Probability. Part II - Counting Rules. Eli Gurarie. January 30, Stat Lecture 12 University of Washington - Seattle
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1 Probability Part II - Counting Rules Eli Gurarie Stat Lecture 12 University of Washington - Seattle January 30, 2013
2 Review of Friday s Lecture Definition of random process: The outcome of any single event X is not predictable The outcome of many repeated events gives rise to a probability distribution: f (k) = p(x = k) We can describe / characterize / predict / model the probability distribution.
3 Review of Friday s Lecture Definition of random process: The outcome of any single event X is not predictable The outcome of many repeated events gives rise to a probability distribution: f (k) = p(x = k) We can describe / characterize / predict / model the probability distribution. Elements of probability: Outcome: the result of a single trial (H or Ace) We also say that the outcome is the realization of a random process. Sample space: the list of all possible outcomes S Event: a subset of the sample space (A, B) An outcome is an event, but not all events are outcomes. (Note: This is technical and not very important)
4 Types Types of events of events Overlap? S Disjoint no outcomes in common Not disjoint some elements in common S Influence? Independent outcomes do not influence each other Dependent outcome of an event influences another Coin flips Card draws Stat 311 Spring 2011 : M. Morris 6
5 Unions, Unions, Intersections, Intersections Complements and Complements Events can be combined in different ways Union is or : (A U B) A B S Intersection is and : (A B) A B S Complement is not : A c Stat 311 Spring 2011 : M. Morris 7
6 Basic Basic rules ofrules probability of probability Valid range: 0 P(A) 1 for any event A in S Total probability: P(S) = 1 Complement rule: For any event A, where A c is the event not A P(A c ) = 1 P(A) Addition rule: If A and B are disjoint events, then P(A U B) = P(A) + P(B) Multiplication rule: If A and B are independent events, then P(A B) = P(A)*P(B) A A A B B B S S Stat 311 Spring 2011 : M. Morris 8
7 Uniform probability spaces There is a class of random processes for which each outcome has equal probability, for example: Coin flips Dice rolls Cards from a shuffled deck
8 Uniform probability spaces There is a class of random processes for which each outcome has equal probability, for example: Coin flips Dice rolls Cards from a shuffled deck But not: Free throws
9 A surprising fact A great amount of the theory underlying classical statistical inference can be derived from considering (in great detail) independent events from equal probability sample spaces!
10 Consider rolling 2 dice Question: What is the probability that the sum is 5?
11 What is the probability that the sum is 5? The sample space consists of 36 equally probable events: S = {(1, 1), (1, 2), (1, 3),...}
12 What is the probability that the sum is 5? The sample space consists of 36 equally probable events: S = {(1, 1), (1, 2), (1, 3),...} How do we know? We counted: N S = 6 6 Note: A and B are independent, so P(A B) = P(A)P(B).
13 What is the probability that the sum is 5? The sample space consists of 36 equally probable events: S = {(1, 1), (1, 2), (1, 3),...} How do we know? We counted: N S = 6 6 Note: A and B are independent, so P(A B) = P(A)P(B). How many sum to 5?
14 What is the probability that the sum is 5? The sample space consists of 36 equally probable events: S = {(1, 1), (1, 2), (1, 3),...} How do we know? We counted: N S = 6 6 Note: A and B are independent, so P(A B) = P(A)P(B). How many sum to 5? We counted: (1,4), (2,3), (3,2), (4,1) N A = 4
15 What is the probability that the sum is 5? The sample space consists of 36 equally probable events: S = {(1, 1), (1, 2), (1, 3),...} How do we know? We counted: N S = 6 6 Note: A and B are independent, so P(A B) = P(A)P(B). How many sum to 5? We counted: (1,4), (2,3), (3,2), (4,1) N A = 4 P(D 1 + D 2 = 5) = N A /N S = 4/36 = 0.111
16 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5
17 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5 What s the probability of 2 dice giving a sum > 9? S has 36 outcomes, A (> 9) has six outcomes, N A /N S = 6/36 = 0.166
18 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5 What s the probability of 2 dice giving a sum > 9? S has 36 outcomes, A (> 9) has six outcomes, N A /N S = 6/36 = What s the probability that at least 2 people in a class of 23 people have the same birthday?
