Graded Homework

P. 163, #29P. 170, #35, 37

Graded Homework, cont.

P. 163, #29U. Of Pennsylvania•1,033 admitted early (E)•854 rejected outright (R)•964 deferred (D)•Typically 18% of deferred early admission are admitted in the regular admission process (173.5)•Total number of students admitted = 2,375

Graded Homework, cont.

a. P(E) = 1033/2851 = 0.362P(R) = 854/2851 = 0.300P(D) = 964/2851 = 0.338

b. Yes, a student cannot be both admitted and deferred, so P(E∩D)=0

c. 1033/2375 = 0.435d. 1033/2851 + (964/2851)(.18) = 0.423

Graded Homework, cont.P. 170, 35Occupation Male Female TotalManagerial/Professional 19079 19021 38100Tech./Sales/ Administrative 11079 19315 30394Service 4977 7947 12924Precision Production 11682 1138 12820Operators/Fabricators/Labor 10576 3482 14058Farming/Forestry/Fishing 1838 514 2352Total 59231 51417 110648

Graded Homework, cont.P. 170, 35Occupation Male Female TotalManagerial/Professional 0.17 0.17 0.34Tech./Sales/ Administrative 0.10 0.17 0.27Service 0.04 0.07 0.12Precision Production 0.11 0.01 0.12Operators/Fabricators/Labor 0.10 0.03 0.13Farming/Forestry/Fishing 0.02 0.00 0.02Total 0.54 0.46 1.00

b. P(Manager|Female) = 0.17/0.46 = 0.37c. P(Precision production|Male) = 0.11/0.54 = 0.20d. No, P(Manager|Female) = 0.37, P(Manager) = 0.34

Graded Homework, cont.P. 170, 37P(PC) = .37 P(Y) = .14P(Y|PC) = .19 P(O|PC) = .81

a. P(PC|Y) = P(Y ∩ PC)/P(Y) = [P(PC)P(Y|PC)]/P(Y) = [(.37)(.19)]/(.14) = 0.5

b. P(PC|O) = P(O ∩ PC)/P(O) = [P(PC)P(O|PC)]/P(O) = [(.37)(.81)]/(.86) = 0.35

c. People under 24 years old are more likely to use credit cards.

d. Yes, otherwise they can’t establish a credit history and the companies want customers who will make heavy use of the cards. They could put strict limits on the maximum balance for the card.

Multiplication Law

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

Multiplication Rule, cont.

If events A and B are independent then P(A|B) = P(A)P(B). In this special case the multiplication rule reduces from:

P(A ∩ B) = P(B)P(A|B)to:

P(A ∩ B) = P(B)P(A)

Tree Diagram

Red, 2/4Red, 2/4

Black, 2/4

Black, 2/4Red, 2/4

Black, 2/4

Assume we take the four aces out of a deck of cards and we draw twice with replacement:

Are A and B statistically independent in this case?

Sampling and Statistical Independence

If we sample without replacement the outcomes will not be statistically independent.

However, if we are drawing from a large population the change in probability will be so small we can treat the draws as being statistically independent.

Bayes’ Theorem

A technique used to modify a probability given additional information.

Prior probab

ility

New information

Application of

Bayes’ Theorem

Posterior Probabili

ties

)|()()|()()|()()|(

2211

111 ABPAPABPAP

ABPAPBAP

Bayes’ Theorem, cont.

Assume that 10% of the population has a disease. Assume there is a test to see if someone has the disease but it is not very accurate.

Bayes’ Theorem, cont.

Has disease, .1

Test: Has disease, .8

Test: No disease, .2

No disease, .9

Test: Has disease, .3

Test: No disease, .7

Assume we want to calculate the probability that someone has the disease if the test says they have the disease.

Bayes’ Theorem, cont.

Has disease, .1

Test: Has disease, .8

Test: No disease, .2

No disease, .9

Test: Has disease, .3

Test: No disease, .7

.08

.02

.27

.63

A1 = Has the diseaseB = Test says the patient has the diseaseP(A1) = .1 P(A2) = .9P(B|A1) = .8 P(B|A2) = .3

Bayes’ Theorem, cont.

229.027.08.

08.3.9.8.1.

8.1.)|()()|()(

)|()()|(2211

111

ABPAPABPAPABPAPBAP

Bayes’ Theorem, cont.

