Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28CHAPTER 3Probability Theory•3.1 - Basic Definitions and Properties•3.2 - Conditional Probability and Independence•3.3 - Bayes’ Formula•3.4 - Applications (biomedical)POPULATIONP(POPULATION) = 12A = “Lung Cancer”• P(A) corresponds to the ratio of the probability of A, relative to the entire population. A = lung cancer (sub-)populationB= “Smoker”B = smoking (sub-)populationA ∩ BProbability of lung cancerProbability of lung cancer and smokerInformal Description…• P(A ⋂ B) = the probability that both events occur simultaneously in the popul.That is,3A = “Lung Cancer”B= “Smoker”A ∩ B• P(A | B) corresponds to the ratio of the probability of A ∩ B,relative to the probability of B. A = lung cancer (sub-)population• P(A) corresponds to the ratio of the probability of A, relative to the entire population. B = smoking (sub-)populationProbability of lung cancerProbability of lung cancer, given smokerCONDITIONAL PROBABILITYInformal Description…I( ).( )P A BP BProbability of lung cancer and smoker• P(A ⋂ B) = the probability that both events occur simultaneously in the popul.Probability of “Primary Color,” given “Hot Color” = ? E ECF 0.30 0.15 0.45FC0.30 0.25 0.550.60 0.40 1.0Outcome ProbabilityRed 0.10Orange 0.15Yellow 0.20Green 0.25Blue 0.301.00POPULATIONE = “Primary Color” = {Red, Yellow, Blue}F = “Hot Color” = {Red, Orange, Yellow}15.00%20.00%25.00%30.00%10.00%P(E) = 0.60P(F) = 0.45Probability TableVenn Diagram0.150.300.250.30EFBlueGreenOrangeRedYellow0.150.300.250.30EFBlueGreenOrangeRedYellowProbability of “Primary Color,” given “Hot Color” = ? E ECF 0.30 0.15 0.45FC0.30 0.25 0.550.60 0.40 1.0E = “Primary Color” = {Red, Yellow, Blue}F = “Hot Color” = {Red, Orange, Yellow}P(E) = 0.60P(F) = 0.45Probability TableVenn Diagram0.150.300.250.30EFBlueGreenOrangeRedYellowOutcome ProbabilityRed 0.10Orange 0.15Yellow 0.20Green 0.25Blue 0.301.00POPULATION15.00%20.00%25.00%30.00%10.00%P(E | F)I( )( )P E FP F= =0.300.450.667Conditional Probability=P(F | E)=I( )( )P F EP EE ECF 0.30 0.15 0.45FC0.30 0.25 0.550.60 0.40 1.0E = “Primary Color” = {Red, Yellow, Blue}F = “Hot Color” = {Red, Orange, Yellow}P(E) = 0.60P(F) = 0.45Probability TableVenn Diagram0.150.300.250.30EFBlueGreenOrangeRedYellowProbability of “Primary Color,” given “Hot Color” = ?P(E | F)I( )( )P E FP F= =0.300.450.667Outcome ProbabilityRed 0.10Orange 0.15Yellow 0.20Green 0.25Blue 0.301.00POPULATION15.00%20.00%25.00%30.00%10.00%Conditional ProbabilityP(F | E)= =0.300.600.5=I( )( )P F EP E=E ECF 0.30 0.15 0.45FC0.30 0.25 0.550.60 0.40 1.0E = “Primary Color” = {Red, Yellow, Blue}F = “Hot Color” = {Red, Orange, Yellow}P(E) = 0.60P(F) = 0.45Probability TableVenn Diagram0.150.300.250.30EFBlueGreenOrangeRedYellowProbability of “Primary Color,” given “Hot Color” = ?P(E | F) P(EC | F)=I( )( )P E FP F= =0.300.450.667= 1 – 0.667 = 0.333Outcome ProbabilityRed 0.10Orange 0.15Yellow 0.20Green 0.25Blue 0.301.00POPULATION15.00%20.00%25.00%30.00%10.00%Conditional ProbabilityP(F | E)= =0.300.600.5=I( )( )P F EP EE ECF 0.30 0.15 0.45FC0.30 0.25 0.550.60 0.40 1.0E = “Primary Color” = {Red, Yellow, Blue}F = “Hot Color” = {Red, Orange, Yellow}P(E) = 0.60P(F) = 0.45Probability TableVenn Diagram0.150.300.250.30EFBlueGreenOrangeRedYellowProbability of “Primary Color,” given “Hot Color” = ?P(E | F) P(EC | F)P(E | FC)=I( )( )P E FP F= =0.300.450.667= 1 – 0.667 = 0.333Outcome ProbabilityRed 0.10Orange 0.15Yellow 0.20Green 0.25Blue 0.301.00POPULATION15.00%20.00%25.00%30.00%10.00%Conditional ProbabilityP(F | E)= =0.300.600.5=I( )( )P F EP E= =0.300.550.545RedYellow9Example:Women MenFractures 952 343 1295No Fractures1293 1417 27102245 1760 4005Women Fracturesn = 40059521293 343141710P(Fracture, given Woman) ≈P(Fracture, given Man) ≈ P(Man, given Fracture) ≈ P(Woman, given Fracture) =952 / 2245 = 0.424343 / 1760 = 0.195343 / 1295 = 0.2651 – 343 / 1295 = 952 / 1295 = 0.735 P(Fracture) ≈ 1295 / 4005 = 0.323 P(Fracture and Woman) ≈952 / 4005 = 0.238Def: The conditional probability of event A, given event B, is denoted by P(A|B), and calculated via the formula Def: The conditional probability of event A, given event B, is denoted by P(A|B), and calculated via the formula I( )( | ) = .( )P A BP A BP B11Thus, for any two events A and B, it follows that P(A ⋂ B) = P(A | B) × P(B). B occurs with prob P(B) Given that B occurs, A occurs with prob P(A | B)Both A and B occur, with prob P(A ⋂ B) Example: Randomly select two cards with replacement from a fair deck. P(Both Aces) = ? Example: P(Live to 75) × P(Live to 80 | Live to 75) = P(Live to 80) P(Ace1) = 4/52 P(Ace2 | Ace1) = 4/52 P(Ace1 ∩ Ace2) = (4/52)2 P(Ace2 | Ace1) = 3/51 P(Ace1 ∩ Ace2) = (4/52)(3/51) Exercises: P(Neither is an Ace) = ? P(Exactly one is an Ace) = ? P(At least one is an Ace) = ?Exercises: P(Neither is an Ace) = ? P(Exactly one is an Ace) = ? P(At least one is an Ace) = ?A B Example: Randomly select two cards without replacement from a fair deck. P(Both Aces) = ?12Def: The conditional probability of event A, given event B, is denoted by P(A|B), and calculated via the formula Def: The conditional probability of event A, given event B, is denoted by P(A|B), and calculated via the formula Thus, for any two events A and B, it follows that P(A ⋂ B) = P(A | B) × P(B). B occurs with prob P(B) Given that B occurs, A occurs with prob P(A | B)Both A and B occur, with prob P(A ⋂ B) Example: P(Live to 75) × P(Live to 80 | Live to 75) = P(Live to 80) Tree DiagramsP(B)P(Bc)P(A | B)P(Ac | B)P(A | Bc)P(Ac | Bc)P(A ⋂ B)P(Ac ⋂ B)P(A ⋂ Bc)P(Ac ⋂ Bc)Event A AcB P(A ⋂ B) P(Ac ⋂ B)BcP(A ⋂ Bc) P(Ac ⋂ Bc)A BA ⋂ BA ⋂ BcAc ⋂ BAc ⋂ BcMultiply together “branch probabilities” to obtain “intersection probabilities”A B .)()()|(BPBAPBAP13Example: Bob must take two trains to his home in Manhattan after work: the A and the B, in either order. At 5:00 PM…• The A train arrives first with probability 0.65, and takes 30 mins to reach its last stop at Times Square. • The B train arrives first with probability 0.35, and takes 30