POD #50 12/2/2011 2007B #6b

Is this model appropriate?

Write the equation of the LSRL

Stats: Modeling the WorldChapter 15

Probability Rules!

ExampleOn a large college campus, the RA’s randomly selected one dorm room per floor for a total of 100 rooms. This survey revealed that 38 had refrigerators, 52 had TVs, and 21 had both a TV and a fridge How many of the dorm rooms had a TV or a refrigerator? (either or both)

fridge TV

2117 31

How many students were interviewed?

How many students liked Rock?

How many students liked Rap OR Country?

How many students liked 2 out of 3 types?

How many students liked all three types?

How many students liked 1 type of music only?

Using a roster of all students in an elementary school, 35 students were randomly selected. Ten of the students had blonde hair, 14 had brown eyes, and 4 had both blonde hair and brown eyes.

If a child is selected at random, find the probability that the child has blonde hair or brown eyes. (either or both)

Blonde hair Brown eyes

46 10

As 400 college students walked by our interviewer, we learned that 120 are enrolled in math, 220 are enrolled in English, and 55 are enrolled in both.

Find the probability that…

a. the student is enrolled in mathematics.

b. the student is enrolled in mathematics or English.

c. the student is enrolled in either mathematics or English, but not both.

Math English

5565 165

A survey of 100 couples (taken by calling every 50th person from the voter rolls) found the that the husband was employed in 85 of the couples. The wife was employed in 60 of the couples, and both spouses were employed in 55 of the couples.

Find the probability thata. at least one of them is employed.b. neither is employed.

Husband Wife

5530 5

6

12

46

32 21

19

ExampleOn a large college campus, the RA’s randomly selected one dorm room per floor for a total of 100 rooms. This survey revealed that 38 had refrigerators, 52 had TVs, and 21 had both a TV and a fridge How many of the dorm rooms had a TV or a refrigerator? (either or both)

fridge TV

2117 31

General Addition RuleFor any two events A and B,

P(A or B) = P(A) + P(B) – P(A and B)

When the two events are disjoint or mutually

exclusive, then

P(A and B)=0

POD #51 12/5/2011 Vocab Review

Please number your paper 1-10.

Reassessments and makeup work should be done by Wednesday, Dec 14!!!

Comparing Venns to TablesA check of dorm rooms on a large college campus revealed that 38% had refrigerators, 52% had TVs, and 21% had both a TV and a fridge What’s the probability that a randomly selected dorm room has:

Prior to graduation a high school class was surveyed about their plans. The table below displays the results for white and minority students.

a) What percent of the graduates are white?

b) What percent of the graduates are planning to attend a 2-year college?

c) What percent of the graduates are minority or planning to attend a 4-year college?

268 57

2424251719

325

325

268

325

42

325

255

Prior to graduation a high school class was surveyed about their plans. The table below displays the results for white and minority students.

Give the conditional distributions of plans for white and minority students.

Making Connections…

268 57

2424251719

325

White: 4-year college: 74% 2-year college: 13% Military: 2% Employment: 5% Other: 6%

Minority: 4-year college: 77% 2-year college: 11% Military: 2% Employment: 5% Other: 5%

Conditional ProbabilitiesA probability that takes into account a given condition is called a conditional probability

This is written as P(A|B) and read as ‘the probability of A given B.

Prior to graduation a high school class was surveyed about their plans. The table below displays the results for white and minority students.

a) What percent of the white graduates are planning to attend a 2-year college?

c) What percent of the minority graduates are planning to attend a 4-year college?

d) What percent of the graduates pursuing the military are white?

e) What percent of the graduates pursuing a 4-year college are minority?

268 57

2424251719

325

268

36

57

44

5

4

242

44

Note: P(A) cannot equal 0, since we know that A has occurred.

(A and B)( | )( )

PPP

B AA

More Formally…

General Multiplication Rule

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

Note: There’s nothing special about which one we write as A or B, so P(A and B) = P(B) x P(A|B)

Replacement… or notSampling without replacement means that once one individual is drawn it doesn’t go back into the pool.

Drawing without replacement is just another instance of working with conditional probabilities.

132

20

11

4

12

5

132

6

11

2

12

3

132

12

11

3

12

4

132

38

132

12620

132

20

11

5

12

4

132

20

11

4

12

5

132

15

11

3

12

5

132

0

11

0

12

4

POD #52 12/6/2011 2004 #5p

1. P(Satisfied) = ?

2. P(Male OR Not Satisfied) = ?

3. P(Satisfied | Female) = ?

4. P (Male AND Satisfied) = ?

ExampleA junk box in your room contains a dozen old batteries, five of which are totally dead. You start picking batteries one at a time and testing them. Find the probability of each outcome.a. The first two you choose are both good.

b. The first four you pick all work.

c. You have to pick 5 batteries in order to find one that works.

d. At least one of the first three works.

