CH9 & 10

The California lottery has offered a number of games over the years - One such game is Decco, in which a player chooses 1 card from each of the 4 suits in a regular deck of playing cards - For example, the player might choose "4 of hearts," "3 of clubs," "10 of diamonds," & "jack of spades" - A winning card is then drawn from each suit - If one or more of the choices match the winning cards drawn, a prize is awarded - It costs $1 for each play - Let X be the net gain for any single ticket - How much would you win or lose per ticket in this game, over the long run?

# of Matches Prize Net Gain Probability 4 $5000 $4999 0.000035 3 $50 $49 0.00168 2 $5 $4 0.0303 1 Free ticket 0 0.242 0 None -1 0.726 E(X) = ($4999 x 0.000035) + ($49 x 0.00168) + ($4 x 0.0303) + ($0 x 0.242) + (-$1 x 0.726) = -$0.35 This tells us that over many replicates of the game, players will lose an average of 35 cents per play - From the perspective of Lottery Commission, this means that they pay out 65 cents in prizes for each $1 ticket they sell for this game

Normal Probability Distribution of a Random Variable X

When the Density Curve is: - f(x) = 1/(σ√2π) x e^(-1/2)[(x - µ)/σ]²

Continuous Random Variable

A Random Variable that can take any value in an interval or collection of intervals - Usually defined on continuous sample spaces

Complete Independence DOES NOT Imply Pairwise Independence

An urn contains one red, one blue, one white, two yellow, & three black balls A ball is drawn randomly from the urn Let A be the Event that a red, yellow, or white ball is drawn; B the Event that a blue, yellow, or white ball is drawn; & C the Event that a black or white ball is drawn It is straightforward to show that: - P(A) = P(B) = P(C) = 1/2 - P(A & B) = 3/8 - P(A & C) = P(B & C) = 1/8 - P(A & B & C) = 1/8 Thus, A, B, & C are COMPLETELY Independent but not Pairwise Independent (Crow, E.L., 1957, A counterexample in independent events, American Mathematical Monthly 74, 716-717)

Multiplication Rule for Independent Events

If 2 Events A & B are INDEPENDENT, then: - P(A & B) = P(A) x P(B) - The inverse is also true

Sampling WITH Replacement

Individuals are RETURNED TO the ELIGIBLE POOL for EACH SELECTION

A Triangular Distribution

Let X be the sum of two Random Numbers between 0 & 1 - X can take any value between 0 & 2 The Density Curve for X is: - Height = 1 - We know this because the base = 2, & the area under the curve has to equal 1 by definition - The area of this isosceles triangle is ½ (base x height) What is the Probability that X is < 1? What is the Probability that X < 0.5?

Sample Event

ONE OUTCOME in the Sample Space

What is the Probability of randomly drawing an ace & a heart from a pack of 52 playing cards? There are 4 aces in the pack & 13 hearts.

P(ace) = 4/52 P(heart | ace) = 1/4 P(ace & heart) = P(ace) x P(heart|ace) (4/52) x (1/4) 1/52

Assume that there are equal chance of selecting boys & girls. What is the Probability that a random selected teenager is a boy & a gambler?

P(boy & gambler) = P(boy) x P(gambler | boy) 0.5 x 0.2 0.1

Two light bulbs are connected in the following two ways. If the reliability is the probability of functioning, P(A) = 0.9 & P(B) = 0.8 - Which circuit is more reliable?

Parallel P(A or B) = P(A) + P(B) - P(A & B) 0.9 + 0.8 - (0.9)(0.8) 1.7 - 0.72 0.98; this series is better Series P(A & B) = P(A) x P(B) 0.9 x 0.8 0.72

Probability Distributions of Continuous Random Variables

Since a Continuous Random Variable X can take an infinite number of values, its distribution is described by DENSITY CURVE - We only consider the PROBABILITY VALUES it takes in an INTERVAL Convention: - The Probability of a Continuous Random Variable taking a SINGLE VALUE is always ZERO Uniform Random Variable: - The Probability of a single event is zero: P(X = 1) = (1 − 1)*1 = 0 - P(0 ≤ X ≤ 0.5) = (0.5 − 0)*1 = 0.5 - P(0 < X < 0.5) = (0.5 − 0)*1 = 0.5 - P(0 ≤ X < 0.5) = (0.5 − 0)*1 = 0.5 - P(X < 0.5 or X > 0.8) = P(X < 0.5) + P(X > 0.8) = 1 − P(0.5 < X < 0.8) = 0.7

A quality-control procedure for testing Ready-Flash disposable cameras consists of drawing 2 cameras at random from each lot of 100 without replacing the first camera before drawing the second. If both are defective, the entire lot is rejected. Find the Probability that both cameras are defective if the lot contains 10 defective cameras.

