Probabilities What is the probability that among 23 people (this class) there will be a shared birthday?

# people required(i.e. n) when m=365 The number of people required so that the probability that some pair will have a birthday separated by k days or fewer will be higher than 50% is: k # people required(i.e. n) when m=365 23 1 14 2 11 3 9 4 8 5 6 7

Patrick's Casino

What is the probability of picking an ace?

Probability =

What is the probability of picking an ace? 4 / 52 = .077 or 7.7 chances in 100

Every card has the same probability of being picked

What is the probability of getting a 10, J, Q, or K?

(.077) + (.077) + (.077) + (.077) = .308 16 / 52 = .308

What is the probability of getting a 2 and then after replacing the card getting a 3 ?

(.077) * (.077) = .0059

What is the probability that the two cards you draw will be a black jack?

10 Card = (.077) + (.077) + (.077) + (.077) = .308 Ace after one card is removed = 4/51 = .078 (.308)*(.078) = .024 But you could also get an Ace (.077) and then a ten (.078*4 = .312) or .077*.312 = .024 So the prob of either of these occurring and getting blackjack is .024 + .024 = .048

Practice What is the probability of rolling a “1” using a six sided dice? What is the probability of rolling either a “1” or a “2” with a six sided dice? What is the probability of rolling two “1’s” using two six sided dice?

Practice What is the probability of rolling a “1” using a six sided dice? 1 / 6 = .166 What is the probability of rolling either a “1” or a “2” with a six sided dice? What is the probability of rolling two “1’s” using two six sided dice?

Practice What is the probability of rolling a “1” using a six sided dice? 1 / 6 = .166 What is the probability of rolling either a “1” or a “2” with a six sided dice? (.166) + (.166) = .332 What is the probability of rolling two “1’s” using two six sided dice?

Practice What is the probability of rolling a “1” using a six sided dice? 1 / 6 = .166 What is the probability of rolling either a “1” or a “2” with a six sided dice? (.166) + (.166) = .332 What is the probability of rolling two “1’s” using two six sided dice? (.166)(.166) = .028

Cards What is the probability of drawing an ace? What is the probability your first 2 cards are aces? What is the probability your first 4 cards are aces? What is the probability that out of 4 cards, at least one will be an ace?

Cards What is the probability of drawing an ace? 4/52 = .0769 What is the probability of drawing another ace? 4/52 = .0769; 3/51 = .0588; .0769*.0588 = .0045 What is the probability the next four cards you draw will each be an ace? .0769*.0588*.04*.02 = .000003 What is the probability that an ace will be in the first four cards dealt? .0769+.078+.08+.082 = .3169

Probability .00 1.00 Event must occur Event will not occur

Probability In this chapter we deal with discreet variables i.e., a variable that has a limited number of values Previously we discussed the probability of continuous variables (Z –scores) It does not make sense to seek the probability of a single score for a continuous variable Seek the probability of a range of scores

Key Terms Independent event Mutually exclusive When the occurrence of one event has no effect on the occurrence of another event e.g., voting behavior, IQ, etc. Mutually exclusive When the occurrence of one even precludes the occurrence of another event e.g., your year in the program, if you are in prosem

Key Terms Joint probability The probability of the co-occurrence of two or more events The probability of rolling a one and a six p (1, 6) p (Blond, Blue)

Key Terms Conditional probabilities The probability that one event will occur given that some other vent has occurred e.g., what is the probability a person will get into a PhD program given that they attended Villanova p(Phd l Villa) e.g., what is the probability that a person will be a millionaire given that they attended college p($$ l college)

Does not own a video game Example Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

What is the simple probability that a person will own a video game? Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game What is the simple probability that a person will own a video game? 35 / 100 = .35 Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game What is the conditional probability of a person owning a video game given that he or she has children? p (video l child) Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game What is the conditional probability of a person owning a video game given that he or she has children? 25 / 55 = .45 Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game What is the joint probability that a person will own a video game and have children? p(video, child) Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game What is the joint probability that a person will own a video game and have children? 25 / 100 = .25 Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game 25 / 100 = .25 .35 * .55 = .19 Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Does not own a video game The multiplication rule assumes that the two events are independent of each other – it does not work when there is a relationship! Owns a video game Does not own a video game Total No Children 10 35 45 Children 25 30 55 65 100

Practice Republican Democrat Total Male 52 27 79 Female 18 65 83 70 92 162

p (republican). p(female) p (republican, male) p (republican) p(female) p (republican, male) p(female, republican) p (republican l male) p(male l republican) Republican Democrat Total Male 52 27 79 Female 18 65 83 70 92 162

p (republican) = 70 / 162 =. 43 p (republican, male) = 52 / 162 = p (republican) = 70 / 162 = .43 p (republican, male) = 52 / 162 = .32 p (republican l male) = 52 / 79 = .66 Republican Democrat Total Male 52 27 79 Female 18 65 83 70 92 162

p(female) = 83 / 162 =. 51 p(female, republican) = 18 / 162 = p(female) = 83 / 162 = .51 p(female, republican) = 18 / 162 = .11 p(male l republican) = 52 / 70 = .74 Republican Democrat Total Male 52 27 79 Female 18 65 83 70 92 162

Foot Race Three different people enter a “foot race” A, B, C How many different combinations are there for these people to finish?