132

42

11

6

12

7

11880

840

9

4

10

5

11

6

12

7

95040

840

8

7

9

2

10

3

11

4

12

5

955.010

3

11

4

12

51)(1

noneP

Working with the FormulasAn aerospace company has submitted bids on two separate federal defense contracts A and B. The company feels that it has a 60% chance of winning contract A and a 30% chance of winning contract B. Given that it wins contract B, the company believes it has an 80% chance of winning contract A. P(A) = 0.6 P(B) = 0.3 P(A|B) = 0.8a. What is the probability that the company will win both contracts?

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

P(A and B) = 0.3 x 0.8 = 0.24

b. What is the probability that the company will win at least one of the two contracts? (That means A OR B)

P(A or B) = P(A) + P(B) – P(A and B) P(A or B) = 0.6 + 0.3 – 0.24 = 0.66

Vocab Alert…The “rules” rely on two important ideas…

Disjoint events (Mutually Exclusive)- No overlap in the circles… the events do not occur together… P(A and B) = 0

Independent events- Event A does not affect Event B

Disjoint???A university requires its biology majors to take a course called BioResearch. The prerequisite for this course is that students must have taken either a Stat course or a computer course. By the time they are juniors, 52% of Biology majors have take Stat, 23% a computer course, and 7% both.

Are taking these two courses disjoint events? Explain.

Not disjoint – you can take both courses

Independent???A university requires its biology majors to take a course called BioResearch. The prerequisite for this course is that students must have taken either a Stat course or a computer course. By the time they are juniors, 52% of Biology majors have take Stat, 23% a computer course, and 7% both.

Are taking these two courses independent events?

In other words… does taking a computer course CHANGE your probability of taking a stat course?

Independent??? How can we tell???A university requires its biology majors to take a course called BioResearch. The prerequisite for this course is that students must have taken either a Stat course or a computer course. By the time they are juniors, 52% of Biology majors have take Stat, 23% a computer course, and 7% both.

23

52

100

748

77

4516 32

If you took computers, what’s the chance of stat???

General chance of taking a stat course???

Since computers did affect your chances of stat, these are Dependent Events!!!!

52%30%

Formal IndependenceThe Rule…

Events A and B are independent whenever P(B|A) = P(B).

(Again, note that it does not matter which event is A and which is B.)

Independent??? How can we tell???A university requires its biology majors to take a course called BioResearch. The prerequisite for this course is that students must have taken either a Stat course or a computer course. By the time they are juniors, 52% of Biology majors have take Stat, 23% a computer course, and 7% both.

23

52

100

748

77

4516 32

P(B|A) = P(B) P(Stat|Computer) = P(Stat) 30% = 52%

Disjoint vs. IndependenceIn the real estate ads, 64% of homes have garages, 21% have swimming pools, and 17% have both features.

a) Are having a garage and a pool disjoint events?

21

64

100

1736

79

474 32

Not disjoint – you can have both

Disjoint vs. IndependenceIn the real estate ads, 64% of homes have garages, 21% have swimming pools, and 17% have both features.

b) Are having a garage and a pool independent events?

21

64

100

1736

79

474 32

Dependent Events!!!!

P(B|A) = P(B) P(Garage|Pool) = P(Garage) 81% = 64%

POD #53 12/7/2011 2011 #2ab

POD #53 12/7/2011 2011 #2ab

I am NOT your mother!!!

-If you get it out, put it back where you got it-If you make a mess, clean it up-If you write on your desk, erase it (we now have washcloths – use them!)

Disjoint or Not…

Ace Hearts

352

152

1252

Not Disjoint

Ace King

452

452

Disjoint

Rolling Dice…

Roll a 6P(A)=

Not a 6P(B)=

Roll a 6P(A|A)=

Not a 6P(B|A)=

Roll a 6P(A|B)=

Not a 6P(B|B)=

Drawing Cards…

AceP(A)=

Not AceP(B)=

AceP(A|A)=

Not AceP(B|A)=

AceP(A|B)=

Not AceP(B|B)=

Tree DiagramsA tree diagram helps us think through conditional probabilities by showing sequences of events as paths that look like branches of a tree.

Dan’s Diner employs three dishwashers. Al washes 40% of dishes and breaks only 1%. Betty and Chuck each wash 30% with Betty breaking only 1% and Chuck breaking 3% of his. You go to Dan’s for supper and hear a dish break at the sink. What’s the probability that Chuck is on the job?

A private college report contains these stats:70% of incoming Freshmen attended public schools.75% of public school students who enroll as freshmen eventually graduate.90% of other freshmen eventually graduate.

What percent of freshmen eventually graduate?What percent of students who graduate from college attended a public high

school?

What can go wrong? Don’t use the wrong formula! Many people use the

simple rules from Ch 14 when they should use the general rules. They (incorrectly) assume the events are disjoint or independent when they are not.

Watch out for replacement! When dealing with small populations, your denominator will change!

P(B|A) is NOT the same as P(A|B) Disjoint does NOT mean Independent!