Step 1: What is the probability of getting a defective camera on the first draw? P(1st is defective) = 10/100 = 1/10 Step 2: What is the probability of getting a defective camera on the second draw if the first camera was defective? P(2nd is defective | 1st is defective) = 9/99 = 1/11 Step 3: What is the probability of getting two defective camera? P(1st is defective & 2nd is defective) = P(1st is defective) x P(2nd is defective | 1st is defective) = (1/10)*(1/11) = 1/110 = 0.009

Sample Space

The COLLECTION of UNIQUE, NONOVERLAPPING POSSIBLE OUTCOMES of a Random Circumstance

Prevalence

The Probability BEFORE the TEST is carried out that the SUBECT HAS the DISEASE - Also known as the PRIOR PROBABILITY OF THE DISEASE

Odds of any Outcome of a Random Circumstance

The RATIO OF the PROBABILITY OF that OUTCOME OCCURRING OVER the PROBABILITY OF that OUTCOME NOT occurring - If the Outcome has Probability p of occurring then odds(A) = p/(1-p)

Coin Toss

The result of any single coin toss is random - But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin toss is not influenced by the result of the previous toss) The Probability of heads is 0.5 = the Relative Frequency of times you get heads in many repeated trials Coin tossers - Count Buffon 4040 (2048 H) - Karl Pearson 24,000 (12,012 H) - John Kerrich 10,000 (5,067 H)

Density Curves

They are always ON/ABOVE the HORIZONTAL AXIS - Have an AREA OF exactly ONE underneath it The area under the curve for a range of values is the Probability that a randomly selected value from the distribution belongs to the range

Independence of 3 Events

Three events A, B & C are Independent if A, B, & C are INDPENDENT IN PAIRS & P(A & B & C) = P(A) x P(B) x P(C) (called COMPLETE INDEPENDENCE)

Summary of Probability Rules

1. "Not the Event" or Complementary): - To find the Probability of Ac , use P(Ac) = 1 - P(A) 2. Addition Rule for "Either or"): - To find the Probability that either A or B happens 2a. General: - P(A or B) = P(A) + P(B) - P(A & B) 2b. Disjoint Events: - If A & B are Mutually Exclusive then P(A or B) = P(A) + P(B) 3. Multiplication Rule for "&": - To find the Probability that two Events both occur simultaneously 3a. General: - P(A & B) = P(A) x P(B | A) 3b. Independent Events: - If A & B are Independent Events, then P(A & B) = P(A)P(B) - Extension of rule 3b to more than two Independent Events: P(A1 & A2 ... & An) = P(A1 ) x P(A2) x ...P(An) 4 Conditional Probability: - To find the probability that B occurs given that A has occurred - P(B I A) = [P(A & B)]/[P(A)]

Steps to Find the Probability Distribution of a Discrete Random Variable from a Sample Space

1. List all Simple Events in the Sample Space 2. Find the Probability for each Simple Event (often all Simple Events are equally likely but this is not true in general) 3. List the possible values for the Random Variable X & identify the value for each Simple Event 4. Find all Simple Events for which X = x for each possible value of x 5. P(X = x) is the sum of the Probabilities for all Simple Events for which X = x Ex #1: - Let X be the sum of 2 fair dice; find the Probability Distribution of X - x: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 - P(X = x): 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36, 4/36, 3/36, 2/36, 1/36 Ex #2: - Let X be the number of girls in a family of 3 children, what is the Probability that the family will have at least 1 child of each gender? - P(X = 1 or 2) = P(X = 1) + P(X = 2) = ¾

4 Steps to Find Probabilities

1. State: - List the Random Circumstance involved in the problem 2. Plan: - Express the given information in terms of Events & their Probabilities - This includes: list the possible outcomes for each Random Circumstance; find the Probabilities of Simple Events; specify the Event for which you want to determine the Probability 3. Solve: - Determine which of the Probabilities in the Plan stage are needed & which Probability Rules can be combined to find the Probability of interest 4. Conclude: - Answer the question being asked

Complimentary Events

2 Events DO NOT CONTAIN ANY OF the SAME SIMPLE EVENTS & TOGETHER THEY COVER the ENTIRE SAMPLE SPACE - 2 Events whose Probability add up to 1 P(Ac) = 1 - P(A)

How bad is Alicia's day? - Event A = Alicia has disease D - Event B = Alicia is selected to answer the 1st question in her statistics class - Event C = Alicia is selected to answer the 2nd question in her statistics class Are the Events Independent or Dependent?

A & B are Independent B & C are Dependent

Consider rolling a six-sided (numbered 1 to 6) die once & assume that each side is equally likely to face above. Let A be the event that an even number shows up, B = {1, 2, 3}, & C = {1, 2, 3, 4}. The P(A) = 1/2, P(A | B) = 1/3, & P(A | C) = 1/2. - Are these Events Independent? Dependent?

A = {2, 4, 6} B = {1, 2, 3} C = {1, 2, 3, 4} A & B = {2} A & C = {2, 4} P(A) = 1/2 P(B) = 1/2 P(C) = 4/6 = 2/3 P(A & B) + 1/6 P(A & C) = 2/6 = 1/3 P(A I B) = [P(A & B)]/[P(B)] = (1/6)/(1/2) = 1/3 P(A I C) = [P(A & C)]/[P(C)] = (2/6)/(4/6) = 1/2 That is, Events A & B are Dependent, whereas Events A & C are Independent

Event

A COLLECTION of ONE/MORE SAMPLE EVENTS IN the SAMPLE SPACE - Any collection of outcomes contained in the Sample Space - Denoted by A, B, C, & so on Ex: - Freeway exit: A = {RLL, LRL, LLR} = exactly one of the three vehicles turn right; B = {LLL, RLL, LRL, LLR} = at most one of the vehicles turn right; C = {LLL, RRR} = all three vehicles turn in the same direction - Battery examination: A = {S, FS, FFS} = at most three batteries are examined; B = {FS, FFFS, FFFFFS...} = an even number of batteries are examined; C = { S, FFS, FFFFS...} = an odd number of batteries are examined - Gas pumps: A = {(0,0), (1,1), (2,2), (3,3), (4,4), (5,5)} = the number of pumps in use is the same for both stations; B = {(0,4), (1,3), (2,2), (3,1), (4,0)} = total number of pumps in use is four; C = {(0,0), (0,1), (1,0), (1,1)} = at most one pump is in use at each station

