chapter 14 probability basics laws of probability odds and probability probability trees

Chapter 14 Probability Basics• Laws of Probability• Odds and

Probability• Probability Trees

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

Probability

•Formal study of uncertainty•The engine that drives Statistics

• Primary objective of lecture unit 4:

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 certain• We gauge the chances of

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

Tomorrow's Weather

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 probabilityRandomness ≠ 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.

4.1 The Laws of 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

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 FCertain 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(Tail) = 0.5

Probability rules (cont’d)

Venn diagram:

Sample space made up of an event

A and its complement 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 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: , orIntersections: , and

AÇB

AÈB

Mutually Exclusive (Disjoint) Events

• Mutually exclusive ordisjoint events-no outcomesfrom S in common

A and B disjoint: A B=

A and B not disjoint

AÈB

AÇB

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 (cont.)

Multiplication Rule

6. For two independent events A and B

P(A and B) = P(A) × P(B)

Note: assuming events are independent doesn’t make it true.

Multiplication Rule• The probability that you encounter a

green light at the corner of Dan Allen and Hillsborough is 0.35, a yellow light 0.04, and a red light 0.61. What is the probability that you encounter a red light on both Monday and Tuesday?

• It’s reasonable to assume that the color of the light you encounter on Monday is independent of the color on Tuesday. So

P(red on Monday and red on Tuesday) = P(red on Monday) × P(red on Tuesday) = 0.61 × 0.61= 0.3721

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,

thenP(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)

6. For two independent events A and B

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

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

P({2})= 3/6 + 3/6 - 1/6 = 5/6

Chapter 14 (cont.)

Odds and ProbabilitiesProbability Trees

ODDS AND PROBABILITIES

World Series Odds

From probability to oddsFrom 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

Chapter 14 (cont.) Probability Trees

A Graphical Method for Complicated Probability Problems

Example: Prob. of playing pro baseball

Data shows the following probabilities concerning high school baseball players:• 6.1% of hs players go on to play at the college level• 9.4% of hs players that play in college go on to play

professionally• 0.2% of hs players that do not compete in college play

pro baseball What is the probability that a high school

baseball ultimately plays professional baseball?

Probability Tree Approach

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

Q1: probability that a high school baseball ultimately plays professional baseball?

P(play college and play professionally)

= .061*.094= .005734

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

P(do not play college and play professionally)

= .939*.002= .001878

P(high school baseball player plays professional baseball) = .005734 + .001878 = .007612

Question 2 Given that a high school player

played professionally, what is the probability he played in college?

.005734.7533

.005734 .001878

P(play college and play professionally)

= .061*.094= .005734

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

P(do not play college and play professionally)

= .939*.002= .001878

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 TreeMultiply

branch probsclinical reliability

clinical reliability

Question 1 Answer

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

Pr(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.

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

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