Warm-upWarm-up Define the sample space of each of the Define the sample space of each of the

following situations…following situations… Choose a student in your class at random. Choose a student in your class at random.

Ask how much time that student spent Ask how much time that student spent studying during the past 24 hours.studying during the past 24 hours.

The Physician’s Health Study asked 11,000 The Physician’s Health Study asked 11,000 physicians to take an aspirin every other physicians to take an aspirin every other day and observed how many of them had a day and observed how many of them had a heart attack in a five-year period.heart attack in a five-year period.

In a test of new package design, you drop a In a test of new package design, you drop a carton of a dozen eggs from a height of 1 carton of a dozen eggs from a height of 1 foot and count the number of broken eggs.foot and count the number of broken eggs.

Section 6.2 Section 6.2

Independence and the Independence and the Multiplication RuleMultiplication Rule

Two Special RulesTwo Special Rules

We’ve learned the We’ve learned the addition rule for addition rule for disjoint eventsdisjoint events: If A and B are : If A and B are disjointdisjoint, , then P(A or B) = P(A) + P(B).then P(A or B) = P(A) + P(B).

Now we’ll learn the Now we’ll learn the multiplication rule multiplication rule for independent eventsfor independent events: that if A and B : that if A and B are are independentindependent, then P(A and B) = , then P(A and B) = P(A)●P(B)P(A)●P(B) Remember that two events are independent if Remember that two events are independent if

knowing that one occurs does not change the knowing that one occurs does not change the probability that the other occurs.probability that the other occurs.

Examples of Independent EventsExamples of Independent Events

Toss a coin twice. Let A = first toss Toss a coin twice. Let A = first toss is a head and B = second toss is a is a head and B = second toss is a head. Events A and B are head. Events A and B are independent. Thus, the P(Aindependent. Thus, the P(A∩B) = ∩B) = P(A)*P(B).P(A)*P(B). P(Head and Tail) = P(Head) * P(Tail) = ½ * ½ = ¼ P(Head and Tail) = P(Head) * P(Tail) = ½ * ½ = ¼

Draw 3 cards from a deck, replacing and Draw 3 cards from a deck, replacing and shuffling in between each draw. This is shuffling in between each draw. This is called “with replacement.” called “with replacement.”

Without ReplacementWithout Replacement

If you draw three cards from a deck If you draw three cards from a deck without replacing, the probabilities without replacing, the probabilities change on each draw. Therefore, change on each draw. Therefore, drawing without replacement is NOT drawing without replacement is NOT independent.independent.

CautionsCautions The addition rule The addition rule

for disjoint events for disjoint events and the and the multiplication rule multiplication rule for independent for independent events only work events only work when the criteria when the criteria are met. Resist the are met. Resist the temptation to use temptation to use them for events them for events that are that are not not disjoint or disjoint or not not independent.independent.

You must be told or You must be told or have prior have prior knowledge that an knowledge that an event is disjoint or event is disjoint or independent.independent.

Do not confuse Do not confuse disjoint with disjoint with independent.independent. Disjoint can be Disjoint can be displayed in a Venn displayed in a Venn Diagram. Diagram. Independence can Independence can not.not.

Sample QuestionsSample Questions

Suppose that among the 6000 Suppose that among the 6000 students at a high school, 1500 are students at a high school, 1500 are taking honors courses and 1800 taking honors courses and 1800 prefer watching basketball to prefer watching basketball to watching football. If taking honors watching football. If taking honors courses and preferring basketball are courses and preferring basketball are independent, how many students are independent, how many students are both taking honors courses and both taking honors courses and prefer basketball to football?prefer basketball to football?

Sample QuestionsSample Questions

Suppose that for any given year, the Suppose that for any given year, the probabilities that the stock market probabilities that the stock market declines is .4, and the probability declines is .4, and the probability that women’s hemlines are lower that women’s hemlines are lower is .35. Suppose that the probability is .35. Suppose that the probability that both events occur is .3. Are the that both events occur is .3. Are the two events independent?two events independent?