Probability Model

A Model that MATHEMATICALLY DESCRIBES the OUTCOME OF a RANDOM CIRCUMSTANCE They consist of 2 parts: 1. Sample Space (S) 2. A way of assigning probabilities to events Ex: - Coin Toss - S = {Head, Tail} - Probability of heads = 0.5 - Probability of tails = 0.5

Discrete Random Variable

A Random Variable that can either take a FINITE SET or else can be listed in an INFINITE SQUENCE OF DISTINCT VALUES such as INTEGERS (DISCRETE Sample Spaces)

Random Variable

A Variable that ASSIGNS A NUMBER TO EACH OUTCOME of a Random Circumstance 2 different broad classes: 1. Discrete 2. Continuous Ex: - The number of heads in three times random coin tossing - The number of boys in a randomly selected family of three children - The number of blood tests in the experiment that people randomly select married couples & do a blood test on each person until they find a husband & wife who both have the Rhesus factor - The number of batteries examined before a good one is obtained - The height of a person randomly picked on campus - The waiting time to see a red Mercedes-Benz at the cross of Walton BLVD & Squirrel Road

Venn Diagram

A diagram that SHOWS the SIMILARITIES & DIFFERENCES for different SETS - It is a visual way to REPRESENT SETS along with THIER UNIONS & INTERSECTIONS

Continuous Sample Space

A whole INTERVAL - S = {all numbers between 0 & 1} - It contains an INFINITE NUMBER OF POSSIBLE OUTCOMES We cannot assign Probabilities to each individual value in the Sample Space - Use a mathematical function to assign Probabilities for any range of values within the Sample Spac - These mathematical functions are called Density Curves Ex: - Length of a randomly chosen cellphone call S = {all X such that X≥0} = {X | X≥0} - Randomly chosen student's GPA S = {X | 0.00 ≤ X ≤ 4.00} - 2 gas stations are located at certain intersection; each one has 5 gas pumps; consider the experiment in which the number of pumps in use at a particular time of the day is recorded for each of the stations - If a new type-D flashlight battery has a voltage that is outside certain limits, that battery is characterized as a failure (F); if the battery has a voltage within the prescribed limits, it is a success (S); an experiment is set up to test each battery as it comes off an assembly line until a success is observed

Bernoulli Random Variable

Any Random Variable whose ONLY POSSIBLE VALUES are ZERO & ONE

Probability Tree Diagram

Conditional probabilities can get complex - it is often a good strategy to build one of these Diagram that represents all possible outcomes graphically & assigns Conditional Probabilities to subsets of events

Probability in a Continuous Sample Space

Contains an INFINITE NUMBER OF EVENTS - They typically are intervals of possible, continuously distributed outcomes Ex: - A bus arrives at a bus stop every hour - If a person arrives at the bus stop at a random time, how long will he or she have to wait for the next bus? - S = {interval containing all numbers between 0 & 1} = [0, 1] How do we describe Probabilities to Events in an infinite Sample Space? - We use DENSITY CURVES & compute Probabilities for INTERVALS - Uniform Density Curve - The Probability of the uniformly distributed Variable X to be within 0.3 & 0.7 is the area under the Density Curve corresponding to that interval - Thus: P(0.3 ≤ X ≤ 0.7) = (0.7 − 0.3)*1 = 0.4

Probability in a Discrete Sample Space

Determined by PROBABILITIES OF SIMPLE EVENTS - A valid Probability assignment must be between 0 & 1 - The sum of Probability over all possible Simple Events = 1 Ex: - You toss 2 dice - What is the Probability of the Outcomes summing to 5? - There are 36 possible outcomes in S, all of them are equally likely (given fair dice) - Sample Space S = {(1,1), (1,2), (1,3), (1,4), (1,5),(1,6) ... (6,1), (6,2), (6,3), (6,4), (6,5),(6,6)} - Thus, the Probability of any one of them to appear is 1/36 - P(the roll of two dice sums to 5) = P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 * 1/36 = 1/9 = 0.111

Pairwise Independence DOES NOT Imply Complete Independence

Ex. provided by Russian mathematician Sergei Natanovich Bernstein (Gnedenko, B. V., 1963, The Theory of Probability, P. 62) Suppose that a regular tetrahedron has one red face, one green face, one blue face, & its fourth face colored with red, green, & blue stripes Toss the tetrahedron & observe the face that appears on the bottom Let A, B, & C, be the Events that red, green, & blue colors appear, respectively Then: - P(A) = P(B) = P(C) = 1/2 - P(A & B) = P(A & C) = P(B & C) = 1/4 Therefore, A, B, & C are PAIRWISE Independent - However, P(A & B & C) = 1/4 > P(A)P(B)P(C)

General Addition Rule

For any 2 events A & B: - The probability that A occurs, or B occurs, or both events occur is P(A or B) = P(A) + P(B) - P(A & B) - Especially, if A & B are disjoint P(A or B) = P(A) + P(B) Ex 1: - A card is randomly drawn from a 52-card deck of playing cards - What is the probability the card is either a "face" or a heart? - Event A = the card is a "face" card - There are 12 face cards in the deck (jack, queen, & king in each of the four suits), so P(A) = 12/52 - Event B = the card is a heart - There are 13 cards in each suit so P(B) = 13/52 - There are three face cards that are hearts so P(A & B) =3/52 - Thus, P(face or heart)= P(face) + P(heart) - P(face & heart) = 12/52 + 13/52 - 3/52 = 22/52 = 0.423

