# 7-3-1 - 7-3 Part I Conditional Probability Intersection of...

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Unformatted text preview: 7-3 Part I: Conditional Probability, Intersection of EventsCONDITIONALEVENTSExperiment: I flip a coin twiceE1 = two heads = {HH}P(E1) = ¼ BUT…suppose I tell you that the outcome was the samefor both flips, that is, that eventE2 = {HH, TT} has occurredHow would you assess the probability that E1 occurred GIVEN THATE2 has occurred?written: P(E1 GIVENE2) or P(E1 | E2)S = Sample SpaceNote: since we know that E2 has occurred, our sample space has been reduced to just by that assumption:sample space is now E2 = {TT, HH}and thus P(E1 | E2) = 1 out of 2 = ½7-3 Part Ip. 1HHHTTHE2TTE1In general:P(E1 GIVENE2) = P(E1 | E2) = )2E(n)2E1E(n∩= )S(n/)2E(n)S(n/)2E1E(n∩= )2E(P)2E1E(P∩= 2141= ½P(A GIVENB)For any two events A and BP(A | B) = )()(BPBAP∩THEAND(INTERSECTION) OFTWOEVENTSRandom experiment: flip coin twiceEvent 1: E1 = get head on first throw = {HH, HT}Event 2: E2 = get the same on both throws = {HH, TT}Event E3: head on the 1stthrowANDsame on both throwsWe already know: P(E2 | E1) =...
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## This note was uploaded on 12/11/2011 for the course MATH 1324 taught by Professor Staff during the Spring '11 term at Austin Community College.

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7-3-1 - 7-3 Part I Conditional Probability Intersection of...

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