Probabilities and Proportions C 4 L 1 S

Probabilities and Proportions C 4, L 1, S 1

Lotto I am offered two lotto cards: – Card 1: has numbers – Card 2: has numbers Which card should I take so that I have the greatest chance of winning lotto? C 4, L 1, S 2

Roulette In the casino I wait at the roulette wheel until I see a run of at least five reds in a row. I then bet heavily on a black. I am now more likely to win. C 4, L 1, S 3

Coin Tossing I am about to toss a coin 20 times. What do you expect to happen? Suppose that the first four tosses have been heads and there are no tails so far. What do you expect will have happened by the end of the 20 tosses ? C 4, L 1, S 4

Coin Tossing • Option A – Still expect to get 10 heads and 10 tails. Since there already 4 heads, now expect to get 6 heads from the remaining 16 tosses. In the next few tosses, expect to get more tails than heads. • Option B – There are 16 tosses to go. For these 16 tosses I expect 8 heads and 8 tails. Now expect to get 12 heads and 8 tails for the 20 throws. C 4, L 1, S 5

TV Game Show • In a TV game show, a car will be given away. – 3 keys are put on the table, with only one of them being the right key. The 3 finalists are given a chance to choose one key and the one who chooses the right key will take the car. – If you were one of the finalists, would you prefer to be the 1 st, 2 nd or last to choose a key? C 4, L 1, S 6

Let’s Make a Deal Game Show • You pick one of three doors – two have booby prizes behind them – one has lots of money behind it • The game show host then shows you a booby prize behind one of the other doors • Then he asks you “Do you want to change doors? ” – Should you? ? ! (Does it matter? ? !) • See the following website: • http: //www. stat. sc. edu/~west/javahtml/Lets. Makea. Deal. html C 4, L 1, S 7

Game Show Dilemma Suppose you choose door A. In which case Monty Hall will show you either door B or C depending upon what is behind each. No Switch Strategy ~ here is what happens Result A B C Win Car Goat Lose Goat Car P(WIN) = 1/3 C 4, L 1, S 8

Game Show Dilemma Suppose you choose door A, but ultimately switch. Again Monty Hall will show you either door B or C depending upon what is behind each. Monty will show Monty B show Switch Strategy ~ here is what happens eitherwillor C. Result A B Lose Car Goat Win Goat Car Win Goat door C, you You switch to the switch to shown one not B and C win. and lose. Goat Monty will show door B, you Goat switch to C and win. Car P(WIN) = 2/3 !!!! C 4, L 1, S 9

Matching Birthdays • In a room with 23 people what is the probability that at least two of them will have the same birthday? • Answer: . 5073 or 50. 73% chance!!!!! • How about 30? • . 7063 or 71% chance! • How about 40? • . 8912 or 89% chance! • How about 50? • . 9704 or 97% chance! C 4, L 1, S 10

Probability In this section we will… – Introduce us to basic ideas about probabilities: • what they are and where they come from • simple probability models • conditional probabilities • independent events – Teach us how to calculate probabilities: • through tables of counts and probability tables for independent events C 4, L 1, S 11

Probability I toss a fair coin (where fair means ‘equally likely outcomes’) • What are the possible outcomes? Head and tail • What is the probability it will turn up heads? 1/2 I choose a person at random and check which eye she/he winks with • What are the possible outcomes? Left and right • What is the probability they wink with their left eye? ? ? ? C 4, L 1, S 12

What are Probabilities? • A probability is a number between 0 & 1 that quantifies uncertainty • A probability of 0 identifies impossibility • A probability of 1 identifies certainty C 4, L 1, S 13

Where do probabilities come from? • Probabilities from models: The probability of getting a four when a fair dice is rolled is 1/6 (0. 1667 or 16. 7%) C 4, L 1, S 14

Probabilities and Proportions Probabilities and proportions are numerically equivalent. (i. e. they convey the same information. ) e. g. The proportion of U. S. citizens who are left handed is 0. 1; a randomly selected U. S. citizen is left handed with a probability of approximately 0. 1. C 4, L 1, S 15

Where do probabilities come from? • Probabilities from data – In a survey conducted by students in a STAT 110 course there were 348 WSU students sampled. – 212 of these students stated they regularly drink alcohol. – The estimated probability that a randomly chosen Winona State students drinks alcohol is 212/348 (0. 609, 60. 9%) C 4, L 1, S 16

