111, section 7.5 Conditional Probability notes by Tim Pilachowski In conditional probability an outcome or event B is dependent upon another outcome or event A. Formally, P(B | A) = P(B given A) = probability that B will happen given the prior condition that A has already happened. 7.1, 7.2 & 7.4 Example D revisited: Suppose that a box contains 3 blue blocks and 2 yellow blocks. If we picked blocks and then put them back (“with replacement”) the probability of picking a blue block would never change, it would always be the same. There would always be 3 blue blocks in the box, no matter what had been picked previously. ( ) ( ) ( )53outcomes possible ofnumber happencan waysofnumber 321====iBBPBPBP The picks are independent, like two coins tossed or two dice rolled. The tree diagram for randomly picking three blocks without replacement, would look like this: B2 B2 B1 Y2 Y2 Y1 Y3 B3 Y3 B3 B3 Y3 B3 Experiment: Pick three blocks without replacement. Let B1 = blue on the first pick, B2 = blue on the second pick, B3 = blue on the third pick, Y1 = yellow on the first pick, Y2 = yellow on the second pick, Y3 = yellow on the third pick. You can fill in the probabilities on the tree diagram above as we calculate them in class. Those marked with * below will be left for you to find on your own. P(picking blue second given a blue was picked first) =()=12| BBP ()=12| BYP ()=12| YBP ()=12| YYP *P(picking blue third given a blue was picked first and second) =()=∩213| BBBP ()=∩213| BBYP ()=∩213| YBBP *()=∩213| YBYP ()=∩213| BYBP *()=∩213| BYYP *()=∩213| YYBP *()=∩213| YYYPIn an “intersection”, i.e. an “and” situation, moving left-to-right on the tree diagram, multiply probabilities. P(blue first and blue second and yellow third) = *()()()()213121321|| BBBPBBPBPBBBP ∩∗∗=∩∩ = ()=∩∩321YBBP ()=∩∩321BYBP *()=∩∩321YYBP ()=∩∩321BBYP *()=∩∩321YBYP *()=∩∩321BYYP In a “union”, i.e. an “or” situation, moving up-and-down on the tree diagram, add probabilities. *P(3 blue blocks) = ()=∩∩321BBBP P(2 blue blocks and 1 yellow block) = *P(1 blue block and 2 yellow blocks) = *P(3 yellow blocks) = P(S) = observations about Lecture 7.4 Example D compared to Lecture 7.5 Example D: For Example D, we were easily able to count how many blocks were left. In other situations, we won’t be able to count so easily. So we have a formal definition and formula. Earlier, we used the multiplication principle. ()()()FEPEFPEP ∩=∗ | With algebraic manipulation, we get a formula for conditional probability. ( )()( )EPFEPEFP∩=| ( )( )( )()( )( )( )( )( )FFEFPFEPFEPSFSFEin events ofnumber in events ofnumber |in events ofnumber in events ofnumber in events ofnumber in events ofnumber ∩==∩=∩ In words, when we’re considering only event F, the conditional probability is that portion/fraction that also includes event E ?Example A: Given two events C and D in a sample space S, if we know that P(C) = 0.3 and P(D | C) = 0.2, then what is ()DCP ∩ ? Example B: Given two events E and F in a sample space S, if we know that ( ) ( )31,43== FPEP and ( ),41=∩ FEP then a) what is P(E | F)? b) what is P(F | E)? 7.1, 7.2 & 7.4 Example D revisited: Suppose that a box contains 3 blue blocks and 2 yellow blocks. ()=12| BBP ()=12| YBP Notice the mathematical relationship. Formally, if two events E and F are independent then (1) P(E | F) = P(E) and (2) P(F | E) = P(F). Using the multiplication principle, ()()().| EFPEPFEP ∗=∩ Then, if the two events are independent, (3) ()()()FPEPFEP ∗=∩ . Any one of these three formulas is sufficient to prove or disprove independence of events. Example A revisited: Given two events C and D in a sample space S, if we know that P(C) = 0.3, P(D) = 0.4, and ()64.0=∪ DCP , are events C and D independent? Example B revisited: Given two events E and F in a sample space S, if we know that ( ) ( )31,43== FPEP and ( ),41=∩ FEP then are events E and F independent?Example C: For two events G and H in a sample space S, we know that they are independent and that P(G) = 0.7 and P(H) = 0.4. ()=∩ HGP ()=HGP | ()=cGHP | ()=ccGHP | Example E: You toss two standard six-sided dice. S = { (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6) } A = at least one of the dice is a 4 P(A) = B = the sum of the two dice is 9 P(B) = =∩BA ()=∩ BAP A | B = P (A | B) = B | A = P (B | A) = Are events A and B independent? How do you know?Example F-1: You pick a card from a standard deck of 52. n(S) = 52. S = { A-S, A-H, A-D, A-C, K-S, K-H, K-D, K-C, Q-S, Q-H, Q-D, Q-C, … , 2-S, 2-H, 2-D, 2-C }. C = the card is an Ace P(C) = D = the card is a Spade P(D) = =∩DC ()=∩ DCP C | D = P (C | D) = D | C = P (D | C) = Are events C and D independent? How do you know? Example F-2: You pick two cards from a standard deck of 52. n(S) = E = at least one card is an Ace P(E) = Ec = P(Ec) = F = picking a pair P(F) = =∩FE ()=∩ FEP Are events E and F independent? How do you know?Example F-2 revisited: You pick two cards from a standard deck of 52. n(S) = C(52, 2) E = at least one card is an Ace, F = picking a pair, =∩FE picking a pair of Aces P(Ace on first pick) = P(not-Ace on first pick) = P(Ace | Ace) = P(not-Ace | Ace) = P(Ace | not-Ace) = P (not-Ace | not-Ace) = P(two Aces) = ()=∩ FEP P(Ace then not-Ace) = P(not-Ace then Ace) = P (neither card is an Ace) = P (E) = P (F) = Example G: Silver Springs, Florida, has a snack bar and a gift shop. The management observes 100 visitors, and counts 55 who make a purchase in the gift shop (event G), 65 who eat in the snack bar (event H), and 40 who do both. G | H = P (G | H) = interpretation of G | H: H | G = P (H | G) = interpretation of H | G: A ce not Ace Ac e A ce not A ce not A ceExample H: At Matriarch University (U. Mama), 35% of the students have an academic scholarships. Of the students who have …