Personal Belief Interpretation of Probability

From SUBJECTIVE CONSIDERATIONS, typically ABOUT UNIQUE EVENTS (e.g., Probability of a large meteorite hitting the Earth; Probability of life on Mars) - These do not make sense in terms of Frequency A PERSONAL PROBABILITY represents an INDIVIDUAL'S PERSONAL DEGREE OF BELIEF BASED ON PRIOR KNOWLEDGE - It is also called BAYESIAN Probability for the mathematician who developed the concept Ex: - We may say "there is a 40% chance of life on Mars" - In fact, either there is or there isn't life on Mars - The 40% Probability is our degree of belief, how confident we are about the presence of life on Mars based on what we know about life requirements, pictures of Mars, & probes we sent - If you decide to drive to UM central campus this Saturday afternoon, you will have a 30% Probability to find a good parking spot - The Probability that you will have a better time if you go to Cancun or to Florida for your next spring break is 80%

Relative Frequency Interpretation of Probability

In situations that we can envision REPEATEDLY OBSERVING the RESULTS OF RANDOM CIRCUMSTANCE, the PROBABILITY OF a SPECIFIC OUTCOME is the PORTION OF TIMES IT WOULD OCCUR OVER a LONG-RUN OBSERVATION Ex: - Buying lottery tickets regularly & calculate the Probability you win - Commuting to work daily & calculate the Probability of a certain traffic signal is red when you encounter it - Surveying 1,000 adults & determining what portion smokes - The Probability of rain today - Observing births & noting how often the baby is a boy; the Long-Run Relative Frequency of males born in the United States is about 0.512; over the long run, there are 512 male babies & 488 female babies born per 1000 births

How good is the Liver Scan at Diagnosis of Abnormal Pathology (Altman, D. G. and Bland J.M., 1994, Diagnostic tests. 1: Sensitivity and specificity, BMJ 308, 1552)?

Liver Scan Pathology Abnormal + Normal - Total Abnormal 231 27 258 Normal 32 54 86 Total 263 81 344 Sensitivity - P(Abnormal LS | Abnormal Pathology) - 231/258 - 0.90 Specificity - P(Normal LS | Normal Pathology) - 54/86 - 0.63 Prevalence - 258/344 - 0.75 Positive Predictive Value - P(Abnormal Pathology | Abnormal LS) - 231/263 - 0.88 Negative Predictive Value - P(Normal Pathology | Normal LS) - 54/81 - 0.59

Disjoint Events

MUTUALLY EXCLUSIVE - When 2 Events (A & B) have NO OUTCOMES IN COMMON & CAN NEVER HAPPEN TOGETHER

Random Circumstance

One in which the OUTCOME is UNPREDICTABLE - In many cases, the outcome is not determined until we observe it - In other cases, the outcome is already determined, but our knowledge of it is uncertain Ex: - Human eye color is controlled by a single pair of genes (one from the father & the other from the mother); brown eye color, B, is dominant over blue color, A - Randomly pick a person who can wiggle his or her ears - Randomly pick a person who can raise one eyebrow at a time

Discrete Sample Space

One made up of a FINITE/COUNTABLE LIST OF INDIVDUAL OUTCOMES To assign Probabilities: - List the Probabilities of all the individual outcomes - These Probabilities must be numbers between 0 & 1; must have a sum 1 - The Probability of any Event = sum of the Probabilities of the outcomes making up the Event Ex: - Blood Types S = {O+, O-, A+, A-, B+, B-, AB+, AB-} - Walmart payment methods S = {cash, debit card, credit card, check} - Check 3 different fuses in sequence & record the result of each examination by N: not defective & D: defective - Consider an experiment in which each of three vehicles taking a particular freeway exit turns left (L) or right (R) at the end of the exit ramp

Simple Event

One that consists of exactly ONE OUTCOME IN the SAMPLE SPACE Ex: - Fuse check: S = { NNN, NND, NDN, NDD, DNN, DND, DDN, DDD} - Freeway exit: S = {LLL, RLL, LRL, LLR, LRR, RLR, RRL, RRR} - Battery examination S = {S, FS, FFS, FFFS...} - Gas pumps: S = { (0,0), (0,1), (0,2), (0,3), (0,4), (0,5)...(5,5)}

Assuming birth months are equally likely, what is the Probability that the next 2 strangers you meet share your birth month? Event A = the first stranger shares your birth - P(A) = 1/12 Event B = the second stranger shares your birth - P(B) = 1/12

P(A & B) = P(A) x P(B) 1/12 x 1/12 1/144 0.007

A study of hearing impairment in dogs examined over 5000 Dalmatians for both hearing impairment and iris color. Hearing "impaired" was defined as deafness in either one or both ears. Dogs with one or both irises blue were labeled "blue" eyed. The study found that 28% of Dalmatians were hearing impaired, 11% were blue eyed, & 5% were hearing impaired and blue eyed. - What is the probability for a randomly selected Dalmatian to be "blue" eyed or hearing impaired?

P(A or B) = P(A) + P(B) - P(A & B) - A = hearing impaired - B = blue eyed - P(A) = 0.28 - P(B) = 0.11 -P(A & B) = 0.05 P(blue or impaired) = P(blue) + P(impaired) - P(blue & impaired) 0.11 + 0.28 - 0.05 0.39 - 0.05 0.34

Professor Jackson is in charge of a program to prepare people for a high school equivalency exam. Records show that 80% of the students need work in math, 70% need to work in English, & 55% need to work both areas. - What is the probability that a student selected at random needs math or English?