Foot Race A, B, C A, C, B B, A, C B, C, A C, B, A C, A, B 6 different permutations of these three names taken three at a time

Foot Race Six different people enter a “foot race” A, B, C, D, E, F How many different permutations are there for these people to finish?

Permutation Ingredients: N = total number of events r = number of events selected

Permutation Ingredients: N = total number of events r = number of events selected A, B, C, D, E, F Note: 0! = 1

Foot Race Six different people enter a “foot race” A, B, C, D, E, F How many different permutations are there for these people to finish in the top three? A, B, C A, C, D A, D, E B, C, A

Permutation Ingredients: N = total number of events r = number of events selected

Permutation Ingredients: N = total number of events r = number of events selected

Foot Race Six different people enter a “foot race” If a person only needs to finish in the top three to qualify for the next race (i.e., we don’t care about the order) how many different outcomes are there?

Combinations Ingredients: N = total number of events r = number of events selected

Combinations Ingredients: N = total number of events r = number of events selected

Note: Use Permutation when ORDER matters Use Combination when ORDER does not matter

Practice There are three different prizes 1st $1,000 2nd $500 3rd $100 There are eight finalist in a drawing who are going to be awarded these prizes. A person can only win one prize How many different ways are there to award these prizes?

Practice 336 ways of awarding the three different prizes

Practice There are three prizes (each is worth $200) There are eight finalist in a drawing who are going to be awarded these prizes. A person can only win one prize How many different ways are there to award these prizes?

Combinations 56 different ways to award these prizes

Practice A shirt comes in four sizes and six colors. One also has the choice of a dragon, alligator, or no emblem on the pocket. How many different kinds of shirts can you order?

Practice A shirt comes in four sizes and six colors. One also has the choice of a dragon, alligator, or no emblem on the pocket. How many different kinds of shirts can you order? 4*6*3 = 72 Don’t make it hard on yourself!

Practice In a California Governor race there were 135 candidates. The state created ballots that would list candidates in different orders. How many different types of ballots did the state need to create?

Practice 2.6904727073180495e+230 Or

26,904,727,073,180,495,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

Bonus Points Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice? 

http://www.shodor.org/interactivate/activities/SimpleMontyHall/

You pick #1 Door 1 Door 2 Door 3 Results GAME 1 AUTO GOAT Switch and you lose. GAME 2 Switch and you win. GAME 3 GAME 4 Stay and you win. GAME 5 Stay and you lose. GAME 6

Practice The probability of winning “Blingoo” is .30 What is the probability that you will win 20 of the next 30 games of Blingoo ? Note: previous probability methods do not work for this question

Binomial Distribution Used with situations in which each of a number of independent trials results in one of two mutually exclusive outcomes

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Game of Chance The probability of winning “Blingoo” is .30 What is the probability that you will win 20 of the next 30 games of Blingoo ? Note: previous probability methods do not work for this question

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events p = .000029

What does this mean? p = .000029 This is the probability that you would win EXACTLY 20 out of 30 games of Blingoo

Game of Chance Playing perfect black jack – the probability of winning a hand is .498 What is the probability that you will win 8 of the next 10 games of blackjack?

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events p = .0429

Binomial Distribution What is this doing? Its combining together what you have learned so far! One way to fit our 8 wins would be (joint probability): W, W, W, W, W, W, W, W, L, L = (.498)(.498)(.498)(.498)(.498)(.498)(.498)(.498)(.502)(.502)= (.4988)(.5022)=.00095 pX q(N-X)

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Other ways to fit our question W, L, L, W, W, W, W, W L, W, W, W, W, L, W, W W, W, W, L, W, W, W, L L, L, W, W, W, W, W, W W, L, W, L, W, W, W, W W, W, L, W, W, W, L, W

Binomial Distribution Other ways to fit our question W, L, L, W, W, W, W, W = .00095 L, W, W, W, W, L, W, W = .00095 W, W, W, L, W, W, W, L = .00095 L, L, W, W, W, W, W, W = .00095 W, L, W, L, W, W, W, W = .00095 W, W, L, W, W, W, L, W = .00095 Each combination has the same probability – but how many combinations are there?

Combinations Ingredients: N = total number of events r = number of events selected 45 different combinations

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Any combination would work . 00095+ 00095+ 00095+ 00095+ 00095+ 00095+ 00095+ 00095+. . . . . . 00095 Or 45 * . 00095 = .04

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events