Sample QuestionsSample Questions

In a 1974 “Dear Abby” letter, a woman In a 1974 “Dear Abby” letter, a woman lamented that she had just given birth to lamented that she had just given birth to her eighth child, and all were girls! Her her eighth child, and all were girls! Her doctor had assured her that the chance of doctor had assured her that the chance of the eighth child being a girl was only 1 in the eighth child being a girl was only 1 in 100.100. A) What was the real probability that the A) What was the real probability that the

eighth child would be a girl?eighth child would be a girl? B) Before the birth of the first child, what was B) Before the birth of the first child, what was

the probability that the woman would give the probability that the woman would give birth to eight girls in a row?birth to eight girls in a row?

Section 6.3Section 6.3General Probability RulesGeneral Probability Rules

Special Probability RulesSpecial Probability Rules

If A and B are disjoint, then P(AUB) = If A and B are disjoint, then P(AUB) = P(A) + P(B).P(A) + P(B).

This rule relies on a condition to be This rule relies on a condition to be met.met.

So, what if the events are NOT So, what if the events are NOT disjoint?disjoint?

What if there are more than 2 What if there are more than 2 disjoint events?disjoint events?

General Probability RulesGeneral Probability Rules

P(AUB) = P(A) + P(B) – P(A∩B)P(AUB) = P(A) + P(B) – P(A∩B)

This is how the formula will appear This is how the formula will appear on your formula sheet for the exam on your formula sheet for the exam (and thus my tests).(and thus my tests).

If A and B are disjoint, then what does P(A∩B) equal?

Sample Question: Student Sample Question: Student SurveySurvey

(making an A in Statistics only) = 18 students (making an A in Statistics only) = 18 students (making an A in Calculus only) = 63 students (making an A in Calculus only) = 63 students (making an A in Stats and in (making an A in Stats and in Calculus) = 27 students Calculus) = 27 students

S=150 studentsS=150 students Draw a Venn diagram.Draw a Venn diagram. Find the probability of making an A in Stats but not in Find the probability of making an A in Stats but not in

Calc.Calc. Find the probability of making an A in Calc but not in Find the probability of making an A in Calc but not in

Stats.Stats. Find the probability of not making an A in either Find the probability of not making an A in either

subject.subject. Find the probability of making an A in either Calc or Find the probability of making an A in either Calc or

Stats.Stats. Are these events independent? Are these events independent?

Extending the Rules to Three Extending the Rules to Three EventsEvents

If you have three disjoint events, If you have three disjoint events, then P(A or B or C) = P(A) + P(B) + then P(A or B or C) = P(A) + P(B) + P(C)P(C)

If they are not disjoint, then P(A or B If they are not disjoint, then P(A or B or C) = P(A) + P(B) + P(C) – P(A∩B) – or C) = P(A) + P(B) + P(C) – P(A∩B) – P(A∩C) – P(B∩C)P(A∩C) – P(B∩C)

Often it is easier to draw a Venn Often it is easier to draw a Venn Diagram than to use the formula.Diagram than to use the formula.

Look at p365 6.51Look at p365 6.51

Sample QuestionsSample Questions Suppose that the probability that you will Suppose that the probability that you will

receive an A in AP Statistics is .35, the receive an A in AP Statistics is .35, the probability that you will receive As in both probability that you will receive As in both AP Statistics and AP Biology is .19, and the AP Statistics and AP Biology is .19, and the probability that you will receive an A in AP probability that you will receive an A in AP Bio but not in Stats is .17. Which is a Bio but not in Stats is .17. Which is a proper conclusion?proper conclusion? A) P(A in AP Bio) = .36A) P(A in AP Bio) = .36 B) P(you didn’t take Bio) = .01B) P(you didn’t take Bio) = .01 C) P(not making an A in AP Stat or Bio) = .38C) P(not making an A in AP Stat or Bio) = .38 D) The given probabilities are impossible.D) The given probabilities are impossible.