P(A or B) = P(A) + P(B) - P(A & B) - A = math - B = English - P(A) = 0.8 - P(B) = 0.7 - P(A & B) = 0.55 P(Math or English) = P(Math) + P(English) -P(Math & English) 0.80 + 0.70 - 0.55 1.5 - 0.55 0.95

Forests are complex, evolving ecosystems. For example, pioneer tree species can be displaced by succession species better adapted to the changing environment. Ecologists mapped a large Canadian forest plot dominated by pioneer Douglas fir with an under-story of the invading success species western red cedar. The following Two-Way Table records the distribution of all 2050 trees in the plot by species & by life stage. The distinction between live & sapling tree is that one is taller than & the other is shorter than 1.3 meters, respectively. Dead Live Sapling Total Western red cedar (WRC) 41 (0.02) 205 (0.10) 164 (0.08) 410 (0.20) Douglas fir (DF) 328 (0.16) 328 (0.16) 0 (0.00) 656 (0.32) Western hemlock (WH) 472 (0.23) 430 (0.21) 82 (0.04) 984 (0.48) Total 841 (0.41) 963 (0.47) 246 (0.12) 2050 (1) - Given that a randomly selected tree is a sapling, what is the Probability that the tree is a western hemlock?

P(WH & sapling) = 0.04 Conditional Probability P(WH | sapling) = [P(WH & sapling)]/[P(sapling)] 82/246 (82/2050)/(246/2050) 0.04/0.12 0.3333

State: - Intense or repetitive sun exposure can lead to skin cancer - Patterns of sun exposure, however, differ between men & women - A study of cutaneous malignant melanoma in the Italian population found that 15% of skin cancers are located on the head & neck, another 41% on the trunk, & the remaining 44% on the limbs - Moreover, 44% of individuals with skin cancer on the head are men, as are 63% of those with a skin cancer on the trunk & only 20% of those with skin cancer on the limbs - What is the Probability that random selected skin cancer patient in this study is a woman?

Plan: - P(head) = 0.15 - P(trunk) = 0.41 - P(limbs) = 0.44 - P(man | head) = 0.44 - P(man | trunk) = 0.63 - P(man |limbs) =0.20 - Want to find P(woman) Solve: - Venn diagram or tree diagram Conclude: - The Probability of a random selected patient in this study is a woman is 0.588

State: - A charity raises funds by calling a list of prospective donors to ask for pledges - It is able to talk to 40% of the names on its list - Of those the charity reaches, 30% make a pledge - But only half of those who pledge actually make a contribution - What is the probability that a randomly selected donor is reached, makes a pledge, & contributes?

Plan: - Express the given information in terms of Events & their Probabilities - Let A = {a randomly selected prospect is reached}, then P(A) = 0.4 - Let B = {a randomly selected prospect makes a pledge}, then P(B | A) = 0.3 - Let C = {a randomly selected prospect contributes}, then P(C | A & B) = 0.5 - It is asking P(A & B & C) Solve: - Use the Multiplication Rule - P(A & B & C) = P(A & B) x P(C | A & B) - P(A) x P(B | A) x P(C | A & B) - 0.4 x 0.3 x 0.5 - 0.06 Conclude: - The Probability that a randomly selected donor is reached, makes a pledge, & contributes is 0.06

State: - Video sharing sites, led by YouTube, are popular destinations on the internet - The Pew Internet & American Life Project finds that, among adult internet users of age 18 & over, 27% are 18 to 29 years old, 45% are 30 to 49 years old, & the remaining 28% are 50 & over - In the user group of 18 to 29 years old, 70% have visited a video sharing site, while 51% of those aged 30 to 49 & 36% of those 50 or older have visited a video sharing site - What is the Probability that randomly selected internet user in the study has visited a video sharing site?

Plan: - P(age 18 to 29) = 0.27 - P(age 30 to 49) = 0.45 - P(age 50 or older) = 0.28 - P(video yes | age 18 to 29) = 0.70 - P(video yes| age 30 to 49) = 0.51 - P(video yes | age 50 or older) = 0.26 - Need to find P(video yes) Solve: - Tree diagram Conclude: - The Probability that a randomly selected internet user in the study has visited a video site is 0.4914

State: - Breast cancer occurs most frequently among older women - Of all age groups, women in their 60s have the highest rate of breast cancer - The National Cancer Institute (NCI) compiles U.S. epidemiology data for a number of different cancers - The NCI estimate that 3.65% of women in their 60s get breast cancer - Mammograms are X-ray images of the breast used to detect breast cancer - A mammogram can typically identify correctly 85% of the cancer cases (i.e., 85% sensitivity) & 95% of cases without cancer (i.e., 95% specificity) - If a randomly selected woman in her 60s gets a positive mammogram, what is the Probability that she indeed has breast cancer?