Where do probabilities come from? • Subjective Probabilities – The probability that there will be another outbreak of ebola in Africa within the next year is 0. 1. – The probability of rain in the next 24 hours is very high. Perhaps the weather forecaster might say a there is a 70% chance of rain. – A doctor may state your chance of successful treatment. C 4, L 1, S 17

Simple Probability Models Terminology: • a random experiment is an experiment whose outcome cannot be predicted – E. g. Draw a card from a well-shuffled pack • a sample space is the collection of all possible outcomes – 52 outcomes (AH, 2 H, 3 H, …, KH, …, AS, …, KS) C 4, L 1, S 18

Simple Probability Models • an event is a collection of outcomes – E. g. A = card drawn is a heart • an event occurs if any outcome making up that event occurs – drawing a 5 of hearts • the complement of an event A is denoted as , it contains all outcomes not in A Eg = card drawn is not a heart = card drawn is a spade, club or diamond C 4, L 1, S 19

Simple Probability Models “The probability that an event A occurs” is written in shorthand as P(A). For equally likely outcomes, and a given event A: Number of outcomes in A P(A) = Total number of outcomes C 4, L 1, S 20

Example: Sum of two die C 4, L 1, S 21

Example: Sum of two die C 4, L 1, S 22

Example: Sum of two die C 4, L 1, S 23

Example: Sum of two die C 4, L 1, S 24

1. House Sales Example Below is a table containing some information for a sample of 600 sales of single family houses in 1999. Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 25

1. House Sales Example Let A be the event that a sale over $400, 000 • – is is the event that a sale is NOT over $400, 000 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 26

1. House Sales Example B be the event that a sale is made within 45 days • – So is the event that a sale takes longer than 45 days Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 27

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: a) over $400, 000, i. e. event A occurs. P(A) = 61/600 = 0. 102 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 28

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: b) not over $400, 000, i. e. occurs. P( ) = (155+384)/600 = 539/600 = 0. 898 * Note that P(A) + P( ) = 1 and that P( ) = 1 – P(A) Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 29

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: c) made in 45 days or more, i. e. occurs. P( ) = (300 + 54)/600 = 354/600 = 0. 59 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 30

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: d) made within 45 days and sold for over $400, 000, i. e. both B and A occur. P(B and A) = 20/600 = 0. 033 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 31

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: e) made within 45 days and/or sold for over $400, 000, i. e. either A or B occur. Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 32

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: e) made within 45 days and/or sold for over $400, 000. P(B and/or A) = (246 +Days 20)/600 Market 61 – on the = 287/600 = 0. 478 Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 33

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: f ) on the market for less than 45 days given that it sold for over $400, 000 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 34

1. House Sales Example For a sale selected at random from these 600 sales, find the probability that the sale was: f ) on the market for less than 45 days given that it sold for over $400, 000 P(B given A) = P(B|A) = 20/61 = 0. 328 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 35

Conditional Probability • The sample space is reduced. • Key words that indicate conditional probability are: “given that”, “of those”, “if …”, “assuming that” C 4, L 1, S 36

Conditional Probability “The probability of event A occurring given that event B has already occurred” is written in shorthand as P(A|B) It is defined as follows: P(A|B) = P(A and B)/P(B) C 4, L 1, S 37

1. House Sales Example For a sale selected at random from these 600 sales, g) What proportion of the houses that sold in less than 45 days, sold for more than $400, 000? Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 38

1. House Sales Example For a sale selected at random from these 600 sales, g) What proportion of the houses that sold in less than 45 days, sold for more than $400, 000? P (A|B) = 20/246 = 0. 081 Days on the Market Less than 45 Days Under $250, 000 - 400, 000 Over $400, 000 45 - 90 Days More than 90 Days 75 151 20 77 192 31 3 41 10 155 384 61 246 300 54 600 C 4, L 1, S 39

1. Heart Disease In 1996, 6631 New Zealanders died from coronary heart disease. The numbers of deaths classified by age and gender are: Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 40

1. Heart Disease Let A be the event of being under 45 B be the event of being male C be the event of being over 64 Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 41

1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: P(A) = 92/6631 = 0. 0139 a) under 45 Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 42

1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: b) male assuming that the person was younger than 45. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 43

2. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: b) male given that the person was younger than 45. P(B|A) = 79/92 = 0. 8587 Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 44

1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: c) male and was over 64. P(B and C) = (1081 + 1795)/6631= 2876/6631 Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 45

1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: d) over 64 given they were female. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 46

1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: d) over 64 given they were female. P(C|B) = (499+2176)/2904 = 0. 9211 Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 C 4, L 1, S 47