Probability TheoryConditional ProbabilitySection 5.2General multiplication rule A conditional probability, P(B | A), is the probability of some event, B, under the condition that some other event, A, has definitely occurred General multiplication rule:P(A and B)= P(A) P(B | A)P(A and B)= P(A) P(B | A) Extended version for multiple events:P(A and B and C) = P(A) P(B | A) P(C | A and B)etc.Implications of the general multiplication rule A definition of conditional probability is P(B | A) = P(A and B) / P(A)To derive: rearrange the general multiplication rule If A and B are independent, P(A and B) = P(A) P(B),henceP(B | A) = P(B)Example: Unemployment ratesDraw a single individual from a certain population(cnt./1000) Employment status RowGender Emp. Unemp. NA totalsFemale 4313 446 4357 9116Male392752046619058P(“female”) = 9116 / 18174 = 0.5002P(“employed” and “female”) = 4313 / 18174 = 0.2367P(“employed” | “female”) = 4313 / 9116 = 0.4731P(“employed”) = 8240 / 18174 = 0.4534Male392752046619058Column totals 8240 966 8968 18174Example: Prosperity and educationDraw a single household from a certain populationA = household is “educated”B = household is “prosperous”Known profile of the population:P(A) = 0.254, P(B) = 0.134, P(Aand B) = 0.080P(A) = 0.254, P(B) = 0.134, P(Aand B) = 0.080Are A and B independent?P(B | A) = P(A and B) / P(A) = 0.080 / 0.254 = 0.315 ≠ 0.134 = P(B)Not independent.Tree diagramsTree diagrams can help to organize thinking about conditional probabilitiesExample: High-school to professional athleteCollegeProfessionalHigh schoolCollegeAAcBBcBBc0.050.95S0.0170.9830.00010.9999P(A)P(B|Ac)General multiplication rule from a tree diagram Multiply along a complete path for the general multiplication rule, P(A and B) = P(A) P(B | A)Example: High-school to professional athleteP(Aand B) = P(A) P(B | A)AB0.017P(Aand B) = P(A) P(B | A)= (0.05)(0.017) = 0.00085P(Acand B) = P(Ac) P(B | Ac)= (0.95)(0.0001) = 0.000095AAcBcBBc0.050.95S0.0170.9830.00010.9999A total probability rule from a tree diagram Probability of a terminal event is the sum of the product of probabilities along all complete paths leading to itExample: High-school to professional athleteP(B) = P(Band A) + P(Band Ac)AB0.017P(B) = P(Band A) + P(Band A)= P(A) P(B | A) + P(Ac) P(B | Ac)= (0.05)(0.017)+ (0.95)(0.0001)= 0.000945AAcBcBBc0.050.95S0.0170.9830.00010.9999“Reverse” conditional probabilities Bayes’s rule:Example: High-school to professional athleteP(A) P(B | A) = (0.05)(0.017) = 0.00085P(Ac) P(B | Ac) = (0.95)(0.0001) = 0.000095⇒Example: Medical screeningTree diagram for breast cancer screening of women in twentiesCancerIncidenceCancerDiagnosisAB0.8“Sensitivity”False Womenin 20sAcBcBBc0.00040.9996S0.20.10.9“Specificity”False negativeFalse positiveExample: Medical screening (continued)Given a cancer diagnosis, what is the probability of a cancer incidence?ABBcB0.00040.9996S0.80.20.1AcBc0.99960.9Example: Defective partsTree diagram for parts from two different suppliersSupplierDefectiveABBc0.650.020.98PartAcBcBBc0.35S0.980.050.95ABBcB0.650.35S0.020.980.05Example: Defective parts (continued)Given a