Plan: - P(cancer) = 0.0365 - P(no cancer) = 0.9636 - P(test + | cancer) = 0.85 - P(test - | cancer) = 0.15 - P(test + | no cancer) = 0.05 - P(test - | no cancer) = 0.95 - What we want to know is P(cancer | test +) Solve: - Tree diagram - P(cancer | test+) = [P(cancer) x P(test+ I cancer)]/[P(cancer) x P(test+ I cancer) x P(no cancer) x P(test+ I no cancer)] - (0.0365 x 0.85)/(0.0365 x 0.85 x 0.963 x 0.05) - 0.392 Conclude: - If a randomly selected woman in her 60s gets a positive mammogram, there is a 39% chance that she indeed has breast cancer - This value is called the Positive Predictive Value, or PPV - It is an important piece of information but, unfortunately, is rarely communicated to patients

Probability

QUANTIFICATION OF RANDOMNESS - Related to a Random Circumstance A number BETWEEN ZERO & ONE (inclusive) that DESCRIBES the LIKELIHOOD OF A POSSIBLE OUTCOME OF A RANDOM CIRCUMSTANCE - Those closer to ONE indicate that the outcome is more LIKELY to appear - Those closer to ZERO indicate that the outcome is LESS likely to occur Ex: - Henry figures that if he guesses on a true-false question, the likelihood of getting it right is ½ - The Right to Health Lobby claims the likelihood is 0.40 of getting an erroneous report from a medical laboratory in one low-cost medical center - A college counselor claims that 70% of the first-year students receive counseling to help plan their schedule

Probabilities for Some Lottery Events

Random Circumstance - A 3-digit winning lottery number is selected Sample space - S = {000, 001, 002 ...997, 998, 999} - There are 1000 Simple Events Probabilities for Simple Events - Assume that all 3-digit numbers are equally likely - That is, the Probability that any specific 3-digit number is a winner is 1/1000 Probabilities for some Events - Event A = the last digit is a 9 - A = {009, 019, 029...999} - P(A) = 1/10 - Event B = the 3 digits are all the same - B = {000, 111, 222, 333, 444, 555, 666, 777, 888, 999} - P(B) = 10/1000 = 1/100 - Are A & B Mutually Exclusive? One event is the complement of another if the two events do not contain any of the same Simple Events & together they cover the entire Sample Space - P(Ac) = 1 - P(A) = 9/10 - P(Bc ) = 1 - P(B) = 99/100

For a family with 3 children, what are possible sequences of boys (B) & girls (G)?

Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} Assume all 8 outcomes in the Sample Space are equally likely - The probability of each sequence is thus 1/8 What are the numbers of girls (X) they could have? - P(X = 0) = P(BBB) = 1/8 - P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 3/8 - P(X = 2) = P(BGG or GGB or GBG) = P(BGG) + P(GGB) + P(GBG) = 3/8 - P(X = 3) = P(GGG) = 1/8 Probability Distribution: - Value of X: 0, 1, 2, 3 - Probability: 1/8, 3/8, 3/8, 1/8 Probabilities: {0.13, 0.13, 0.13, 0.13, 0.12, 0.12, 0.12, 0.12} What are the numbers of girls (X) they could have? - P(X = 0) = P(BBB) = 0.13 - P(X = 1) = P(BBG or BGB or GBB) = P(BBG) + P(BGB) + P(GBB) = 0.39 - P(X = 2) = P(BGG or GGB or GBG) = P(BGG) + P(GGB) + P(GBG) = 0.36 - P(X = 3) = P(GGG) = 0.12 Probability Distribution: - Value of X: 0, 1, 2, 3 - Probability: 0.13, 0.39, 0.36, 0.12

Sampling WITHOUT Replacement

Sampled individuals are NOT returned to the eligible pool for each subsequent selection

Ex. of Risk & Odds in terms of Probability

Sickle-cell anemia is a serious, inherited blood disease affecting the shape of red blood cells - Individuals with both genes causing the defect suffer pain from blocked arteries & can have their life shortened from organ damage - Individuals carrying only one copy of the defective gene ("sickle-cell trait") are generally healthy but may pass on the gene to their offspring - If a couple learns from blood tests that they both carry the sickle-cell trait, the genetic laws of inheritance tell us that there is a 25% chance that they could conceive a child suffering from sickle-cell anemia - That is, the risk of conceiving a child who will suffer from sickle-cell anemia is 0.25 - The odds of this are odds = 0.25/(1 - 0.25) = 1/3, or 1:3

Discrete Random Variable Standard Deviation & Variance

Standard Deviation - Describes the spread of values of X around the expected value µ - σ = √σ² Variance - Var(X) = σ² = ∑(x - µ)²[P(x)]

Diagnostic Test

State Test Results Test Results Total Present + (POS) - (NEG) Probability Yes (S) Sensitivity P(POS I S) False - rate P(NEG I S) 1.0 No (Sc) False + rate P(POS I Sc) Specificity P(NEG I Sc) 1.0

Optimism for Alicia - She Is Probably Healthy

Steps 1 to 3: Random Circumstances, possible outcomes, & known Probabilities - Random Circumstance 1: Alicia's disease status - Possible Outcomes: D = disease; DC = no disease - Probabilities: P(D) = 1/1000 = 0.001; P(DC) = 999/1000 = 0.999 - Random Circumstance 2: Alicia's test results - Possible Outcomes: P = test is positive; N = test is negative - Probabilities: We only know Conditional Probabilities from her physician: P(P|D) = 0.95; P(N|D) = 0.05; P(P|DC) = 0.05; P(N|DC) = 0.95 Step 4: Specify the Event for which you want to determine the Probability - We want to know P(disease | positive test) = P(D | P) Step 5: Determine which of the Probabilities from step 3 & which Probability Rules can be combined to find the Probability of interest - P(P) = P(P & D) + P(P & DC); - P(P & D) = P(P | D) x P(D) = 0.95 x 0.001 = 0.00095 - P(P & DC) = P(P|DC)P(DC) = 0.05 x 0.999 = 0.04995 - P(P) = 0.0509 - P(D|P) = P(D & P)/P(P) = 0.00095/0.0509 = 0.019