19 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5 What s the probability of 2 dice giving a sum > 9? S has 36 outcomes, A (> 9) has six outcomes, N A /N S = 6/36 = What s the probability that at least 2 people in a class of 23 people have the same birthday? Yikes!
20 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5 What s the probability of 2 dice giving a sum > 9? S has 36 outcomes, A (> 9) has six outcomes, N A /N S = 6/36 = What s the probability that at least 2 people in a class of 23 people have the same birthday? Yikes! What s the probability that after 20 coin flips, you ll get exactly 10 heads?
21 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5 What s the probability of 2 dice giving a sum > 9? S has 36 outcomes, A (> 9) has six outcomes, N A /N S = 6/36 = What s the probability that at least 2 people in a class of 23 people have the same birthday? Yikes! What s the probability that after 20 coin flips, you ll get exactly 10 heads? Yikes!
22 Lots of probability problems are just counting problems! What s the probability of 1 die giving an odd number? S has 6 outcomes, A (Odds) had 3 outcomes, N A /N S = 3/6 = 0.5 What s the probability of 2 dice giving a sum > 9? S has 36 outcomes, A (> 9) has six outcomes, N A /N S = 6/36 = What s the probability that at least 2 people in a class of 23 people have the same birthday? Yikes! What s the probability that after 20 coin flips, you ll get exactly 10 heads? Yikes!
23 Counting is not always easy!
24 Counting Rules What s for lunch? Food: Sushi, Teriyaki, Udon noodle Drink: Fanta, Green Tea, H 2 0 How many different meals can I make?
25 Counting Rules Fundamental counting rule Let A 1 be a set with n 1 elements and A 2 be a set with n 2 elements. If one element is taken from A 1 and one element is taken from A 2, there are: possible unique outcomes. n 1 n 2
26 Counting Rules Fundamental counting rule Let A 1 be a set with n 1 elements and A 2 be a set with n 2 elements. If one element is taken from A 1 and one element is taken from A 2, there are: possible unique outcomes. n 1 n 2 Answer 3 3 = 9
27 Counting Rules What s for dinner? Food: Escargots, Fondue, Grenouilles Drink: Bordeaux, Burgundy, Beaujolais Dessert: Crème fraîche, Tarte aux pommes, Sorbet aux pêches How many people at a table can have a unique meal?
28 Counting Rules Multiplicative rule Let A 1, A 2, A 3 be k sets with n 1, n 2,... n k elements (respectively) in each set. If one element is taken from each set, then there are possible unique outcomes. n 1 n 2... n k
29 Counting Rules Multiplicative rule Let A 1, A 2, A 3 be k sets with n 1, n 2,... n k elements (respectively) in each set. If one element is taken from each set, then there are possible unique outcomes. n 1 n 2... n k Answer = 27
30 Counting Rules How do I rank my favorite animals? Some animals: Aardvark, Baboon, Cheetah, Dolphin How many different ways can I rank them according to how cool I think they are?
31 Counting Rules Factorial rule for permutations A set of n elements can be ordered n! different ways Definition of factorial n! = n(n 1)(n 2)(n 3)...1 And: 1! = 0! = 1
32 Counting Rules Factorial rule for permutations A set of n elements can be ordered n! different ways Definition of factorial n! = n(n 1)(n 2)(n 3)...1 And: 1! = 0! = 1 Answer 4! = = 24
33 Counting Rules How do I rank my favorite four out of eleven animals? Some animals: Aardvark, Baboon, Cheetah, Dolphin, Egret, Flamingo, Giraffe, Hippo, Iguana, Jackal, Kangaroo How many different ways can I rank the 4 coolest ones?
34 Counting Rules Permutations (order matters) A selection of r elements from a set of n total elements can be rank ordered in n! (n r)! different ways. Answer 11! (11 4)! = 11! 7! = = 7920
35 Counting Rules How do I pick four animals I want to study? Some animals: Aardvark, Baboon, Cheetah, Dolphin, Egret, Flamingo, Giraffe, Hippo, Iguana, Jackal, Kangaroo How many different ways can I separate this group into 4 that I want to study and 7 that I don t?