EventsAi

Prior ProbabilitiesP(Ai)

Conditional ProbabilitiesP(B|Ai)

Joint ProbabilitiesP(Ai∩B)

Posterior ProbabilitiesP(Ai|B)

A1 .1 .8 .08 .229

A2 .9 .3 .27 .771

Total 1.00 .35 1.00

Practice

Assume that 40% of a company’s parts are produced in Boston and 60% are produced in Chicago. Also assume that 20% of the parts produced in Boston are defective, and 10% of the parts produced in Chicago are bad.

A randomly chosen part is defective. Use Bayes Theorem to find the probability the part came from Boston.

Bayes’ Theorem, cont.

Boston, .4Defective, .2

Ok, .8

Chicago, .6Defective, .1

Ok, .9

Bayes’ Theorem, cont.

Boston, .4Defective, .2

Ok, .8

Chicago, .6Defective, .1

Ok, .9

.08

.32

.06

.54

A1 = BostonB = Part is defectiveP(A1) = .4 P(A2) = .6P(B|A1) = .2 P(B|A2) = .1

Bayes’ Theorem, cont.

571.006.08.

08.1.6.2.4.

2.4.)|()()|()(

)|()()|(2211

111

ABPAPABPAPABPAPBAP

Bayes’ Theorem, cont.

EventsAi

Prior ProbabilitiesP(Ai)

Conditional ProbabilitiesP(B|Ai)

Joint ProbabilitiesP(Ai∩B)

Posterior ProbabilitiesP(Ai|B)

A1 .4 .2 .08 .571

A2 .6 .1 .06 .429

Total 1.00 .14 1.00

Counting Rules

• Number of possible outcomes• Combinations• Permutations

Number of Possible Outcomes

Given k steps (or rounds) in an experiment and ni possible outcomes at step i, the total number of possible outcomes is:(n1) (n2)…(nk)

Number of Possible Outcomes, cont.

Assume a diner can choose:• Either soup or salad• One of three main dishes (beef, chicken or vegetarian)• Either potatoes or beans

How many possible meals are there?(2)(3)(2) = 12

Number of Possible Outcomes, cont.

Soup

BeefPotatoes

Beans

ChickenPotatoes

Beans

VeggiePotatoes

Beans

Salad

BeefPotatoes

Beans

ChickenPotatoes

Beans

VeggiePotatoes

Beans

Number of Possible Outcomes, cont.

Assume car buyer can choose:• automatic or standard transmission• 6 different colors• 3 body styles• 4 different accessory packages

How many possible different outcomes are there?(2)(6)(3)(4) = 144

Factorials

N! = (N)(N-1)(N-2)…(2)(1)

5! = (5)(4)(3)(2)(1) = 120What is the value of 4! ?(4)(3)(2)(1) = 24

What is the value of 5!/3! ?[(5)(4)(3)(2)(1)]/[(3)(2)(1)] = (5)(4) = 20

By definition 0! = 1

Permutations

!!!nN

NnN

nPNn

The number of permutations of N objects taken n at a time:

Permutations, cont.

Assume a broker is going to pick 3 stocks from a pool of 10 stocks. Also assume he will invest 60% of his money in one stock, 30% in another, and 10% in another. How many portfolios can be constructed?

720)8)(9)(10(!7!10

)!310(!10

!!!

nNN

nN

nPNn

Combinations

!!!nNn

NnN

C Nn

The number of combinations of N objects taken n at a time:

Combinations, cont.

A bank is constructing a bond based on mortgages. It is going to base the bond from four mortgages, it has ten mortgages to choose from. How many ways can the bond be structured?

210)1)(2)(3)(4()7)(8)(9)(10(!6!4!10

)!410(!4!10

!!!

nNnN

nN

C Nn

Practice1. Three employees will be chosen from an office of 8 workers

for a committee to evaluate a new production technique. How many possible committees could be formed?

2. Assume a club has 5 members and they are going to elect a president, treasurer, and secretary. How many ways can the offices be filled?

3. A magazine is going to recommend two of ten products to its readers. It will identify the rankings of the two products that are selected. How many potential rankings are there?

Graded Homework

P. 151, 1, 3 + redo 3 assuming order is important (counting rules)P. 169, 31 (Multiplication rule)P. 177, 43 (Bayes’ theorem)