Stat 321 – Day 7Quiz 1Keep in MindKeep in MindLast Time – Counting MethodsDon’t always assume equally likely outcomes!Day 6 handout, Question 17PracticeSlide 9Example 1: Top 100 FilmsSlide 11Slide 12Definition: Conditional ProbabilityDirect CalculationConditional ProbabilitySlide 16Slide 17Example 1, questions h and iExample 1, question jExample 1, questions k-mMultiplication Rule if…Example 2: Distributive ComputingFor ThursdayStat 321 – Day 7Conditional Probability (2.4)Independence (2.5)Quiz 1See solutions onlineGood job explaining answers…Keep in MindOften helpful if can define mutually exclusive eventsP(AB)=P(A)+P(B)-P(AB)=P(A'B)+P(AB)+P(AB')A B'A  BA'  BKeep in Mind A  B'P(A) = P(A  B')+ P(A  B) P(A  B')= P(A) – P(A  B)Exactly oneP(A'B  AB') =P(A  B)-P(AB)A B'A  BA B'A'  BLast Time – Counting MethodsP(event) = # in event/# in sample spaceWhen outcomes are equally likelyPermutations vs. CombinationsDoes order matter?Selecting team vs. selecting batting orderCount “consistently”Consider the complement of the eventDon’t always assume equally likely outcomes!Nightline, Oct. 1997TED KOPPEL: Dr. Andrews, I'm sure you have heard such cautionary advice before so on what basis is the assumption being made that this is the one that's going to have the kind of impact on southern California in particular that's being predicted?RICHARD ANDREWS: Well, in the business that I'm in and that local government and state government is in, which is to protect lives and property, we have to take these forecasts very seriously. We have a lot of forecasts about natural hazards in California and we have a lot of natural events here that remind us that we need to take these forecasts seriously. I listen to earth scientists talk about earthquake probabilities a lot and in my mind every probability is 50-50, either it will happen or it won't happen. And so we're trying to take the past historical record, our own recent experience of the last, two of the last three years and make the necessary preparedness measures that can help protect us as much as we can from these events.Day 6 handout, Question 17Choosing two people at random from the committee of six comprised of two women and four men. (e) P(2 women) = C(4,0)C(2,2)/C(6,2)(d) P(1 W, 1 M) = C(4,1)C(2,1)/C(6,2)(f) P(2 men) = C(4,2)C(2,0)/C(6,2)These three probabilities sum to 1.Hypergeometric probabilitiesPractice1) In a league of 12 teams, how many ways are there to select(a) four teams to make the play-offs? C(12,4)(b) the champion and runner-up? 12(11)2) In a class of 30 men and 18 women, how many ways are there to select(a) three students to receive the top three scores in the class?48(47)(46) assuming order matters(b) four students? C(48,4)(c) 2 men and 2 women?C(30,2)C(18,2)Note: Watch for “number of” vs. “probability” in questionLab 3 questions?Example 1: Top 100 FilmsReturn to films, but ask a slightly different questionBeth yes Beth no TotalAllan yes 42 6 48Allan no 17 35 52Total 59 41 100What’s the probability that Allan has not seen a film that you know Beth has not seen?Example 1: Top 100 FilmsReturn to films, but ask a slightly different questionBeth yes Beth no TotalAllan yes 42 6 48Allan no 17 35 52Total 59 41 100What’s the probability that Allan has not seen a film that you know Beth has not seen?Example 1: Top 100 FilmsOf the 41 Beth has not seenBeth yes Beth no TotalAllan yes 42 6 48Allan no 17 35 52Total 59 41 100 Allan has not seen 35 of them as well. Probability = 35/41 = .854Definition: Conditional ProbabilityP(A|B) = “probability of A given B,” the probability that event A happens knowing that event B has happened. Restriction of the sample space to only those outcomes that are contained in event B. In general, to calculate the conditional probability of event A given event B we use the formula: P(A|B) = P(AB) P(B)Direct CalculationP(Allan has not seen the movie, given Beth has not seen the movie)= P(neither seen it) / P(Beth has not)= .35 / .41= .854 = P(A'|B') = P(A'  B')/P(B')match up the event condition onConditional ProbabilityKnowing Beth has not seen it, probability Allan has not = .854Without knowing whether Beth has seen it, probability Allan has not = .52Is a higher probability Allan has not seen it if know Beth has notconditional probabilityunconditionalExample 1: Top 100 FilmsP(Allan|Beth)=P(AB)/P(B)=.42/.59=.712is greater than P(Allan) = .48AB(e) Knowing whether or not Beth has seen it changes the probabilities for Allan.(f) The events are not independentExample 1: Top 100 Films(g) P(A|B') = .06/.41 = .146This is related to P(A'|B') = .854A|B' is the “complement” of A'|B'DBExample 1, questions h and iP(D|B)= .23 .59= .389< P(D)=.60 so they are not independent. Lower chance Dave has seen it knowing Beth has. Of those seen by Beth, fairly small proportion seen by Dave.Example 1, question j(j) P(B|D) = .20/.75 = .267  P(B) again showing not independent  P(D|B) NOT THE SAME•Think carefully about which event you are conditioning on.Example 1, questions k-m(k) P(E|D) = .80(l) P(E  D) = P(E|D)P(D)= .80(.60) = .48(m) If E and D are independentP(E|D) = P(E)EDMultiplication Rule if…P(E  D) = P(E|D)P(D)If D and E are independentP(E|D) = P(E)P(E|D)P(D) = P(E)P(D)Multiplication Rule for Independent EventsP(E  D) = P(E)P(D)Example 2: Distributive ComputingP(all 3 work) = P(WWW) = (.9)3 = .729Much harder to get all 3 to work than just oneP(intersection) gets smaller…P(at least one works) = P(WW  W) = 1-P(W'W'W')= 1-(.1)3 =.999Much easier for system to work with backupsP(union) gets larger…For ThursdayQuiz on HW 2 at end of classSolutions to be posted onlinePassword protectedContinue Lab 3Review problems, including on newer material, to be posted