Chapter2_large

Chapter2_large

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Unformatted text preview: omes •Consider a more complicated example: rolling 2 dice. What is P({2,6})? It is the same as P({1,1}) or P({3,4}). All outcomes are equally likely. We need to count the outcomes or have rules to compute probabilities. Events ORF 245 Prof. Rigollet Fall 2012 An event is a collection of outcomes. It can be described either with words or using formal notation from set theory. Passing from the first one to the second is a necessary skill. Flipping two coins. We know that the outcomes are (H,H), (H,T), (T, H), (T,T). Consider the events {twice the same}={(H,H), (T,T)} {heads first}={(H,T), (H,H)} {no heads}={(T,T)} We want to find rules to compute the probability of events from the probability of outcomes. ORF 245 Prof. Rigollet Fall 2012 Operations on events B A Operations on events ORF 245 Prof. Rigollet Fall 2012 Union of two events A and B: A B = {outcomes that are either in A or in B or in both} Intersection of two events A and B: A B= {outcomes that in A and in B} Complement of the event A: Ac = {outcomes that are not in A} ORF 245 Prof. Rigollet Fall 2012 Operations on events Union B A B A “or” B A[B A Intersection A B B A “and” B A Complement Ac “not” A A\B A A c Operations on events Flipping two coins. A={twice the same}={(H,H), (T,T)} B={heads first}={(H,T), (H,H)} C={no heads}={(T,T)} A B ={(H,H), (T,T), (H,T)} ={twice the same or heads first} A B ={(H,H)} ={twice the same and heads first} ={twice heads} C c ={ } ={at least one heads} ORF 245 Prof. Rigollet Fall 2012 Addition rule of disjoint events ORF 245 Prof. Rigollet Fall 2012 Two events are disjoint if they have no outcome in common. For example: {(H,H), (T,T)} and {(H,T), (H,T)} are disjoint {(H,H), (T,T)} and {(H,T), (H,H)} are not disjoint Equivalently, two events are disjoint if their intersection is empty (no outcome is in in both) For disjoint events A and B we have the addition rule P (A B ) = P (A) + P (B ) ORF 245 Prof. Rigollet Fall 2012 Addition rule of disjoint events More generally, if k events A1 , A2 , . . . , Ak are disjoint P (A1 A2 ... Ak ) = P (A1 ) + P (A2 ) + . . . + P (Ak ) B A A B C D P (A B ) = P (A) + P (B ) P (A B C D) = P (A) + P (B ) + P (C ) + P (D) Addition rule of disjoint events ORF 245 Prof. Rigollet Fall 2012 Example: Tossing five coins at random. What is the probability of at least four heads? Possible outcomes={HHHHH, THHHH, HTHHH, HHTHH, HHHTH, HHHHT, TTHHH,...,TTTTT} There are 25 possible outcomes (2 for each round) P(at least 4 heads) =P(4 heads or 5 heads) =P(4 heads)+P(5 heads) =1/32 + 5/32 = 6/32=3/16 Ge...
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This document was uploaded on 10/14/2013.

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