Counting Rules. Counting operations on n objects. Sort, order matters (perms) Choose k (combinations) Put in r buckets. None Distinct.

Transkrypt

1 Probability

2 Review

3 Counting Rules Counting operations on n objects Sort, order matters (perms) Choose k (combinations) Put in r buckets Distinct n! Some Distinct n! n 1!n 2!... n k Distinct = n! k!(n k)! Distinct None Distinct r n (n + r 1)! n!(r 1)!

4 Sample and Event Spaces Sample Space, S, is set of all possible outcomes of an experiment Event Space, E, is some subset of S (E Í S)

5 What is a Probability What is a probability? P (E) = lim n!1 n(e) n n is the number of trails Hit: 11 Thrown: 24 The event E is that you hit the target P(E) 0.46

6 Axioms of Probability Recall: S = all possible outcomes. E = the event. Axiom 1: 0 P(E) 1 Axiom 2: P(S) = 1 Axiom 3: P(E c ) = 1 P(E) Aside: axiom 3 is often stated as the probability of mutually exclusive events. We ll come back to that later in the lecture

7 Equally Likely Outcomes P(Each outcome) = 1 S number of outcomes in E E number of outcomes in S S In that case, P(E) = =

8 End Review

9 Mandarins and Bananas 4 Mandarins and 3 Bananas in a Bag. 3 drawn. What is P(1 Mandarin and 2 Bananas drawn)? Equally likely sample space? Thought experiment

10 Mandarins and Grapefruit 4 Mandarins and 3 Bananas in a Bag. 3 drawn. What is P(1 Mandarin and 2 Bananas drawn)? Ordered: Pick 3 ordered items: S = 7 * 6 * 5 = 210 Pick Mandarin as either 1st, 2nd, or 3rd item: E = (4 * 3 * 2) + (3 * 4 * 2) + (3 * 2 * 4) = 72 P(1 Mandarin, 2 Grapefruit) = 72/210 = 12/35 Unordered: æ7ö S = ç = 35 è3ø æ4öæ 3ö E = ç ç = 12 è1øè 2ø P(1 Mandarin, 2 Grapefruit) = 12/35

11 Make indistinct items distinct to get equally likely sample space outcomes *You will need to use this trick with high probability

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S = E = What is an example of one outcome? Is each outcome equally likely? P (straight) = E S =

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14 When approaching an equally likely probability problem, start by defining sample spaces and event spaces.

15 Chip Defect Detection n chips manufactured, 1 of which is defective. k chips randomly selected from n for testing. What is P(defective chip is in k selected chips)? S = E = = ænö ç èk ø æ1öæ n ç ç è1øè k -1ö -1ø P(defective chip is in k selected chips) 1 (n 1)! 1 (k 1)!(n k)! n 1 k 1 n k = n! k!(n k)! = k n

16 Target Revisited Screen size = 800x800 Radius of target = 200 The dart is equally likely to land anywhere on the screen. What is the probability of hitting the target? =0.191 S = E = p(e) =

17 Target Revisited Screen size = 800x800 Radius of target = 200 The dart is equally likely to land anywhere on the screen. What is the probability of hitting the target? = S = E = p(e) =

18

19

20 Serendipity Say the population of Stanford is 17,000 people You are friends with? Walk into a room, see 268 random people. What is the probability that you see someone you know? Assume you are equally likely to see each person at Stanford

21 Many times it is easier to calculate P (E C ).

22 Trailing the dovetail shuffle to it s lair Persi Diaconosis

23 Making History What is the probability that in the n shuffles seen since the start of time, yours is unique? S = (52!) n E = (52! - 1) n P(no deck matching yours) = (52!-1) n /(52!) n For n = 10 20, P(deck matching yours) < * Assume 7 billion people have been shuffling cards once a second since cards were invented

24 Back to Axiom 3

25 Axioms of Probability Recall: S = all possible outcomes. E = the event. Axiom 1: 0 P(E) 1 Axiom 2: P(S) = 1 Axiom 3: P(E c ) = 1 P(E) Aside: axiom 3 is often stated as the probability of mutually exclusive events. We ll come back to that later in the lecture

26 Axioms of Probability Recall: S = all possible outcomes. E = the event. Axiom 1: 0 P(E) 1 Axiom 2: P(S) = 1 Axiom 3: If events E and F are mutually exclusive: P (E [ F )=P (E)+P (F )

27 Mutually Exclusive Events 7/50 4/50 If events are mutually exclusive, probability of OR is simple: P (E [ F )=P (E)+P (F )

28 Mutually Exclusive Events 7/50 4/50 If events are mutually exclusive, probability of OR is simple: 7 P (E [ F )=P = 11 50

