Module 3

write it 10 times

Bayes Rule, or Bayes Theorem, gives us a way to relate conditional probabilities. - In its simplest form, it follows from the previous results

A selection of r objects from a set of n objects without regarding to the order of selection is called a combination. - Use the choose function pictured - the (^n _r) part is read "n choose r"

Combination - A selection of r objects from a set of n objects without regarding to the order of selection is called a combination.

Answer in picture

Combination Example: How many ways are there to select 3 candidates from 8 equally qualified recent graduates for openings in an accounting firm?

The subset of all the elements of S (sample space) that are __not in A__

Complement of A (A^- or A^c)

part a) P(A) = n/N = 1/6 part b) P(A) = n/N = 3/6 part c) P(A) = n/N = 3/6

Consider rolling a fair, six-sided die. part a) What is the probability of rolling a 2? part b) What is the probability of rolling an even number? part c) What is the probability of rolling a prime number?

A: A⋂B = ∅ (Events A and B are mutually exclusive) A⋂C = {2} B⋂C = {3,5}

Consider rolling a six sided die. Let A = Rolling a 2 A = {2} Let B = Rolling an odd number A={1,3,5} Let C = Rolling a prime number A={2,3,5} find Intersection: A⋂B A⋂C B⋂C

A: A⋃B = {1,2,3,5} A⋃C = {2,3,5} B⋃C = {1,2,3,5}

Consider rolling a six sided die. Let A = Rolling a 2 A = {2} Let B = Rolling an odd number A={1,3,5} Let C = Rolling a prime number A={2,3,5} find Intersection: A⋃B A⋃C B⋃C

A^c = {1,3,4,5,6} B^c = {2,4,6} C^c = {1,4,6}

Consider rolling a six sided die. Let A = Rolling a 2 A = {2} Let B = Rolling an odd number A={1,3,5} Let C = Rolling a prime number A={2,3,5} find complement: A^c B^c C^c

A⋃C = {2,5,9,13,18,20,22}

Consider the following sets: A={2,20,22,5,13}, B={5,9,20,12,22}, and C={5,9,18}. Find A⋃C.

P(Multiracial | Hamilton)=P(Multiracial ⋂ Hamilton)P(Hamilton)=(73/21278)(3028/21278) = 73/3028

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). a) Calculate the probability that a randomly selected student on the Hamilton campus identifies as Multiracial.

P(Asian ⋂ Middletown) = 26/21278

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). b) Calculate the probability that a randomly selected student identifies as Asian and attends the Middletown campus.

P(Hisp/Lat ⋃ Oxford) = (772/21278) + (16370/21278) - (623/21278) = 16159/21278

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). c) Calculate the probability that a randomly selected student identifies as Hispanic / Latino or attends the Oxford campus.

P(AIAN | Middletown) = 6/1880

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). d) Given that a randomly selected student attends the Middletown campus, calculate the probability that the student identifies as American Indian (AI) or Alaska Native (AN).

P(Oxford) = 16370/21278 = 0.7693 P(Oxford | White) = 12503/16481 = 0.7586 Since P(Oxford) does not equal P(Oxford | White) the events attending the Oxford campus and identifying as White (and unknown) are not independent.

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). e) Are the events attending the Oxford campus and identifying as White (and unknown) independent? Provide a mathematical argument to justify your answer.

P(Hamilton) = 3028/21278 = 0.1423

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). f) Calculate the probability that a randomly selected student attends the Hamilton campus.

P(Oxford | Multi-racial) = 524/633 = 0.8278

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). g) Given that a randomly selected student identifies as Multi-racial, calculate the probability that the student attends the Oxford campus.

P(Middletown Asian) = (1880/21278) + (425/21278) - (26/21278) = 2279/21278

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). h) Calculate the probability that a randomly selected student attends the Middletown campus or identifies as Asian.

P(Non-Resident Alien) = (2061/21278) = 0.0969 P(Non-Resident Alien | Hamilton) = (18/3028) = 0.0059 since P(Non-Resident Alien) does not equal P(Non-Resident Alien | Hamilton) the events identifying as Non-Resident Alien and attending the Hamilton campus are not independent

Contingency Table Introduction: Example: The fall 2015 undergraduate enrollment by race/ethnicity is provided below for all three campuses of Miami University (Hamilton, Middletown, and Oxford). i) Are the events identifying as Non-Resident Alien and attending the Hamilton campus independent? Provide a mathematical argument to justify your answer.

