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Explore the world of probability including randomness, long-run behavior of chance processes, counting rules, permutations, combinations, conditional probability, and more. Gain insights into probability calculations and applications in various scenarios.
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Randomness • Long-Run (limiting) behavior of a chance (non-deterministic) process • Relative Frequency: Fraction of time a particular outcome occurs • Some cases the structure is known (e.g. tossing a coin, rolling a dice) • Often structure is unknown and process must be simulated • Probability of an event E (where n is number of trials) :
Set Notation • S a set of interest • Subset: A S A is contained in S • Union: A B Set of all elements in A or B or both • Intersection: AB Set of all elements in both A and B • Complement: Ā Set of all elements not in A
Probability • Sample Space (S)- Set of all possible outcomes of a random experiment. Mutually exclusive and exhaustive form. • Event- Any subset of a sample space • Probability of an event A (P(A)): • P(A) ≥ 0 • P(S) = 1 • If A1,A2,... is a sequence of mutually exclusive events (AiAj = ):
Counting Rules for Probability (I) • Multiplicative Rule- Brute-force method of defining number of elements in S • Experiment consists of k stages • The jth stage has nj possible outcomes • Total number of outcomes = n1...nk • Tabular structure for k = 2
Counting Rules for Probability (II) • Permutations: Ordered arrangements of r objects selected from n distinct objects without replacement (r ≤ n) • Stage 1: n possible objects • Stage 2: n-1 remaining possibilities • Stage r : n-r+1 remaining possibilities
Counting Rules for Probability (III) • Combinations: Unordered arrangements of r objects selected from n distinct objects without replacement (r ≤ n) • The number of distinct orderings among a set of r objects is r ! • # of combinations where order does not matter = number of permutations divided by r !
Counting Rules for Probability (IV) • Partitions: Unordered arrangements of n objects partitioned into k groups of size n1,...nk (where n1 +...+ nk = n) • Stage 1: # of Combinations of n1 elements from n objects • Stage 2: # of Combinations of n2 elements from n-n1 objects • Stage k: # of Combinations of nk elements from nk objects
Counting Rules for Probability (V) • Runs of binary outcomes (String of m+n trials) • Observe n Successes (S) • Observe m Failures (F) • k = minimum # of “runs” of S or F in the ordered outcomes of the m+n trials
Conditional Probability and Independence • In many situations one event in a multi-stage experiment occurs temporally before a second event • Suppose event A can occur prior to event B • We can consider the probability that B occurs given A has occurred (or vice versa) • For B to occur given A has occurred: • At first stage A has to have occurred • At second stage, B has to occur (along with A) • Assuming P(A), P(B) > 0, we can obtain “Probability of B Given A” and “Probability of A Given B” as follow: If P(B|A) = P(B) and P(A|B) = P(A), A and B are said to be INDEPENDENT
Bayes’ Rule - Updating Probabilities • Let A1,…,Ak be a set of events that partition a sample space such that (mutually exclusive and exhaustive): • each set has known P(Ai) > 0 (each event can occur) • for any 2 sets Ai and Aj, P(Aiand Aj) = 0 (events are disjoint) • P(A1) + … + P(Ak) = 1 (each outcome belongs to one of events) • If C is an event such that • 0 < P(C) < 1 (C can occur, but will not necessarily occur) • We know the probability will occur given each event Ai: P(C|Ai) • Then we can compute probability of Ai given C occurred:
Example - OJ Simpson Trial • Given Information on Blood Test (T+/T-) • Sensitivity: P(T+|Guilty)=1 • Specificity: P(T-|Innocent)=.9957 P(T+|I)=.0043 • Suppose you have a prior belief of guilt: P(G)=p* • What is “posterior” probability of guilt after seeing evidence that blood matches: P(G|T+)? Source: B.Forst (1996). “Evidence, Probabilities and Legal Standards for Determination of Guilt: Beyond the OJ Trial”, in Representing OJ: Murder, Criminal Justice, and the Mass Culture, ed. G. Barak pp. 22-28. Harrow and Heston, Guilderland, NY
Northern Army at Gettysburg • Regiments: partition of soldiers (A1,…,A9). Casualty: event C • P(Ai) = (size of regiment) / (total soldiers) = (Column 3)/95369 • P(C|Ai) = (# casualties) / (regiment size) = (Col 4)/(Col 3) • P(C|Ai) P(Ai) = P(Ai and C) = (Col 5)*(Col 6) • P(C)=sum(Col 7) • P(Ai|C) = P(Ai and C) / P(C) = (Col 7)/.2416
CRAPS • Player rolls 2 Dice (“Come out roll”): • 2,3,12 - Lose (Miss Out) • 7,11 - Win (Pass) • 4,5,6.8,9,10 - Makes point. Roll until point (Win) or 7 (Lose) • Probability Distribution for first (any) roll: • After first roll: • P(Win|2) = P(Win|3) = P(Win|12) = 0 • P(Win|7) = P(Win|11) = 1 • What about other conditional probabilities if make point?
CRAPS • Suppose you make a point: (4,5,6,8,9,10) • You win if your point occurs before 7, lose otherwise and stop • Let P mean you make point on a roll • Let C mean you continue rolling (neither point nor 7) • You win for any of the mutually exclusive events: • P, CP, CCP, …, CC…CP,… • If your point is 4 or 10, P(P)=3/36, P(C)=27/36 • By independence, and multiplicative, and additive rules:
CRAPS • Similar Patterns arise for points 5,6,8, and 9: • For 5 and 9: P(P) = 4/36 P(C) = 26/36 • For 6 and 8: P(P) = 5/36 P(C) = 25/36 Finally, we can obtain player’s probability of winning:
CRAPS - P(Winning) Note in the previous slides we derived P(Win|Roll), we multiply those by P(Win to obtain P(Roll&Win) and sum those for P(Win). The last column gives the probability of each come out roll given we won.
Odds, Odds Ratios, Relative Risk • Odds: Probability an event occurs divided by probability it does not occur odds(A) = P(A)/P(Ā) • Many gambling establishments and lotteries posted odds against an event occurring • Odds Ratio (OR): Odds of A occurring for one group, divided by odds of A for second group • Relative Risk (RR): Probability of A occurring for one group, divided by probability of A for second group
Example – John Snow Cholera Data • 2 Water Providers: Southwark & Vauxhall (S&V) and Lambeth (L)