Independent EventsJuana [email protected] Department of StatisticsJuly 9J. Sanchez Stat 100A. Conditional ProbabilityAnnouncementsHomework 2 is due Today at the beginning of lecture. Insert in themanila envelopes for Hwk 2. Homework turned in at the end oflecture will get points deducted.Quiz 2 is done today. Everybody stay in class to start at 10:05.When I say to stop, everybody stops. I will collect all quizzes fromeach individual student.J. Sanchez Stat 100A. Conditional ProbabilityPlease, you may sit next to each other but only in the evennumbered rows.Odd-numbered rows must be left emptyJ. Sanchez Stat 100A. Conditional ProbabilityTodayReference reading this part of lecture: Ross (8th edition), chapter 3, page79-82 PAges 75, 76, 77 in 9th edition,I. Product rule for joint occurrence of events (review)II. Special case: product rule for independent eventsIII. Comments on simulation done in the TA session today.J. Sanchez Stat 100A. Conditional ProbabilityNotesNotesNotesNotesI. Product rule for joint occurrence of eventsLet E and F be two events. Review: By definition of conditionalprobabilityP(E | F ) =P(EF )P(F )and P(F | E ) =P(EF )P(E)As a consequence of this definition,General Product rule for the occurrence of two eventsP(EF ) = P(E | F )P(F )andP(EF ) = P(F | E)P(E )J. Sanchez Stat 100A. Conditional ProbabilityLet E1, E2, ....., Enbe n events. By definition of conditional probability,P(E1E2E3....En−1| En) =P(E1E2E3.....En)P(En)As a consequence,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 ProbabilityExampleIf we were to draw three cards from a box containing 52 cards, 26 ofwhich are black, and we draw without replacement, then the probabilityof three black cards isP(B1B2B3) = P(B1)P(B2| B1)P(B3| B2B1) = (26/52)(25/51)(24/50).where B1is the event that the first is black, B2is the event that thesecond is black, and B3is the event that the third is black.J. Sanchez Stat 100A. Conditional ProbabilityExampleAn incoming lot of silicon wafers is to be inspected for defective (D) byan engineer in a microchip manufacturing plant. Suppose that, in a traycontaining twenty wafers, four are defective and sixteen are workingproperly Dc. Two wafers are to be selected randomly for inspection. Findthe probabilities of the following events.(a) Neither is defective (Call this event A)Let Dcibe event that ith wafer is working, i=1,2.. The i hereindicates order drawn.P(A) = P(Dc1Dc2) = P(Dc1)P(Dc2| Dc1) =1620×1519= 0.6316(b) At least one of the two is defective (call this event M)P(M) = 1 − P(Dc1Dc2) = 1 − 0.6316 = 0.3684J. Sanchez Stat 100A. Conditional ProbabilityNotesNotesNotesNotesII. Special case: product rule for independent eventsLet E1, E2, ....Enbe any number of events. Here is the definition ofindependence.Definition of IndependenceE1, E2, ....Enare independent ifP(E1E2.....En) = P(E1)P(E2)P(E3)............P(En). and any group ofthose sets are also independent, i.e., P(E1E2) = P(E1)P(E2),P(E1E3E4) = P(E1)P(E3)P(E4) .... and so on.In the case of just two event, say A and B, A and B are independentevents if P(AB)= P(A)P(B).J. Sanchez Stat 100A. Conditional ProbabilityExampleIf we were to draw three cards from a box containing 52 cards, 26 ofwhich are black, and we draw with replacement, then the probability ofthree black cards isIn this case, because the cards are replaced each time one is drawn, theprobability of getting a black is the same from draw to draw. Therefore,P(3 black cards) = (26/52) × (26/52) × (26/52) = 0.125J. Sanchez Stat 100A. Conditional ProbabilityApplications in Industrial ReliabilityFactories produce millions of items. Thus, when we sample them toobserve quality, we can pretend we are sampling with replacement. Thatis the assumption in industrial reliability and other contexts wherepopulations are very large.ExampleWhat is the probability of the event A=having 1 or less defectivecomponents in a system with 3 components where the probability ofdefective is 0.3 and the components are independent of each other?(1=defective, 0=non defective)S = {ddd, dddc, ddcd, ddcdc, dcdd, dcddc, dcdcd, dcdcdc}A = {ddcdc, dcddc, dcdcd, dcdcdc}P(A) = 3(0.3)0.72+ 0.73= 0.784J. Sanchez Stat 100A. Conditional ProbabilityDetermining whether events are independent, givenprobabilities.To determine whether events are independent, use the definition ofindependence.ExampleSuppose a foreman must select one worker from a pool of four availableworkers (numbered 1, 2, 3, 4) for a special job. He selects the worker bymixing the four names and randomly selecting one. Let A denote theevent that worker 1 or 2 is selected, let B denote the event that worker 1or 3 is selected, and let C denote the event that worker 1 is selected. AreA and B independent? Are A and C independent?Randomly selected means that the probability of selecting each of theworkers is 1/4 .P(A)=P(1 or 2) = 1/4 + 1/4 = 1/2P(B)=P(1 or 3) = 1/4 +1/4 = 1/2P( C) = 1/4; P(AB)=P(1)= 1/4Are A and B independent? Yes, 1/4 = P(AB) and P(A)P(B) = 1/4; soP(AB)=P(A)P(B)But A and C are not independent: P(AC) = P(1)=1/4 but P(A)P(C)=1/2 x 1/4 = 1/8J. Sanchez Stat 100A. Conditional ProbabilityNotesNotesNotesNotesDetermining whether events are independent inapplicationsSummary:If the sampling being done is done on a big population, we can assumedrawing with replacement scenario.If the sampling being done is done on a small population, thenindependence holds only if we draw with replacement.J. Sanchez Stat 100A. Conditional ProbabilityIII. Simulation results.You discuss today in the TA session results you obtained for a lie detectortest simulation.What is the conditional probability of being guilty given that the liedetector test says that the suspect is guilty?G= guilty suspect; TRG = Test results say guiltyP(G | TRG ) = P(G and TRG) / P(TRG)We want to find this with simulationJ. Sanchez Stat 100A. Conditional ProbabilityExamples of possible simulation results in a few trialsTrial Suspect Test result ConclusionNumber (G or NG (A or NA ) from testSuspect is G or NG1 G NA NG2 NG A NG3 G A G· · · · · · · · · · · ·10Do 10 trials.Then complete the table ”summary of individual results.” Ask three otherstudents what they got in their 10 trials and complete the table”Summary of group results”. Compute the probability asked using theindividual results and group results separately.
View Full Document