(A⋂B)^C = A^C ⋃ B^C (A⋃B)^C = A^C ∩ B^C

DeMorgan's Laws

A __ collection of possible outcomes _____ from the random experiment - A ___subset___ of the sample space - Typically denoted with a ____capitol letter____ (i.e. A, D, F, etc.) - Ex: A single outcome (i.e. getting heads on a single coin flip), Several outcomes (i.e. rolling a prime number on a standard six-sided die), The entire sample space S, Null set - No outcome at all (Denoted by ∅)

Event

a) P(Buisness) = (0.6)(0.5) + (0.3)(0.6) + (0.1)(0.9) = 0.3 + 0.18 + 0.09 = 0.57 b) P(Private | Business) = (P(Private ⋂ Business)) / (P(Business)) = (0.3)(0.6)(0.57) = 619 = 0.3158

Ex of Bayes Rule: Of the travelers arriving at a small airport, 60% fly on major airlines, 30% fly on privately owned planes, and the remainder fly on commercially owned planes not belonging to a major airline. Of those traveling on major airlines, 50% are traveling for business reasons, whereas 60% of those arriving on private planes and 90% of those arriving on other commercially owned planes are traveling for business reasons. Suppose that we randomly select one person arriving at this airport. Hint: Draw a Tree Diagram. a) What is the probability that the person is traveling on business? b) What is the probability that the person arrived on a privately owned plane, given that the person is traveling for business reasons?

No restrictions: 8! = 8*7*6*5*4*3*2*1 = 40320 Each couple is to sit together: 4!*2!*2!*2!*2! = 384 All the men sit together to the right of all the women: 4!*4! = 576

Four married couples have purchased 8 seats in the same row for a concert. In how many different ways can they be seated with no restrictions? If each couple is to sit together? If all the men sit together to the right of all the women? 1. No restrictions: ... 2. Each couple is to sit together: ... 3. All the men sit together to the right of all the women: ...

If there are m ways to do the first step, and n ways to do the second there are m * n ways to do both steps.

Fundamental Counting Principle

A: (6!)/(3!*2!*1!) = 60 6! = 6 letters in the word PEPPER 3! = 3 P's in PEPPER 2! = 2 E's in PEPPER 1! = 1 R in PEPPER

How many different arrangements can be formed with the letters of the word PEPPER?

A: (^20 _4)(^70 _6)/(^90 _10) (^20 _4) = number of ways to pick the 4 male nurses out of 20 (^70 _6) = number of ways to pick the 6 female nurses out of 70 (^90 _10) = number of ways to pick 10 nurses out of 90

Hypergeometric Distribution Example: A study is to be conducted in a hospital to determine the attitudes of nurses towards various administrative procedures. Twenty of the 90 nurses are male. If 10 nurses are randomly selected from those employed by the hospital, what is the probability that the sample of ten will include exactly 4 male (and 6 female) nurses?

P(A⋃B) =P(A)+P(B)

If A and B are mutually exclusive, then

P(A⋃B) = P(A) + P(B) - P(A⋂B) - think about it as the pictured Venn Diagram.

If A and B are two events, then ...

The event containing all of the elements that are __common to both A and B__

Intersection of A and B (A⋂B)

write that 10 times

Law of Total Probability: If B_1, B_2, ... , B_n constitute a partition of S, then for any event A of S

A: (7!)/(2!*3!*2!) = 210 ways

Multinomial Example: Suppose I have 7 coins: 2 pennies, 3 nickels, and 2 dimes. In how many ways can they be arranged?First assume that they are all distinct, P1, P_2, N_1, N_2, N_3, D_1, D_2. There are 7! = 5040 ways to arrange them. However, in reality, they are not distinct. Now we need to divide out by the number of ways each group can be arranged.

In general, if an experiment involves k steps, and the i^(th) step has n_(i) possible outcomes, there are n_(1)*n_(2)*...*n_(k) possible outcomes in the experiment.

Multiplication Rule of Counting

Two events A and B are said to be mutually exclusive (disjoint) if __A⋂B = ∅__ - In other words, A and B have __no elements__ in common

Mutually Exclusive Events

The observed outcome of an experiment

Observation

P(A) = probability of event A n = number of outcomes that correspond to event A N = number of total outcomes

P(A) = n = N =

area of A (the bean) relative to the area of S

P(A) = ...

called multinomial coefficients

Partitioning n distinct objects into k nonoverlapping groups

The number of ways you can arrange n distinct objects - In general, there are n!=n*(n-1)*...*3*2*1 permutations on n objects - In general, the number of permutations of n distinct objects taken r at a time is, nPr = (n!)/((n-r)!)

Permutation

Pick 1st Person: 4 options Pick 2nd Person: 3 options Pick 3rd Person: 2 options Pick 4th Person: 1 option Final Answer: 4*3*2*1 = 24 ways, or = 4!

Permutation Example: Consider the following four people: Alice, Bob, Carol, and Dave. How many ways can you order the four people?

Answer in picture

Permutation Example: Ryan, Emily, Megan, John, and Charlie want to take a photo in which three of the five friends are lined up in a row. How many different photos are possible?

A ____measure of one's belief____ in the occurrence of a future event. Used to quantify _____uncertainty____

Probability

A: False

Q: -0.25 is an example of a probability T/F?

