Agresti/Franklin Statistics, 1 of 87 Chapter 5 Probability in Our Daily Lives Learn …. About probability – the way we quantify uncertainty How to measure

Agresti/Franklin Statistics, 1 of 87

Chapter 5Probability in Our Daily Lives

Learn …. About probability – the way we quantify

uncertainty

How to measure the chances of the possible outcomes of random phenomena

How to find and interpret probabilities


Section 5.1

How Can Probability Quantify Randomness?


Randomness

Applies to the outcomes of a response variable

Possible outcomes are known, but it is uncertain which will occur for any given observation


Some Popular Randomizers

Rolling dice

Spinning a wheel

Flipping a coin

Drawing cards


Random Phenomena

Individual outcomes are unpredictable

With a large number of observations, predictable patterns occur


Random Phenomena

With random phenomena, the proportion of times that something happens is highly random and variable in the short run but very predictable in the long run.


Jacob Bernoulli: Law of Large Numbers

As the number of trials of a random phenomenon increases, the proportion of occurrences of any given outcome approaches a particular number “in the long run”.


Probability

With a random phenomenon, the probability of a particular outcome is the proportion of times that the outcome would occur in a long run of observations.


Roll a Die

What is the probability of rolling a ‘6’?

a. .22

b. .10

c. .17


Question about Random Phenomena

If a family has four girls in a row and is expecting another child, does the next child have more than a ½ chance of being a boy?


Independent Trials

Different trials of a random phenomenon are independent if the outcome of any one trial is not affected by the outcome of any other trial.


Section 5.2

How Can We Find Probabilities?


Sample Space

For a random phenomenon, the sample space is the set of all possible outcomes


Example: Roll a Die Once

The Sample Space consists of six possible outcomes:

{1, 2, 3, 4, 5, 6}


Example: Flip a Coin Twice

The Sample Space consists of the four possible outcomes:

{(H,H) (H,T) (T,H) (T,T)}


Example: A 3-Question Multiple Choice Quiz

Diagram of the Sample Space


Tree Diagram An ideal way of visualizing sample

spaces with a small number of outcomes

As the number of trials or the number of possible outcomes on each trial increase, the tree diagram becomes impractical


Event

An event is a subset of the sample space


Probabilities for a Sample Space

The probability of each individual outcome is between 0 and 1

The total of all the individual probabilities equals 1


Example: Assigning Subjects to Echinacea or Placebo for Treating Colds

Experiment• Multi-center randomized experiment to compare

an herbal remedy to a placebo for treating the common cold

• 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



Sample Space to receive the herbal remedy:

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

These six possible outcomes are equally likely



What is the probability of the event that the sample chosen to receive the herbal remedy consists of one man and one woman?


Probability of an Event The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event.

When all the possible outcomes are equally likely:

space sample in the outcomes ofnumber

Aevent in outcomes ofnumber )( AP


Example: What are the Chances of a Taxpayer being Audited?

Each year, the Internal Revenue Service audits a sample of tax forms to verify their accuracy





What is the sample space for selecting a taxpayer?

{(under $25,000, Yes), (under $25,000, No),

($25,000 - $49,000, Yes) …}



For a randomly selected taxpayer in 2002, what is the probability of an audit?



For a randomly selected taxpayer in 2002, what is the probability of an income of $100,000 or more?


Basic Rules for Finding Probabilities about a Pair of Events

Complement of an Event

Intersection of 2 Events

Union of 2 Events



Complement of Event A:• Consists of all outcomes in the sample

space that are not in A

• Is denoted by Ac

• The probabilities of A and Ac add to 1

• P(Ac) = 1 – P(A)




Disjoint Events

Two events, A and B, are disjoint if they do not have any common outcomes


Example: Disjoint Events

Pop Quiz: 3 Multiple-Choice Questions• Event A: Student answers exactly 1

question correctly

• Event B: Student answer exactly 2 questions correctly


Example: Disjoint Events


Intersection of Two Events

The intersection of A and B: consists of outcomes that are in both A and B


Union of Two Events

The union of A and B: Consists of outcomes that are in A or B

In probability, “A or B” denotes that A occurs or B occurs or both occur


Intersection and Union of Two Events


How Can We Find the Probability that A or B Occurs?

Addition Rule: Probability of the Union of Two Events• For the union of two events,

P(A or B) = P(A) + P(B) – P(A and B)

• If the events are disjoint, P(A and B) = 0, so P(A or B) = P(A) + P(B)


How Can We Find the Probability that A and B Occurs?

Multiplication Rule: Probability of the Intersection of Independent Events• For the intersection of two independent

events, A and B:

P(A and B) = P(A) x P(B)


Example: Two Rolls of A Die

P(6 on roll 1 and 6 on roll 2):

1/6 x 1/6 = 1/36


Example: Guessing on a Pop Quiz

Pop Quiz with 3 Multiple-choice questions• Each question has 5 options

A student is totally unprepared and randomly guesses the answer to each question



The probability of selecting the correct answer by guessing = 0.20

Responses on each question are independent


Tree Diagram for the Pop Quiz



What is the probability that a student answers at least 2 questions correctly?

