POD #5012/2/ B #6b Is this model appropriate? Write the equation of the LSRL

Stats: Modeling the World Chapter 15 Probability Rules!

Example On 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

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 4610

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. MathEnglish

A survey of 100 couples (taken by calling every 50 th 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 that a. at least one of them is employed. b. neither is employed. Husband Wife 55305

Example On 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

General Addition Rule For 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 #5112/5/2011Vocab Review Please number your paper Reassessments and makeup work should be done by Wednesday, Dec 14!!!

Comparing Venns to Tables A 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?

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… 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 Probabilities A 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? b)What percent of the minority graduates are planning to attend a 4-year college? c)What percent of the graduates pursuing the military are white? d)What percent of the graduates pursuing a 4-year college are minority?

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

POD #5212/6/ #5p 1.P(Satisfied) = ? 2.P(Male OR Not Satisfied) = ? 3.P(Satisfied | Female) = ? 4.P (Male AND Satisfied) = ?

Example A 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.

Working with the Formulas An 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) = 0.8 a. 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) = – 0.24 = 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 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 Independence The 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 P(B|A) = P(B) P(Stat|Computer) = P(Stat) 30% = 52%

Disjoint vs. Independence In 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? Not disjoint – you can have both

Disjoint vs. Independence In 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? Dependent Events!!!! P(B|A) = P(B) P(Garage|Pool) = P(Garage) 81% = 64%

POD #5312/7/ #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… AceHearts Not Disjoint AceKing Disjoint

Rolling Dice… Roll a 6 P(A)= Not a 6 P(B)= Roll a 6 P(A|A)= Not a 6 P(B|A)= Roll a 6 P(A|B)= Not a 6 P(B|B)=

Drawing Cards… Ace P(A)= Not Ace P(B)= Ace P(A|A)= Not Ace P(B|A)= Ace P(A|B)= Not Ace P(B|B)=

Tree Diagrams A 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!