defective part, what is the probability it came from Supplier A?AcBc0.350.95Probability TheoryThe Binomial and Poisson DistributionsSections 5.3 and 5.4Models for count data The binomial distributions provide a theoretical model for count data having a fixed maximumExamples: # girls among three children# defects in a lot of one hundred items# recoveries among ten sick patients The Poisson distributions provide a theoretical model for open-ended countsExamples: # accidents on I81 the day before Thanksgiving# help-line calls on a typical weekday afternoon# flaws in a section of carpetThe binomial setting* Observe a fixed number of trials, n The trials are independent Each trial has the same two possible outcomes: S = “success” or F = “failure” Constant success probability, p = P(S), across trials* Also called a “binomial experiment”Properties of the binomial setting The sample space consists of 2npossible outcomesExample: n = 5 trials ⇒ 2n= 32 possible outcomesS = { SSSSS, SSSSF, SSSFS, SSFSS, SFSSS, FSSSS, SSSFF, SSFSF, SSFFS, SFSSF, SFSFS, SFFSS, FSSSF, FSSFS, FSFSS, FFSSS, SSFFF, SFSFF, FSSSF, FSSFS, FSFSS, FFSSS, SSFFF, SFSFF, SFFSF, SFFFS, FSSFF, FSFSF, FSFFS, FFSSF, FFSFS, FFFSS, SFFFF, FSFFF, FFSFF, FFFSF, FFFFS, FFFFF }Properties of the binomial setting (continued) P(“outcome”) = p#S(1 – p)#FExample: n = 5 trials, p = P(S) = 0.3P(FFSSF) = p2(1 – p)3= (0.3)2(0.7)3= 0.0309P(SFSSF) = p3(1 – p)2= (0.3)3(0.7)2= 0.0132P(SFFFS) = p2(1 – p)3= (0.3)2(0.7)3= 0.0309P(FFFFF) = p0(1 – p)5= (0.3)0(0.7)5= 0.1681Properties of the binomial setting (continued) The number of outcomes with k “successes” iswhere n! = n × (n – 1) × … × 3 × 2 × 1 (and 0! = 1)n “choose” kn “factorial”Example: n = 5 trials, k = 2 “successes”A = { SSFFF, SFSFF, SFFSF, SFFFS, FSSFF, FSFSF, FSFFS, FFSSF, FFSFS, FFFSS }Binomial random variables The random variable X = # “successes” is a binomial random variable Probabilities are:Example: n = 5 trials, p = P(S) = 0.3Binomial distributionk pk(1 – p)n – kP(X = k)0 1 0.1681 0.16811 5 0.0720 0.36022 10 0.0309 0.30873 10 0.0132 0.13234 5 0.0057 0.02845 1 0.0024 0.0024Binomial mean and standard deviation If X is a binomial random variable of n trials with p = P(S) then µX= n p and σX= √{n p (1 – p)}n = 10, p = 0.10 n = 10, p = 0.50n = 10, p = 0.75Observe: Skewed distribution unless p = 0.5µX= 1.0, σX= 0.95 µX= 5.0, σX= 1.58 µX= 7.5, σX= 1.37Example: Color blindness8% of white male population is colorblindSample n = 140 white males, p = P(“color blind”) = 0.08What are µXand σX?µ= n p = (140)(0.08) = 11.2µX= n p = (140)(0.08) = 11.2σX= √{n p (1 – p)} = √{(140)(0.08)(0.92)} = 3.21Options for finding binomial probabilities Use the formula: Use Excel:=binomdist(k, n, p, 0) for P(X = k)=binomdist(k, n, p, 1) for P(X ≤ k) Use a Normal approximationRule of thumb: valid if np ≥ 10 and n(1 – p) ≥ 10Example: Color blindness8% of white male population is colorblindSample n = 140 white males, p = P(“color blind”) = 0.08What is P(X ≤ 5)?=binomdist(5, 140, 0.08, 1) = 0.0284 =binomdist(5, 140, 0.08, 1) = 0.0284 Note: np = (140)(0.08) = 11.2 ≥ 10n(1 – p) = (140)(0.92) = 128.8 ≥