ProbabilityProbabilityThe Study ofThe Study of

Chance!Chance!Part 3

In this In this powerpoint we powerpoint we will continue our will continue our study of study of probability by probability by looking at:looking at:• Conditional Conditional

ProbabilityProbability• IndependenceIndependence• Reversing the Reversing the

ConditionCondition

Conditional ProbabilitiesConditional Probabilities

What about when we know something What about when we know something up front about the individual that we up front about the individual that we have chosen?have chosen?

How does this change our How does this change our calculations?calculations?

Let’s look a the employment/credit Let’s look a the employment/credit card data that we used in Part 2.card data that we used in Part 2.

Conditional ProbabilitiesConditional Probabilities

TotaTotall

NoNo

YesYes

TotaTotall

NoNoYeYess

JJ

OO

BB

Credit CardCredit Card

18 47

46 109

29

28 34 62

63

What is the probability that a respondent has a credit card given that they are employed?

•We write this type of question like this:

P(C E) *the line between is read “given” and indicates that we already know the outcome of the event that comes after the line.

Since we already know that the respondent has a job, this means that our population of interest is only the 47 respondents that have jobs.

18 of those have credit cards.

So P(C E) = 18/47

Conditional ProbabilityConditional Probability Find the probability that a Find the probability that a

respondent does not have a respondent does not have a job given that they have a job given that they have a credit card.credit card.

P(EP(Ec c C) = C) = • Since we already know they Since we already know they

have a credit card, our have a credit card, our population of interest has population of interest has become they 46 individuals become they 46 individuals who have a credit card.who have a credit card.

• Our “successes” are the 28 of Our “successes” are the 28 of those people who do not have those people who do not have a job.a job.

P(EP(Ec c C) = 28/46C) = 28/46

TotaTotall

NoNo

YesYes

TotaTotall

NoNoYeYess

JJ

OO

BB

Credit CardCredit Card

18 47

46 109

29

28 34 62

63

Disjoint vs IndependentDisjoint vs Independent

Two very important concepts in Two very important concepts in probability are disjoint events and probability are disjoint events and independent events. The ability to independent events. The ability to tell the difference between these are tell the difference between these are very important when calculating very important when calculating probabilities.probabilities.

Disjoint EventsDisjoint Events(Mutually Exclusive)(Mutually Exclusive)

We’ll start with disjoint We’ll start with disjoint events.events.• When events are disjoint this When events are disjoint this

means they have no means they have no outcomes in common.outcomes in common.

• Mutually Exclusive is another Mutually Exclusive is another term used to describe disjoint term used to describe disjoint events.events.

Examples:Examples:• The relationships between disjoint The relationships between disjoint

and joint events may be best and joint events may be best shown using Venn diagrams.shown using Venn diagrams.

• Let the event A: Democratic voterLet the event A: Democratic voter• Let the event B: Republican voterLet the event B: Republican voter

Democratic Voter

AB

Republican Voter

Since a voter can not be registered as both a Democratic Voter and a Republican Voter, there is no overlap of the events.

This can either be shown with non-overlapping Venn diagrams or with the overlap being empty.

IndependenceIndependence Events are independent when knowing the Events are independent when knowing the

outcome of one event does NOT change the outcome of one event does NOT change the probability of the other.probability of the other.

For Example:For Example:• Consider the two events: flip a coin, then roll a dieConsider the two events: flip a coin, then roll a die• We wish to know the probability of the event We wish to know the probability of the event

Head and a number less than 2Head and a number less than 2

• The probability of getting heads on a fair coin is ½The probability of getting heads on a fair coin is ½• The probability of getting a number less than 2 on a fair The probability of getting a number less than 2 on a fair

die is 2/6 or 1/3die is 2/6 or 1/3• Knowing the outcome of the coin toss does NOT change Knowing the outcome of the coin toss does NOT change

the probability of getting a number less than 2 on the the probability of getting a number less than 2 on the die.die.

• These two events are independent, since knowing the These two events are independent, since knowing the outcome of the coin toss didn’t change the probability of outcome of the coin toss didn’t change the probability of the dice roll.the dice roll.

IndependenceIndependence

The formal mathematical definition The formal mathematical definition tells us that two events are tells us that two events are independent if:independent if:

P(A B) = P(A)P(A B) = P(A) In other words, this “insider” In other words, this “insider”

information (knowing that “B” has information (knowing that “B” has happened doesn’t gives us NO new happened doesn’t gives us NO new information about “A” happening!information about “A” happening!

Let’s look at an example.Let’s look at an example.