Bayes' Theorem

Suppose that A1, A2, ... Ak form a partition of the Sample Space (S) - A1, A2, ... Ak are Mutually Exclusive Events & S is the union of A1, A2, ... Ak Assume that the Probabilities of A1 , A2 , ..., Ak are not 0 & 1 - Then, for any other Event B whose Probability is not 0 or 1, we have: - P(Ai I B) = [P(B I Ai) x P(Ai)]/[P(B I A1) x P(A1) + P(B I A2) x P(A2) + ... + P(B I Ak) x P(Ak)]

Standard Deviations for Discrete Random Variables

Suppose you decide to invest $100 in your financial plans - There are 2 choices & the possible net gains after one year & their probabilities are given below: Plan 1 Plan 2 X = Net Gain Probability Y = Net Gain Probability $5000 0.001 $20 0.3 $1000 0.005 $10 0.2 $0 0.994 $4 0.5 E(X) = $5000 x 0.001 + $1,000 x 0.005 + $0 x 0.994 = $10.00 E(Y) = $20 x 0.3 + $10 x 0.2 + $4 x 0.5 = $10.00 If a Random Variable X takes possible values of x1, x2, ... with Probabilities p1, p2, ..., respectively, & its Expected Value is E(X) = µ, then... - The Variance of X is Var(X) = σ² = (x1 - µ)²(p1) + (x2 - µ)²(p2) + ... - The Standard Deviation of X is the square root of Var(X) = σ², i.e., σ - The Standard Deviation of a Random Variable X describes the spread of values of X around the expected value µ Plan 1 Plan 2 (x-µ)² p (x-µ)²p (x-µ)² p (x-µ)²p (5000-10)²=24,900,100 0.001 24,900.1 (20-10)²=100 0.3 30 (1000-10)² = 980,100 0.005 4,900.5 (10-10)² = 0 0.2 0 (0 - 10)² = 100 0.994 99.4 (4 - 10)² = 36 0.5 18 For Plan 1: - Var(X) = 24,900.1 + 4,900.5 + 99.4 = 29,900.00 - σ = $172.92 For Plan 2: - Var(Y) = 40 + 18 = 48.00 - σ = $6.93 These values demonstrate that the possible outcomes for Plan 1 are much more variable than those of Plan 2 - If you want to invest cautiously, you will prefer Plan 2 - If you want to have the chance to gain a large amount of money, you will choose Plan 1

Expected Value of a Discrete Random Variable

The MEAN VALUE that would be OBTAINED FROM an INFINITE NUMBER OF OBSERVATIONS of the Random Variable - It describes WHERE the PROBABILITY DISTRIBUTION IS CENTERED If X is a Random Variable with possible values x1, x2, ... occurring with Probabilities p1, p2, ..., then this is calculated as E(X) = (x1)(p1) + (x2)(p2) + ... - That is, calculating "VALUE TIMES PROBABILITY" separately for EACH POSSIBLE VALUE of X & then SUM up these quantities - Greek letter "mu" µ = E(X) Ex: - A gambling game in which on any play, the Probabilities are 0.3 that a player wins $2 & 0.7 that a player loses $1 - In 10 plays, the Relative Frequencies of win & loss will not necessarily correspond exactly with the Theoretical Probabilities, but to develop an intuitive feeling for E, let's suppose that 10 games go exactly according to the theory - That is, 10 games would result in 3 wins & 7 losses - The total money won & lost = $2 x 3 + (-$1) x 7 - The average result per play = [$2 x 3 + (-$1) x 7] /10 = -$0.1 - E(X) = amount won or lost per play = -$0.1

Risk of an Undesired Outcome of a Random Occurrence

The PROBABILITY OF that UNDESIREABLE OUTCOME - If an outcome A has a Probability of p occurring, then risk (A) = p

Negative Predictive Value (NPV)

The Probability of patients with NEGATIVE test results who are correctly diagnosed - P(Normal Pathology | Normal LS) - [Specificity x (1 - Prevalence)]/[(1 - Sensitivity) x Prevalence + Specificity x (1 - Prevalence)]

Positive Predictive Value (PPV)

The Probability of patients with POSITIVE test results who are correctly diagnosed - P(Abnormal Pathology | Abnormal LS) - (Sensitivity x Prevalence)/[(Sensitivity x Prevalence) + (1 - Specificity) x (1 - Prevalence)]

Specificity

The Probability of true NEGATIVE that are correctly identified by the test - P(Normal LS | Normal Pathology)

Sensitivity

The Probability of true POSITIVES that are correctly identified by the test - P(Abnormal LS | Abnormal Pathology)

Conditional Probability

The Probability that ONE EVENT HAPPENED given that ANOTHER EVENT is ALREADY known to have HAPPENED - The long-run relative frequency with which event B occurs when circumstances are such that A has occurred - Written as P(B | A) P(B | A) = [P(A & B)]/[P(A)] - Probability of event B given event A is Conditional - Provided that P(A) ≠ 0 - 0 Probability = Event is impossible P(B | A) > P(A) - Conditional Event > Unconditional Event - A & B are Positively Correlated P(B | A) < P(A) - A & B are Negatively Correlated P(B | A) = P(A) - A & B are Independent - Requirement: B > 0 (i.e., B must be +) - [P(A & B)]/[P(B)] - P(A & B) = P(A) x P(B) Ex: - The Probability that a randomly picked day is cloudy is different if you live in California compared to if you live in Michigan - That is, P(cloudy | Michigan) > P(cloudy | California) - A survey of 120,000 Minnesota teens found that 20% of the boys & 5% of the girls gambled at least once a week during the previous year (Reuters Website, Aug. 18, 1998) - Assuming these figures are representative of all teens - P[teen is a weekly gambler | teen is boy] = 0.20 - P[teen is a weekly gambler | teen is girl] = 0.05