29 OR with Many Mutually Exclusive Events Wahoo! All my events are mutually exclusive X 1 X 2 X 3 P (X 1 [ X 2 [ [ X n )= nx i=1 P (X i )

30 If events are mutually exclusive probability of OR is easy!

31 P(E c ) = 1 P(E)? P(E E c ) = P(E) + P(E c ) P(S) = P(E) + P(E c ) 1 = P(E) + P(E c ) Since E and E c are mutually exclusive Since everything must either be in E or E c Axiom 2 P(E c ) = 1 P(E) Rearrange

32 Stretch! SCPD Code: dovetail

33 Conditional Probability Piech, CS106A, Stanford University

34 Why study probability? Piech, CS106A, Stanford University

35 Piech, CS106A, Stanford University Dice Our Misunderstood Friends Roll two 6-sided dice, yielding values D 1 and D 2 Let E be event: D 1 + D 2 = 4 What is P(E)? S = 36, E = {(1, 3), (2, 2), (3, 1)} P(E) = 3/36 = 1/12 Let F be event: D 1 = 2 P(E, given F already observed)? S = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)} E = {(2, 2)} P(E, given F already observed) = 1/6

36 Piech, CS106A, Stanford University Dice Our Misunderstood Friends Two people each roll a die, yielding D 1 and D 2. You win if D 1 + D 2 = 4 Q: What do you think is the best outcome for D 1?

37 Piech, CS106A, Stanford University Conditional Probability Conditional probability is probability that E occurs given that F has already occurred Conditioning on F Written as P (E F ) Means P(E, given F already observed) Sample space, S, reduced to those elements consistent with F (i.e. S Ç F) Event space, E, reduced to those elements consistent with F (i.e. E Ç F)

38 Piech, CS106A, Stanford University Conditional Probability With equally likely outcomes: P(E F) = = # of outcomes in E consistent with F # of outcomes in S consistent with F EF EF = SF F P (E) = P (E F )=

39 Piech, CS106A, Stanford University Conditional Probability With equally likely outcomes: P(E F) = = # of outcomes in E consistent with F # of outcomes in S consistent with F EF EF Shorthand notation = SF F for set intersection (aka set and ) P (E) = P (E F )=

40 Piech, CS106A, Stanford University Conditional Probability General definition: P ( E F) = P( EF) P( F) Holds even when outcomes are not equally likely Implies: P(EF) = P(E F) P(F) (chain rule) What if P(F) = 0? P(E F) undefined Congratulations! You observed the impossible!

41 Piech, CS106A, Stanford University Generalized Chain Rule General definition of Chain Rule: P( E E E3... E 1 2 n ) = n- P( E1) P( E2 E1) P( E3 E1E2 )... P( En E1E2... E 1)

42 Conditional Paradigm Piech, CS106A, Stanford University

43 + Learn Piech, CS106A, Stanford University

44 Piech, CS106A, Stanford University Netflix and Learn What is the probability that a user will watch Life is Beautiful? P(E) S = {Watch, Not Watch} E = {Watch} P(E) = ½?

45 Piech, CS106A, Stanford University Netflix and Learn What is the probability that a user will watch Life is Beautiful? P(E)

46 Netflix and Learn What is the probability that a user will watch Life is Beautiful? P(E) P (E) = lim n!1 n(e) n #people who watched movie #people on Netflix P(E) = 10,234,231 / 50,923,123 = 0.20 Piech, CS106A, Stanford University

47 Netflix and Learn Let E be the event that a user watched the given movie: P(E) = 0.19 P(E) = 0.32 P(E) = 0.20 P(E) = 0.09 P(E) = 0.23 * These are the actual estimates Piech, CS106A, Stanford University

48 Piech, CS106A, Stanford University Netflix and Learn What is the probability that a user will watch Life is Beautiful, given they watched Amelie? P(E F) P (E F )= P (EF) P (F )

49 Piech, CS106A, Stanford University Netflix and Learn What is the probability that a user will watch Life is Beautiful, given they watched Amelie? P(E F) P (E F )= P (EF) P (F ) = #people who watched both #people on Netflix #people who watched F #people on Netflix

50 Netflix and Learn What is the probability that a user will watch Life is Beautiful, given they watched Amelie? P(E F) P (E F )= P (EF) P (F ) = #people who watched both #people who watched F P(E F) = 0.42 Piech, CS106A, Stanford University

51 Piech, CS106A, Stanford University Netflix and Learn Let E be the event that a user watched the given movie, Let F be the event that the same user watched Amelie: P(E F) = 0.14 P(E F) = 0.35 P(E F) = 0.20 P(E F) = 0.72 P(E F) = 0.49

52 Piech, CS106A, Stanford University Machine Learning Machine Learning is: Probability + Data + Computers