Tests for Drunk DrivingTests for Drunk Driving Police report that 78% of drivers are given a Police report that 78% of drivers are given a

breath test, 36% a blood test, and 22% both breath test, 36% a blood test, and 22% both teststests

We wish to know if giving the two test are We wish to know if giving the two test are independent?independent?

Let’s construct a contingency table for this Let’s construct a contingency table for this situationsituation

Breath Test

Yes No Total

BloodTest

Yes .22 .14 .36

No .56 .08 .64

Total .78 .22 1

Independent????Independent????Breath Test

Yes No Total

BloodTest

Yes .22 .14 .36

No .56 .08 .64

Total .78 .22 1

•If the two tests are independent then the following should be true:

•P(blood test breath test ) = P(blood test)

•Let’s find out if they are.

Breath Test

Yes No Total

BloodTest

Yes .22 .14 .36

No .56 .08 .64

Total .78 .22 1

P(blood test breath test ) = P(blood test)

•Remember from our study of conditional probabilities that the P(blood test breath test) = .22/.78= .28

•P(blood test) = .36

•Since .28 ≠ .36, these events are NOT independent.

•Knowing the outcome of one gives changes the probability of the other

Disjoint vs IndependentDisjoint vs Independent

Can disjoint events ever be independent?Can disjoint events ever be independent?• Remember that for events to be independent, Remember that for events to be independent,

knowing the outcome of one does NOT change knowing the outcome of one does NOT change the outcome of the other.the outcome of the other.

• Also remember that disjoint events can NOT Also remember that disjoint events can NOT happen at the same time, which means happen at the same time, which means knowing that one happened makes the knowing that one happened makes the probability of the other happening zero probability of the other happening zero (regardless of its original probability)(regardless of its original probability)

In short: Disjoint events can NEVER be In short: Disjoint events can NEVER be independent!!!!!independent!!!!!

Reversing the conditionReversing the condition

Sometimes the information we know Sometimes the information we know and the information we are looking and the information we are looking for doesn’t come in such a way as to for doesn’t come in such a way as to be a straight forward question.be a straight forward question.

Let’s find out how we can use tree Let’s find out how we can use tree diagrams to answer these kinds of diagrams to answer these kinds of questions.questions.

Binge drinking and AccidentsBinge drinking and Accidents Our question of interest is this:Our question of interest is this:

• What is the probability that if a college student has been What is the probability that if a college student has been involved in an alcohol-related accident, s/he is a binge involved in an alcohol-related accident, s/he is a binge drinker?drinker?

Here is what we know:Here is what we know:• A study on binge drinking on Campus found that 44% of A study on binge drinking on Campus found that 44% of

college students engage in binge drinking (5+ for men, and 4+ college students engage in binge drinking (5+ for men, and 4+ for women), 37% drink moderately, and 19% abstain entirely.for women), 37% drink moderately, and 19% abstain entirely.

• Another study found that among binge drinkers aged 21-34, Another study found that among binge drinkers aged 21-34, 17% have been involved in an alcohol-related accident, while 17% have been involved in an alcohol-related accident, while among non-bingers of the same age, only 9% have been among non-bingers of the same age, only 9% have been involved in such accidents.involved in such accidents.

Let’s draw a tree diagram to help answer our Let’s draw a tree diagram to help answer our questionquestion

Binge Drinkers

.44

Moderate

.37

Abstain

.19

Accident

.17

Accident

.09

Accident

.00

None

.83

None

.91

None

1.0

Binge and Acc

.44(.17)=.0748

Binge and None

.44(.83)=.3652

Mod and Acc

.37(.09)=.0333

Mod and None

.37(.91)=.3367

Abstain and Acc

.19(.00) = 0

Abstain and None

.19(1.0) = .19

Remember that our question can be written:

•P(binge accident)

What we know is the second event, (accident). This essentially changes our sample space (our denominater)

We are only concerned with the branches where accidents occurred.

Our numerator is the branch among these in which the driver is a binge drinker.

P(Binge Acc) = .0748 / (.0748 + .0333 + 0) = .0748/.1081 = .6920

What you learnedWhat you learned

Disjoint EventsDisjoint Events Independent EventsIndependent Events Disjoint vs IndependentDisjoint vs Independent Reversing the ConditionReversing the Condition

Additional ResourcesAdditional Resources

The Practice of Statistics—YMMThe Practice of Statistics—YMM• Pg 341 -363Pg 341 -363

The Practice of Statistics—YMSThe Practice of Statistics—YMS• Pg 359-387Pg 359-387

Against All Odds—Video #15Against All Odds—Video #15• http://www.learner.org/resources/series6http://www.learner.org/resources/series6

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