegestotal number of possiblepairs = 4 x 3 = 12

4 junior colleges3 state collegestotal number of possiblepairs = 4 x 3 = 12

Product Rule for Ordered Pairs

• junior colleges: 1, 2, 3, 4• state colleges a, b, c• possible pairs:(1, a) (1, b) (1, c) (2, a) (2, b) (2, c)(3, a) (3, b) (3, c)(4, a) (4, b) (4, c)

In general, if there are n1 waysto choose the first element ofthe pair, and n2 ways to choosethe second element, then the number of possible pairs isn1n2. Here n1 = 4, n2 = 3.

In general, if there are n1 waysto choose the first element ofthe pair, and n2 ways to choosethe second element, then the number of possible pairs isn1n2. Here n1 = 4, n2 = 3.

Counting in “Either-Or” Situations• NCAA Basketball Tournament: how

many ways can the “bracket” be filled out?

1. How many games?2. 2 choices for each game3. Number of ways to fill out the bracket:

263 = 9.2 × 1018

• Earth pop. about 6 billion; everyone fills out 1 million different brackets

• Chances of getting all games correct is about 1 in 1,000

Counting Example

• Pollsters minimize lead-in effect by rearranging the order of the questions on a survey

• If Gallup has a 5-question survey, how many different versions of the survey are required if all possible arrangements of the questions are included?

Solution• There are 5 possible choices for the

first question, 4 remaining questions for the second question, 3 choices for the third question, 2 choices for the fourth question, and 1 choice for the fifth question.

• The number of possible arrangements is therefore

5 4 3 2 1 = 120

Efficient Methods for Counting Outcomes

• Factorial Notation:n!=12 … n

• Examples1!=1; 2!=12=2; 3!= 123=6; 4!

=24;5!=120;• Special definition: 0!=1

Factorials with calculators and Excel

• Calculator: non-graphing: x ! (second function)graphing: bottom p. 9 T I Calculator Commands(math button)

• Excel:Paste: math, fact

Factorial Examples• 20! = 2.43 x 1018

• 1,000,000 seconds?• About 11.5 days• 1,000,000,000 seconds?• About 31 years• 31 years = 109 seconds• 1018 = 109 x 109

• 31 x 109 years = 109 x 109 = 1018 seconds

• 20! is roughly the age of the universe in seconds

Permutations

A B C D E• How many ways can we choose 2

letters from the above 5, without replacement, when the order in which we choose the letters is important?

• 5 4 = 20

Permutations (cont.)

20)!25(

!5:

45!3

!5

)!25(

!52045

25

PNotation

Permutations with calculator and Excel

• Calculatornon-graphing: nPr

• Graphingp. 9 of T I Calculator Commands(math button)

• ExcelPaste: Statistical, Permut

Combinations

A B C D E• How many ways can we choose 2

letters from the above 5, without replacement, when the order in which we choose the letters is not important?

• 5 4 = 20 when order important• Divide by 2: (5 4)/2 = 10 ways

Combinations (cont.)

!)!(

!

102

20

21

45

!2!3

!5

!2)!25(

!525

52

rrn

nC

C

rnnr

ST 101 Powerball Lottery

From the numbers 1 through 20,choose 6 different numbers.

Write them on a piece of paper.

Chances of Winning?

760,38!6)!620(

!20

ies?possibilit ofNumber

important.not order t,replacemen

without 20, from numbers 6 Choose

620206

C

North Carolina Powerball Lottery

Prior to Jan. 1, 2009 After Jan. 1, 2009

:

55!3,478,761

5!50!

:

42!42

1!41!

3,478,761*42

146,107,962

5 from 1- 55

1 from 1- 42 (p'ball #)

:

59!5,006,386

5!54!

:

39!39

1!38!

5,006,386*39

195,249,054

5 from 1- 59

1 from 1- 39 (p'ball #)

Visualize Your Lottery Chances

• How large is 195,249,054?• $1 bill and $100 bill both 6” in length

• 10,560 bills = 1 mile• Let’s start with 195,249,053 $1 bills

and one $100 bill …• … and take a long walk, putting

down bills end-to-end as we go

Raleigh to Ft. Lauderdale…

… still plenty of bills remaining, so continue from …

… Ft. Lauderdale to San Diego

… still plenty of bills remaining, so continue from…

… still plenty of bills remaining, so continue from …

… San Diego to Seattle

… still plenty of bills remaining, so continue from …

… Seattle to New York

… still plenty of bills remaining, so …

… New York back to Raleigh

Go around again! Lay a second path of bills

Still have ~ 5,000 bills left!!

Chances of Winning NC Powerball Lottery?

• Remember: one of the bills you put down is a $100 bill; all others are $1 bills

• Your chance of winning the lottery is the same as bending over and picking up the $100 bill while walking the route blindfolded.

Example: Illinois State Lottery

balls) pong pingmillion 16.5 house, ft (1200

months) 10in second 1about (

165,827,25!6!48

!54

importantnot order t;replacemen

withoutnumbers 54 from numbers 6 Choose

2

654 C

Virginia State Lottery

969000,52!1!24

!25760,118,2

760,118,2

760,118,2!5!45

!50: 5Pick

125

550

C

C

Probability Trees

A Graphical Method for Complicated Probability

Problems

Example: AIDS Testing• V={person has HIV}; CDC: P(V)=.006• +: test outcome is positive (test

indicates HIV present)• -: test outcome is negative• clinical reliabilities for a new HIV test:

1. If a person has the virus, the test result will be positive with probability .999

2. If a person does not have the virus, the test result will be negative with probability .990

Question 1

• What is the probability that a randomly selected person will test positive?

Probability Tree Approach

• A probability tree is a useful way to visualize this problem and to find the desired probability.

Probability Treeclinical reliability

clinical reliability

Probability TreeMultiply

branch probsclinical reliability

clinical reliability

Question 1 Answer

• What is the probability that a randomly selected person will test positive?

• P(+) = .00599 + .00994 = .01593

Question 2

• If your test comes back positive, what is the probability that you have HIV?(Remember: we know that if a person has the virus, the test result will be positive with probability .999; if a person does not have the virus, the test result will be negative with probability .990).

• Looks very reliable

Question 2 Answer

Answertwo sequences of branches lead to positive test; only 1 sequence represented people who have HIV.

P(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

Summary• Question 1:• P(+) = .00599 + .00994 = .01593• Question 2: two sequences of

branches lead to positive test; only 1 sequence represented people who have HIV.

P(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

Recap• We have a test with very high clinical

reliabilities:1. If a person has the virus, the test result will be

positive with probability .9992. If a person does not have the virus, the test

result will be negative with probability .990

• But we have extremely poor performance when the test is positive:

P(person has HIV given that test is positive) =.376

• In other words, 62.4% of the positives are false positives! Why?

• When the characteristic the test is looking for is rare, most positives will be false.

examples1. P(A)=.3, P(B)=.4; if A and B are

mutually exclusive events, then P(AB)=?

A B = , P(A B) = 02. 15 entries in pie baking contest at

state fair. Judge must determine 1st, 2nd, 3rd place winners. How many ways can judge make the awards?

15P3 = 2730