P(CCC) + P(CCI) + P(CIC) + P(ICC) =

0.008 + 3(0.032) = 0.104


Events Often Are Not Independent

Example: A Pop Quiz with 2 Multiple Choice Questions• Data giving the proportions for the actual

responses of students in a class

Outcome: II IC CI CC

Probability: 0.26 0.11 0.05 0.58



Define the events A and B as follows:

• A: {first question is answered correctly}

• B: {second question is answered correctly}



P(A) = P{(CI), (CC)} = 0.05 + 0.58 = 0.63

P(B) = P{(IC), (CC)} = 0.11 + 0.58 = 0.69

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

If A and B were independent,

P(A and B) = P(A) x P(B) = 0.63 x 0.69 = 0.43


Question of Independence

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


Example: A family has two children

If each child is equally likely to be a girl or

boy, find the probability that the family has

two girls.

a. 1/2

b. 1/3

c. 1/4

d. 1/8


Section 5.3

Conditional Probability: What’s the Probability of A, Given B?


Conditional Probability

For events A and B, the conditional

probability of event A, given that event B has occurred is:

)(

) ()|P(

BP

BandAPBA


Conditional Probability




Example: Probabilities of a Taxpayer Being Audited



What was the probability of being audited, given that the income was ≥ $100,000?

• Event A: Taxpayer is audited

• Event B: Taxpayer’s income ≥ $100,000



007.01334.0

0010.0

P(B)

B) andP(A B)|P(A


Example: The Triple Blood Test for Down Syndrome

A positive test result states that the condition is present

A negative test result states that the condition is not present



False Positive: Test states the condition is present, but it is actually absent

False Negative: Test states the condition is absent, but it is actually present



A study of 5282 women aged 35 or over analyzed the Triple Blood Test to test its accuracy





Assuming the sample is representative of the population, find the estimated probability of a positive test for a randomly chosen pregnant woman 35 years or older



P(POS) = 1355/5282 = 0.257


Given that the diagnostic test result is positive, find the estimated probability that Down syndrome truly is present




035.0257.0

009.0

5282/1355

5282/48

P(POS)

POS) and P(D POS)|P(D


Summary: Of the women who tested positive, fewer than 4% actually had fetuses with Down syndrome



Multiplication Rule for Finding P(A and B)

For events A and B, the probability that A and B both occur equals:

• P(A and B) = P(A|B) x P(B)

also

• P(A and B) = P(B|A) x P(A)


Example: How Likely is a Double Fault in Tennis?

Roger Federer – 2004 men’s champion in the Wimbledon tennis tournament• He made 64% of his first serves

• He faulted on the first serve 36% of the time

• Given that he made a fault with his first serve, he made a fault on his second serve only 6% of the time



Assuming these are typical of his serving performance, when he serves, what is the probability that he makes a double fault?



P(F1) = 0.36 P(F2|F1) = 0.06

P(F1 and F2) = P(F2|F1) x P(F1)

= 0.06 x 0.36 = 0.02


Sampling Without Replacement

Once subjects are selected from a population, they are not eligible to be selected again


Example: How Likely Are You to Win the Lotto?

In Georgia’s Lotto, 6 numbers are randomly sampled without replacement from the integers 1 to 49

You buy a Lotto ticket. What is the probability that it is the winning ticket?



P(have all 6 numbers) = P(have 1st and 2nd and 3rd and 4th and 5th and 6th)

= P(have 1st)xP(have 2nd|have 1st)xP(have 3rd| have 1st and 2nd) …P(have 6th|have 1st, 2nd, 3rd, 4th, 5th)



6/49 x 5/48 x 4/47 x 3/46 x 2/45 x 1/44

= 0.00000007


Independent Events Defined Using Conditional Probabilities

Two events A and B are independent if the probability that one occurs is not affected by whether or not the other event occurs


Independent Events Defined Using Conditional Probabilities

Events A and B are independent if:

P(A|B) = P(A)

If this holds, then also P(B|A) = P(B)

Also, P(A and B) = P(A) x P(B)


Checking for Independence

Here are three ways to check whether events A and B are independent:

• Is P(A|B) = P(A)?

• Is P(B|A) = P(B)?

• Is P(A and B) = P(A) x P(B)?

If any of these is true, the others are also true and the events A and B are independent


Example: How to Check Whether Two Events are Independent

The diagnostic blood test for Down syndrome:

POS = positive result

NEG = negative result

D = Down Syndrome

DC = Unaffected



Blood Test:

Status POS NEG Total

D 0.009 0.001 0.010

Dc 0.247 0.742 0.990

Total 0.257 0.743 1.000


Are the events POS and D independent or dependent?

• Is P(POS|D) = P(POS)?




Is P(POS|D) = P(POS)?

P(POS|D) =P(POS and D)/P(D)

= 0.009/0.010 = 0.90

P(POS) = 0.256

The events POS and D are dependent

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Agresti/Franklin Statistics, 1 of 87 Chapter 5 Probability in Our Daily Lives Learn …. About probability – the way we quantify uncertainty How to measure