A: (0.46)(0.46)(0.46) = 0.097 = 9.7%

Q: A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.

A: P(Starbucks or Crazy) = P(Starbucks)+P(Crazy)= 3/9 + 4/9 = 7/9

Q: A city has 9 coffee shops: 3 Starbucks, 2 Caribou Coffees, and 4 Crazy Mocho Coffees. If a person selects one shop at random to buy a cup of coffee, find the probability that it is either a Starbucks or Crazy Mocho Coffees.

A: _(12)C_(4) = 495

Q: A committee of 4 students is to be selected from among the 12 engineering students enrolled in a senior capstone course. How many different committees are possible?

A: S = {1,2,3,4,5,6}

Q: Consider a simple 6-sided die roll. Determine the sample space.

A: S = { (x, y) | x^2 + y^2 < 1}

Q: Consider all the points (x , y) within the unit circle. Determine the sample space. Note: We are thinking of a sample space in terms of an area for this example.

A: P(HH)=(1/2)(1/2)=1/4 P(HT)=(1/2)(1/2)=1/4 P(T1)=(1/2)(1/6)=1/12 P(T2)=(1/2)(1/6)=1/12 P(T3)=(1/2)(1/6)=1/12 P(T4)=(1/2)(1/6)=1/12 P(T5)=(1/2)(1/6)=1/12 P(T6)=(1/2)(1/6)=1/12 S={HH,HT,T1,T2,T3,T4,T5,T6}

Q: Consider flipping a fair two-sided coin. If you flip a head, we flip a different fair two-sided coin. If you flip a tail, we roll a 6-sided die. Determine the sample space below to count the total number of outcomes for this experiment.

A: A = {2} B = {1, 3, 5} C = {2, 3, 5}

Q: Consider rolling a six-sided die. Define the following events: a) Rolling a 2 b) Rolling an odd number c) Rolling a prime number.

A: S = {x | 48 ≤ x ≤ 84}

Q: Consider the height (in inches) of the next person to walk into the classroom. Determine the sample space below. Note: The biologically plausible range for adult height is 48 to 84 inches.

A: a) P(Math or History) = P(Math) + P(History) - P(Math and History) P(Math or History) = 54/100 + 69/100 - 35/100 = 88/100 Using Venn Diagram: P(Math or History) = 0.19+0.35+0.34=0.88 Or P(Math or History) = 1-0.12=0.88 b) P[(Math or History)C] = 1-P(Math or History) =1-0.88=0.12 c) P(History Only) = P(History)-P(History and Math) = 0.69-0.35=0.34

Q: In a high school graduating class of 100 students, 54 studied mathematics, 69 studied history, and 35 studied both mathematics and history. If one of these students is selected at random, find the probability that: a) The student took mathematics or history b) The student did not take either of these subjects c) The student took history but not mathematics

A: 4*3*3*2=72

Q: In how many different ways can you configure a computer system if there are 4 choices of processor, 3 choices of memory, 3 choices of disk drive, and 2 choices of monitor?

A: In the picture.

Q: The probability that a patient recovers from a stomach disease is 0.8. Suppose 3 people are known to have contracted this disease. What is the probability that exactly 2 recover? Hint: Draw a Tree Diagram

A: P(Heart or Queen) = P(Heart) + P(Queen) - P(Heart and Queen) P(Heart or Queen) = 1352 + 452 - 152=1652 Since P(Heart and Queen) does not equal 0, the two events are NOT mutually exclusive.

Q: What is the chance of drawing a card from a full deck and getting either a heart or a queen? Are the 2 events mutually exclusive?

A: 4*10*2 = 80

Q: You own 4 pairs of jeans, 10 T-shirts, and 2 pairs of shoes. How many outfits (jeans, T-shirts, and shoes) can you create?

Any process that generates an unpredictable or __uncertain__ (random) outcome Ex: flipping a coin, rolling a six-sided die

Random Experiment

Probabilities correspond to the __long run__ frequency of an event occurring based on repeated experiments

Relative Frequency Approach to Probability

The ___set of all possible outcomes____ of an experiment - Typically denoted by ___S___ - Can be written using mathematical notation (known as the rule method)

Sample Space

Use intuition or experience to assign probabilities. - An "__expert__ opinion" - Ex: A doctor assigning a patient's chance of recovery from a surgery.

Subjective Approach to Probability

That order matters for a permutation but order does not matter for a combination.

The main difference between a permutation and a combination is ...

Explicitly count all of the possible outcomes and those of interest - Assume a ___mathematical model___ and derive probabilities from the model

Theoretical (Classical) Approach to Probability

1. Theoretical (classical) Approach to Probability 2. Relative Frequency Approach to Probability 3. Subjective Approach to Probability

Three Types of Probability

The event containing all of the elements that belong to ____either A or B or both____

Union of A and B (A⋃B)

P(A) = n/N = 1/52

What is the chance of drawing the queen of spades from a deck of 52 cards?