36 Counting Rules Combinations (order doesn t matters) A selection of r elements from a set of n total elements can be chosen in ( ) n n! = r r!(n r)! different ways. Choose function We call this creature: ( n r) n choose r It is the number of ways we can pick r unique cases from a set of n It is also written: r C n, and called: the binomial coefficient.
37 Counting Rules Combinations (order doesn t matters) A selection of r elements from a set of n total elements can be chosen in ( ) n n! = r r!(n r)! different ways. Choose function We call this creature: ( n r) n choose r It is the number of ways we can pick r unique cases from a set of n It is also written: r C n, and called: the binomial coefficient. Answer 11! 4!(11 4)! = = 330
38 Blaise Pascal ( ) Great French mathematician - described Pacal s triangle in Treatise on the Arithmetical Triangle (1653).
39 Zhu Shijie ( ) Great Chinese mathematician - described the triangle in The Precious Mirror of the Four Elements (1303).
40 Zhu Shijie ( ) Great Chinese mathematician - described the triangle in The Precious Mirror of the Four Elements (1303). Attributes it to Jia Xian (1050).
41 Zhu Shijie ( ) Great Chinese mathematician - described the triangle in The Precious Mirror of the Four Elements (1303). Attributes it to Jia Xian (1050). Also attributed to Omar Khayyam (Persia: ) - Khayyam s Triangle
42 Zhu Shijie ( ) Great Chinese mathematician - described the triangle in The Precious Mirror of the Four Elements (1303). Attributes it to Jia Xian (1050). Also attributed to Omar Khayyam (Persia: ) - Khayyam s Triangle Who attributes it to Al-Karaji (Persia: )
43 Zhu Shijie ( ) Great Chinese mathematician - described the triangle in The Precious Mirror of the Four Elements (1303). Attributes it to Jia Xian (1050). Also attributed to Omar Khayyam (Persia: ) - Khayyam s Triangle Who attributes it to Al-Karaji (Persia: ) Though it was known by Pingala (India: 2nd century)
44 Back to the Birthday Problem What is the probability that in a class of 23 students, at least 2 will have a matching birthday?
45 Back to the Birthday Problem What is the probability that in a class of 23 students, at least 2 will have a matching birthday? What is S? All possible sequences of birthdays (multiplicative rule): N S =
46 Back to the Birthday Problem What is the probability that in a class of 23 students, at least 2 will have a matching birthday? What is S? All possible sequences of birthdays (multiplicative rule): N S = What is A? All possible sequences where at least 2 people have the same birthday. That s a bit tricky.
47 Back to the Birthday Problem What is the probability that in a class of 23 students, at least 2 will have a matching birthday? What is S? All possible sequences of birthdays (multiplicative rule): N S = What is A? All possible sequences where at least 2 people have the same birthday. That s a bit tricky. What is A c? All possible sequences where NO ONE shares a birthday (permutations rule). N A = = 365! (365 23)!
48 Back to the Birthday Problem What is the probability that in a class of 23 students, at least 2 will have a matching birthday? What is S? All possible sequences of birthdays (multiplicative rule): N S = What is A? All possible sequences where at least 2 people have the same birthday. That s a bit tricky. What is A c? All possible sequences where NO ONE shares a birthday (permutations rule). N A = = 365! (365 23)! P(A c ) = N A c N S = 365!/342!
49 Back to the Birthday Problem What is the probability that in a class of 23 students, at least 2 will have a matching birthday? What is S? All possible sequences of birthdays (multiplicative rule): N S = What is A? All possible sequences where at least 2 people have the same birthday. That s a bit tricky. What is A c? All possible sequences where NO ONE shares a birthday (permutations rule). N A = = 365! (365 23)! P(A c ) = N A c N S = 365!/342! P(A) = 1 P(A c ) = 1 365!/342!
50 Back to the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads?
51 Back to the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? Sample size (multiplicative rule) N S = 2 10 What is the event size: N A?
