Conditional ProbabilityJuana [email protected] Department of StatisticsJuly 2nd, Lecture part IIJ. Sanchez Stat 100A. Conditional ProbabilityThis lectureReference reading: Ross (8th edition), chapter 3, sections 3.1 - 3.3 and3.5 (page 93).I. Introduction to Conditional ProbabilityII. Properties of Conditional ProbabilitiesIII. Product rule. Using conditional probability to compute theprobability of two events happening simultaneously.IV. Special case: product rule for independent events (more on thisnext day)V. Law of total probability of an event.VI. Bayes theoremJ. Sanchez Stat 100A. Conditional ProbabilityNew concepts introduced in this lecture.Let G and B be two events.Definition of Conditional ProbabilityP(G | B) =P(GB )P(B ), where P(G | B)= conditional probability of G givenB. | means ”given that”; P(GB) = joint probability; P(B)=marginalprobabilityLaw of total probability of event GP(G ) = P (G | B )P(B ) + P (G | Bc)P(Bc),Product rule.P(GB ) = P(G | B )P(B )Bayes theorem. Inverse probabilityP(B | G ) =P(G |B )P (B )P(G |B )P (B )+P (G |Bc)P(Bc).J. Sanchez Stat 100A. Conditional ProbabilityExampleOf 100 students who completed an introductory statistics course, 20 werebusiness majors. Further, 10 students received A’s in the course, and 3 ofthese were business majors. What is the probability that a business majorgot an A ?Let B=Business major event; A= get A in course; With the informationgiven (in black color in the cross-tab below), we can complete theremaining cells (in red). We want P(A | B ).B BcTotalA 3 7 10Ac17 73 90Total 20 80 100P(A | B ) =P(AB )P(B )=3/10020/100= 3/20B BcTotalA P(AB)= 0.03 P(ABc) = 0.07 P(A)= 0.10AcP(AcB)=0.17 P(AcBc)=0.73 P(Ac) = 0.90Total P(B)=0.20 P(Bc) =0.80 1P(A | B ) =P(AB )P(B )=0.030.2= 3/20J. Sanchez Stat 100A. Conditional ProbabilityWe can do the same thing with trees. But be careful !!! You need tohave clear what each branch of the tree is doing.J. Sanchez Stat 100A. Conditional ProbabilityConditional probabilities satisfy Axioms of ProbabilityFirst, since AB ∈ B , then P(AB ) < P (B). Also, P (AB) > 0 andP(B ) > 0, so0 ≤ P (A | B ) =P(AB )P(B )≤ 1.P(S | B ) =P(SB )P(B )=P(B )P(B )= 1.If A1, A2, ...., are mutually exclusive events, then so areA1B, A2B, .... andP (∪∞i=1Ai| B ) =P ((∪∞i=1Ai) B)P(B )(1)=P (∪∞i=1(AiB))P(B )(2)=∞Xi=1P (AiB)P(B )(3)=∞Xi=1P(Ai|B) (4)J. Sanchez Stat 100A. Conditional ProbabilityI I. Prop erties of Conditional ProbabilitiesThe above axioms imply all the properties of probabilities apply toconditional probabilities. For example:P(A | B ) = 1 − P(Ac| B )P(A ∪ C | B) = P (A | B) + P(C | B ) + P ((A | B )(C | B ))J. Sanchez Stat 100A. Conditional ProbabilityI II. Product rule. Using conditional probability to computeprobability of two events happening simultaneouslyConsider two events A and B happening simultaneously. From thedefinition of conditional probability, it follows that the joint probability ofA and B isP(AB ) = P (A|B)P (B).J. Sanchez Stat 100A. Conditional ProbabilityGeneral Product rule for the intersection of two eventsP(EF ) = P (E | F )P(F )P(EF ) = P (F | E )P(E )General Product rule for the intersection of more than two eventsP(E1E2.....En) = P (E1)P(E2|E1)P(E3|E2E1)............P(En|E1.....En−1).J. Sanchez Stat 100A. Conditional ProbabilityExampleCeline is undecided as to whether to take a French course or a chemistrycourse. She estimates that her probability of receiving an A grade wouldbe 1/2 in a French course and 2/3 in a chemistry course. If Celinedecides to base her decision on the flip of a fair coin, what is theprobability that she gets an A in chemistry?P(A | F ) = 1/2P(A | C ) = 2/3P(C ) = P (F ) = 1/2P(AC ) = P (C )P (A | C ) = 1/2 × 2/3 = 2/6 = 1/3J. Sanchez Stat 100A. Conditional ProbabilityV. Law of total probability of an eventTotal probability of ASuppose we partition the sample space into two sets B and Bc. Supposealso that there is an event A that intersects both B and Bc. Then theevent A can be written asA = AB ∪ ABc. And thus by axiom 3,P(A) = P (AB ) + P(ABc)If conditional probabilities P (A | B ) and P (A | Bc) are known, then wecan apply the product rule to compute P(A) using the conditionalprobabilities.P(A) = P (B )P (A|B ) + P(Bc)P(A|Bc)J. Sanchez Stat 100A. Conditional ProbabilityVI. Bayes theoremBayes TheoremIf B1, B2form a partition of S, and A is any event in S , thenP(Bi| A) =P(A | Bi)P(Bi)Pkj=1P(A | Bj)P(Bj)J. Sanchez Stat 100A. Conditional ProbabilityExampleElectric motors coming off two assembly lines are pooled for storage in acommon stockroom, which contains an equal number of motors fromeach line. Motors from that room are periodically sampled and tested. Itis known that 10% of the motors from line I are defective and that 15%of the motors from line II are defective. If a motors is randomly selectedfrom the stock-room and found to be defective, what is the probabilitythat it came from line I?P(D | I ) = 0.1 ; P (D | II ) = 0.15P(I ) = 0.5 = P (II )By law of total probability,P(D ) = P (DI ) + P(DII ) = (0.1)(0.5) + (0.15)(0.5) = 0.125andP(I | D) =P(DI )P(D )=(0.1)(0.5)0.125= 0.4J. Sanchez Stat 100A. Conditional ProbabilityTo reviewReview these lecture notes and the audio.Recommended problems: Page 103 of textbook, 8th edition: 3.15,3.19, 3.20, 3.21. (These will be discussed in TA session of TuesdayJuly 9th).J. Sanchez Stat 100A. Conditional
View Full Document
Unlocking...