Sample QuestionsSample Questions

If P(A) = .2 and P(B) = .1, what is If P(A) = .2 and P(B) = .1, what is P(AUB) if A and B are independent?P(AUB) if A and B are independent? A) .02A) .02 B) .28B) .28 C) .30C) .30 D) .32D) .32 E) There is insufficient information to E) There is insufficient information to

answer this question.answer this question.

Sample QuestionsSample Questions

Given the probabilities P(A) = .4 and Given the probabilities P(A) = .4 and P(AUB) = .6, what is the probability P(B) if P(AUB) = .6, what is the probability P(B) if A and B are mutually exclusive? If A and B A and B are mutually exclusive? If A and B are independent?are independent? A) .2, .28A) .2, .28 B) .2, .33B) .2, .33 C) .33, .2C) .33, .2 D) .6, .33D) .6, .33 E) .28, .2E) .28, .2

Conditional ProbabilityConditional Probability When events are not independent, the When events are not independent, the

probability of one changes if we know that probability of one changes if we know that the other event occurred.the other event occurred. I will draw two cards without replacement.I will draw two cards without replacement. Let A = {1Let A = {1stst card I draw is an Ace} card I draw is an Ace} Let B = {2Let B = {2ndnd card I draw is an Ace} card I draw is an Ace}

The probability of B occurring changes The probability of B occurring changes depending on whether A occurred.depending on whether A occurred.

The new notation P(B|A) is read “the The new notation P(B|A) is read “the probability of B given A.” It asks you to probability of B given A.” It asks you to find the probability of B knowing that A has find the probability of B knowing that A has occurred.occurred.

Let’s look at education and ageLet’s look at education and age

EducationEducation

AgeAge

25 – 3425 – 34 35 – 5435 – 54 55 +55 + TotalTotalNo high No high schoolschool 4,4744,474 9,1559,155 14,22414,224 27,85327,853

Completed Completed HSHS 11,54611,546 26,48126,481 20,06020,060 58,08758,087

1 to 3 years 1 to 3 years collegecollege 10,70010,700 22,61822,618 11,12711,127 44,44544,445

4+ years of 4+ years of collegecollege 11,06611,066 23,18323,183 10,59610,596 44,84544,845

TotalTotal 37,78637,786 81,43581,435 56,00856,008 175,230175,230

Find these probabilitiesFind these probabilities

Let A = {the person chosen is 25 – 34}Let A = {the person chosen is 25 – 34} Let B = {the person chosen has 4+ Let B = {the person chosen has 4+

years of college}years of college} Find P(A).Find P(A). Find P(A and B).Find P(A and B). Find P(B|A).Find P(B|A). Given that a person has only a HS Given that a person has only a HS

diploma, what is the probability that diploma, what is the probability that the person is 55 or older? the person is 55 or older?

The general rule for The general rule for multiplicationmultiplication

P(A∩B) = P(A)*P(B|A)P(A∩B) = P(A)*P(B|A) This can be rewritten as This can be rewritten as

P(B|A) = P(A∩B)/P(A)P(B|A) = P(A∩B)/P(A)

Example: Two cards are drawn without Example: Two cards are drawn without replacement.replacement.

Let A = {1Let A = {1stst card is a spade} card is a spade} Let B = {2Let B = {2ndnd card is a spade} card is a spade} Find P(B|A). Find P(B|A). Find P(A and B).Find P(A and B).

Next ExampleNext Example

Suppose the probability that the dollar Suppose the probability that the dollar falls in value compared to the yen is 0.5.falls in value compared to the yen is 0.5.

Suppose that the probability that if the Suppose that the probability that if the dollar falls in value, a supplier from Japan dollar falls in value, a supplier from Japan will renegotiate their contract is 0.7.will renegotiate their contract is 0.7.

What is the probability that the dollar will What is the probability that the dollar will fall in value and the supplier demands fall in value and the supplier demands renegotiation?renegotiation?

HomeworkHomeworkChapter 5# 4, 9, 40, 43, 51, Chapter 5# 4, 9, 40, 43, 51,

60, 83, 89, 9460, 83, 89, 94