General Multiplication Rule

The Probability that any two Events, A & B, both occur is: - P(A & B) = P(A) x P(B | A) If A & B are Independent then: - P(A & B) = P(A) x P(B | A) = P(A) x P(B)

Independent Events

The Probability that one EVENT occurs on any given trial of an experiment IS NOT INFLUENCED in any way BY the OCCURRENCE OF THE OTHER EVENT - Otherwise, the 2 Events are Dependent - The outcome of 1 Event does not affect the outcome of the other Event

Sampling from a Very Large Population

The distinction between Sampling with & without replacement is diminished

Basic Probability Rules

These facts follow the idea of Probability being "the long-run portion of repetitions on which an Event occurs" 1. Probabilities range from ZERO (NO CHANCE of the event) to ONE (the EVENT HAS TO HAPPEN) - For any Event A, 0 ≤ P(A) ≤ 1 - Probability of getting heads = 0.5 - We write this as: P(head) = 0.5 2. ALL POSSIBLE OUTCOMES (S) TOGETHER must have PROBABILITY OF ONE or the PROBABILITY OF the complete SAMPLE SPACE MUST EQUAL ONE - P(S) = 1 - P(head) + P(tail) = 0.5 + 0.5 = 1 3. The PROBABILITY OF AN EVENT NOT OCCURRING is 1 minus the Probability that it does occur - P(not A) = 1 - P(A) - P(tail) = 1 - P(head) = 0.5 - P(neither head nor tail) = 1 - (0.5 + 0.5) = 1 - 1 = 0 4. The ADDITION RULE FOR DISJOINT EVENTS - In this case, the Probability that A or B occurs is the sum of their individual Probabilities - P(A or B) = P(A U B) = P(A) + P(B) - P(getting either a head or a tail) = 1 - If you flip two coins & the first flip does not affect the second flip, S = {HH, HT, TH, TT} - The Probability of each of these events is 1/4, or 0.25 - The Probability that you obtain "only heads or only tails" is: P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50

In Mega Millions, for each $2 ticket you buy, you pick 5 numbers between 1 and 70 (the white balls), & then an extra number between 1 & 25 (the gold Mega Ball) - The jackpots start at $40 million & grow by a minimum of $5 million per draw each time the jackpot rolls Match Prize Prize - Cost Probability (Prize - Cost) x Prob 5 + bonus 393,000,000 392,999,998 0.0000000033 1.2968999934 5 1,000,000 999,998 0.0000000826 0.0825998348 4 + bonus 10,000 9,998 0.0000000436 0.0004359128 4 500 498 0.0000010906 0.0005431188 3 + bonus 200 198 0.0000007307 0.0001446786 3 10 8 0.0000182682 0.0001461456 2 + bonus 10 8 0.0000165632 0.0001325056 1 + bonus 4 2 0.0005714286 0.0011428572 bonus 2 0 0.0400000000 0.0000000000 nothing 0 -2 0.9593917892 -1.9187835784

This tells us that over many replicates of the game, players will lose an average of 54 cents per play - From the perspective of Lottery Commission, this means that they pay out $1 & 46 cents in prizes for every $2 ticket they sell for this game E(X) ≈ -$0.54

Hints & Advice for Finding Probabilities

To find P(A & B), define the combined Event in the physical terms & see if you know its Probability - If this fails, try the Multiplication Rule (rule 3) To find the Probability that a series of Independent Events all happen, simply multiply all individual Probabilities (extension of rule 3b) To find the Probability that one of a collection of Mutually Exclusive Events happens, simply add all of the individual Probabilities (rule 2b) It is sometimes easier to find the Probability of the Complement of the Event of interest & then subtract it from one (rule 1) To find the Probability that none of a collection of Mutually Exclusive Events happen, find the Probability that one of them happens & then subtract that from 1 To find a Conditional Probability, define the Event in physical terms & see if you know its Probability - If this fails, try rule 4

In a class of 30 students, there are 3 left-handed students. If the teacher picks up 2 students randomly, what is the Probability that the 2 students are both left-handed? - Think in terms of Sampling with & without replacement - If we have 500 students?

With replacement: - P(left & left) = (3/30) x (3/30) = 1/100 = 0.01 Without replacement: - P(left & left) = (3/30) x (2/29) = 1/145 = 0.007 With replacement: - P(left & left) = (3/500) x (3/500) = 0.000036 Without replacement: - P[left & left] = (3/500) x (2/499) = 0.000024

Probability Distribution of a Discrete Random Variable

X = Discrete Random Variable x = the value that X could have P(X = x) is the Probability that X equals x How many girls are likely in a family of 3 children? - It is assumed that each combination is equally likely - X = the number of girls - x = 0, 1, 2, 3 - P(X = 0) = 1/8 - P(X = 1) = 3/8 - P(X = 2) = 3/8 - P(X = 3) = 1/8 Probability Distribution of X is: - x: 0 1 2 3 - P(X = x): 1/8 3/8 3/8 1/8 It's possible to toss forever - Let X be the number of tosses until the first head occurs, then - x = 1, 2, 3, ... - P(X = 1) = ½ - P(X = 2) = 1/4 - P(X = 3) = 1/8, ... P(X = x) = (½)x