7-3-1 - 7-3 Part I: Conditional Probability, Intersection of

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7-3 Part I: Conditional Probability, Intersection of EventsCONDITIONALEVENTSExperiment: I flip a coin twiceE1 = two heads = {HH}P(E1) = BUTsuppose 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 = 7-3 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= P(A GIVENB)For any two events A and BP(A | B) = )()(BPBAPTHEAND(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) =...
View Full Document

Page1 / 4

7-3-1 - 7-3 Part I: Conditional Probability, Intersection of

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online