52 Back to the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? Sample size (multiplicative rule) N S = 2 10 What is the event size: N A? We can define a sequence of events by 5 numbers chosen from 1 to 10. This is the same as choosing a combination of 5 unique numbers from 10 total, and we don t care about the order (combinations rule): ( 10 ) 10! N A = = 5 5!(10 5)!
53 Back to the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? Sample size (multiplicative rule) N S = 2 10 What is the event size: N A? We can define a sequence of events by 5 numbers chosen from 1 to 10. This is the same as choosing a combination of 5 unique numbers from 10 total, and we don t care about the order (combinations rule): ( 10 ) 10! N A = = 5 5!(10 5)! P(5 heads in 10 tosses) = 10!/(5!(10 5)!)
54 A different way to look at the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads?
55 A different way to look at the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? First: We need 5 heads (H) and 5 tails (T) to happen: P(T ) = P(H) = 1/2 P(HHHHH) = (1/2) 5, P(TTTTT ) = (1/2) 5
56 A different way to look at the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? First: We need 5 heads (H) and 5 tails (T) to happen: P(T ) = P(H) = 1/2 P(HHHHH) = (1/2) 5, P(TTTTT ) = (1/2) 5 But there are many ways in which these sequences can happen! P(5 heads in 10 tosses) = K(1/2) 5 (1/2) 5 What is K?
57 A different way to look at the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? First: We need 5 heads (H) and 5 tails (T) to happen: P(T ) = P(H) = 1/2 P(HHHHH) = (1/2) 5, P(TTTTT ) = (1/2) 5 But there are many ways in which these sequences can happen! P(5 heads in 10 tosses) = K(1/2) 5 (1/2) 5 What is K? Combinations Rule! ( 10 ) 10! K = = 5 5!(10 5)!
58 A different way to look at the Coin Problem What is the probability that after flipping 10 coins, you ll get exactly 5 heads? First: We need 5 heads (H) and 5 tails (T) to happen: P(T ) = P(H) = 1/2 P(HHHHH) = (1/2) 5, P(TTTTT ) = (1/2) 5 But there are many ways in which these sequences can happen! P(5 heads in 10 tosses) = K(1/2) 5 (1/2) 5 What is K? Combinations Rule! ( 10 ) 10! K = = 5 5!(10 5)! P(5 heads in 10 tosses) = 10! 5!(10 5)! (1/2)5 (1/2)
59 More flexible way of thinking about the problem... Example: what is the probability that Shaq will make 8 free throws out of 10? We need 8 successes (H) and 2 failures (M) to happen: P(H) = p = and P(M) = 1 p = P(HHHHHHHH) = p 8, P(MM) = (1 p) 2 How many ways can the sequence of 8 hits happen? Combinations Rule! ( 10 ) K = = 10! 8 8!2!
60 More flexible way of thinking about the problem... Example: what is the probability that Shaq will make 8 free throws out of 10? We need 8 successes (H) and 2 failures (M) to happen: P(H) = p = and P(M) = 1 p = P(HHHHHHHH) = p 8, P(MM) = (1 p) 2 How many ways can the sequence of 8 hits happen? Combinations Rule! ( 10 ) K = = 10! 8 8!2! P(8 hits in 10 FT s) = 10! 8!2! p8 (1 p) %
61 More flexible way of thinking about the problem... Example: what is the probability that Shaq will make 8 free throws out of 10? We need 8 successes (H) and 2 failures (M) to happen: P(H) = p = and P(M) = 1 p = P(HHHHHHHH) = p 8, P(MM) = (1 p) 2 How many ways can the sequence of 8 hits happen? Combinations Rule! ( 10 ) K = = 10! 8 8!2! P(8 hits in 10 FT s) = 10! 8!2! p8 (1 p) % (Note: that number is written in PERCENT!)
62 Binomial Distribution The Binomial Distribution is a discrete probability distribution that tells you the exact probability of k successes out of n tries, if each try is an independent event with probability p: f (k n, p) = Pr(X = k) = n! k!(n k)! pk (1 p) n k where k = {0, 1, 2...n}. Note the following properties: n f (k n, p) = 1 k=0 Note that we derived this distribution from probability rules and counting rules.
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