Chapters 13 and 14

•Probability and counting

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 people23 30 40 60Probability .507.706.891.994We will solve this problem a few slides later using the laws of probability

Probability

•Formal study of uncertainty•The engine that drives Statistics

• Primary objectives:1. use the rules of probability to

calculate appropriate measures of uncertainty.

2. Learn the probability basics so that we can do Statistical Inference

Introduction

Nothing in life is certainWe gauge the chances of successful

outcomes in business, medicine, weather, and other everyday situations such as the lottery or the birthday problem

A phenomenon is random if individual

outcomes are uncertain, but there is

nonetheless a regular distribution of

outcomes in a large number of repetitions.

Randomness and probability

Randomness ≠ chaos

Coin toss The result of any single coin toss is

random. But the result over many tosses

is predictable, as long as the trials are

independent (i.e., the outcome of a new

coin flip is not influenced by the result of

the previous flip).

The result of any single coin toss is

random. But the result over many tosses

is predictable, as long as the trials are

independent (i.e., the outcome of a new

coin flip is not influenced by the result of

the previous flip).

First series of tossesSecond series

The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.

Approaches to Probability

1. 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.)

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

3. 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.Null event: the empty set Certain event: S

Examples1. Toss a coin once

S = {H, T}; A = {H}, B = {T}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}

3.Select 1 card from adeck of 52 cards.S = {all 52 cards}

Laws of Probability

1)(,0)(.2

event any for ,1)(0 1.

SPP

AAP

Coin Toss Example: S = {Head, Tail}Probability of heads = 0.5Probability of tails = 0.5

3) The complement of any event A is the event that A does not occur, written as A.

The complement rule states that the probability

of an event not occurring is 1 minus the

probability that is does occur.

P(not A) = P(A) = 1 − P(A)

Tail = not Tail = Head

P(Tail ) = 1 − P(Head) = 0.5

Probability rules (cont’d)

Venn diagram:

Sample space made up of an event

A and its complementary A , i.e.,

everything that is not A.

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 people23 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: , orIntersections: , and

A

A

Mutually Exclusive (Disjoint) Events

Mutually exclusive ordisjoint events-no outcomesfrom S in common

A and B disjoint: A B=

A and B not disjoint

A

A

Venn Diagrams

Addition Rule for Disjoint Events

4. If A and B are disjoint events, then

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

Laws of Probability (cont.)General Addition Rule

5. For any two events A and B

P(A or B) = P(A) + P(B) – P(A and B)

20

For any two events A and B

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

A

B

P(A) =6/13

P(B) =5/13

P(A and B) =3/13

A or B

+_

P(A or B) = 8/13

General Addition Rule

Laws of Probability: Summary

1. 0 P(A) 1 for any event A2. P() = 0, P(S) = 13. P(A’) = 1 – P(A)4. If A and B are disjoint events, then

P(A or B) = P(A) + P(B)5. For any two events A and B,

P(A or B) = P(A) + P(B) – P(A and B)

M&M candies

Color Brown Red Yellow Green Orange Blue

Probability 0.3 0.2 0.2 0.1 0.1 ?

If you draw an M&M candy at random from a bag, the candy will have one

of six colors. The probability of drawing each color depends on the proportions

manufactured, as described here:

What is the probability that an M&M chosen at random is blue?

What is the probability that a random M&M is any of red, yellow, or orange?

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

P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1 P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)]

= 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1

P(red or yellow or orange) = P(red) + P(yellow) + P(orange)

= 0.2 + 0.2 + 0.1 = 0.5

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 orB) = P(A) + P(B) - P(A andB)

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

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

THE RELATIONSHIP BETWEEN ODDS AND PROBABILITIES

•World Series Odds•The odds at the above link are the odds against a team winning the World Series, though the author claims they’re “odds for winning the World Series”•Odds are frequently a source of confusion. Odds for? Odds against?•From probability to odds•From odds to probability

From Probability to Odds

If event A has probability P(A), then the odds in favor of A are P(A) to 1-P(A). It follows that the odds against A are 1-P(A) to P(A)

If the probability of an earthquake in California is .25, then the odds in favor of an earthquake are .25 to .75 or 1 to 3. The odds against an earthquake are .75 to .25 or 3 to 1

From Odds to Probability

If the odds in favor of an event E are a to b, then

P(E)=a/(a+b)in addition,

P(E’)=b/(a+b)

If the odds in favor of UNC winning the NCAA’s are 3 (a) to 1 (b), thenP(UNC wins)=3/4

P(UNC does not win)=

1/4

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

DiceYou toss two dice. What is the probability of the outcomes summing to 5?

There are 36 possible outcomes in S, all equally likely (given fair dice).

Thus, the probability of any one of them is 1/36.

P(the roll of two dice sums to 5) =

P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 / 36 = 0.111

This is S:

{(1,1), (1,2), (1,3), ……etc.}

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, 4state colleges a, b, cpossible 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, 4state colleges a, b, cpossible 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, 4state colleges a, b, cpossible 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:Insert function: Math and Trig category, FACT function

Factorial Examples20! = 2.43 x 1018

1,000,000 seconds?About 11.5 days1,000,000,000 seconds?About 31 years31 years = 109 seconds1018 = 109 x 109

20! is roughly the age of the universe in seconds

Permutations

A B C D EHow 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)

ExcelInsert function: Statistical, Permut

Combinations

A B C D EHow 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 importantDivide 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 311 Powerball Lottery

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

Write them on a piece of paper.

And the numbers are ...

161121084

wow

scream

Chances of Winning?

760,38!6)!620(

!20

ies?possibilit ofNumber

important.not order t,replacemen

without 20, from numbers 6 Choose

620206

C

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

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 #)

The Forrest Gump Visualization of Your Lottery Chances

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

10,560 bills = 1 mileLet’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.

Put on a blindfold and begin walking along the trail of bills.

Your chance of winning the lottery is the same as your chance of selecting the $100 bill if you stop at a random location along the trail and pick up a bill .

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

Probability Tree Example: probability of playing professional baseball

6.1% of high school baseball players play college baseball. Of these, 9.4% will play professionally.

Unlike football and basketball, high school players can also go directly to professional baseball without playing in college…

studies have shown that given that a high school player does not compete in college, the probability he plays professionally is .002.

Question 1: What is the probability that a high school baseball player ultimately plays professional baseball?

Question 2: Given that a high school baseball player played professionally, what is the probability he played in college?

Question 1: What is the probability that a high school baseball player ultimately plays professional baseball?

P(hs bb player plays professionally)= .061*.094 + .939*.002= .005734 + .001878= .007612

Play coll 0.061

Does not play coll 0.939

Play prof. .094

HS BB Player

.906

Play prof. .002

Does not Play prof. .998

.061*.094=.005734

.939*.002=.001878

Question 2: Given that a high school baseball player played professionally, what is the probability he played in college?

Play coll 0.061

Does not play coll 0.939

Play prof. .094

HS BB Player

.906

Play prof. .002

Does not Play prof. .998

.061*.094=.005734

.939*.002=.001878

P(hs bb player plays professionally) = .005734 + .001878= .007612

(played in college given that played professionally)

.005734= .7533

.007612

P

.061*.094=.005734

Example: AIDS Testing

V={person has HIV}; CDC: Pr(V)=.006 P : test outcome is positive (test

indicates HIV present) N : 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 TreeMultiply

branch probsclinical reliability

clinical reliability

Question 1: What is the probability that a randomly selected person will test positive?

Pr( ) .00599 .00994 .01593P

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: If your test comes back positive, what is

the probability that you have HIV? Pr( ) .00599 .00994 .01593P

(have HIV given that test is positive)

.00599= .376

.00599 .00994

P

Summary

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

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

Pr(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:Pr(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