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Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 c • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(A |B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? B1 B2 (= B1 P ( A | B1 ) P (B1 ) P (B1 | A) = P ( A | B1 ) P (B1 ) + P ( A | B2 ) P (B2 ) c) A P ( A | B2 ) P (B2 ) P (B2 | A) = P ( A | B1 ) P (B1 ) + P ( A | B2 ) P (B2 ) Ac .90 .10 1 P(A) “LAW OF TOTAL PROBABILITY” Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? B1 B2 (= B1 P ( A | B1 ) P (B1 ) P (B1 | A) = P ( A | B1 ) P (B1 ) + P ( A | B2 ) P (B2 ) c) A P ( A | B2 ) P (B2 ) P (B2 | A) = P ( A | B1 ) P (B1 ) + P ( A | B2 ) P (B2 ) Ac .90 .10 1 P(A) “LAW OF TOTAL PROBABILITY” Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? B1 A B2 (= B1 (.2)(.9) P (B1 | A) = (.2)(.9) + (.6)(.1) c) (.2)(.9) (.6)(.1) .90 .10 (.6)(.1) P (B2 | A) = (.2)(.9) + (.6)(.1) Ac 1 P(A) “LAW OF TOTAL PROBABILITY” Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? B1 A B2 (= B1 .18 .06 .90 .10 (.2)(.9) .18 P (B1 | A) = = (.2)(.9) + (.6)(.1) .18 +.06 c) (.6)(.1) .06 P (B2 | A) = = (.2)(.9) + (.6)(.1) .18 +.06 Ac 1 P(A) “LAW OF TOTAL PROBABILITY” Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? B1 A B2 (= B1 (.2)(.9) .18 P (B1 | A) = = (.2)(.9) + (.6)(.1) .24 c) .18 .06 .24 (.6)(.1) .06 P (B2 | A) = = (.2)(.9) + (.6)(.1) .24 .90 .10 1 P(A) “LAW OF TOTAL PROBABILITY” Ac Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? B1 A .18 B2 (= B1 .06 Ac .90 .10 (.2)(.9) .18 P (B1 | A) = = = .75 (.2)(.9) + (.6)(.1) .24 c) .24 (.6)(.1) .06 P (B2 | A) = = = .25 (.2)(.9) + (.6)(.1) .24 In the highlands… •1B1 decreases from a prior of 90% to a posterior of 75% • B2 increases from a prior of 10% to a posterior of 25% Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is posterior probabilities caught in the highlands? In the lowlands? A B1 B2 (= B1c) .18 .06 .72 P (B1 | A ) = = .947 .76 c .24 .04 P (B2 | A ) = = .053 .76 c Ac .72 .04 .90 .10 .76 In the lowlands… •1B1 increases from a prior of 90% to a posterior of 94.7% • B2 decreases from a prior of 10% to a posterior of 5.3% Bayes’ Law (full form, n = 2) P (B | A ) = P ( A | B ) P (B ) P ( A | B ) P ( B ) + P ( A | B c ) P (B c ) Example: Two similar bat species, B1 and B2, occupy both highland (A) and lowland (Ac) areas. prior probabilities • Species B1 makes up 90% of the population; species B2, 10%. P(B1) = .9, P(B2) = .1 • 20% of spec B1 live in the highlands, 80% in the lowlands. P(A|B1) = .2, P(Ac|B1) = .8 • 60% of spec B2 live in the highlands, 40% in the lowlands. P(A|B2) = .6, P(Ac|B2) = .4 What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? posterior probabilities Exercise: Species B1 – with a prior prob of ??? – • ??? to a posterior of ??? in the highlands • ??? to a posterior of ??? in the lowlands. Species B2 – with a prior prob of ??? – • ??? to a posterior of ??? in the highlands • ??? to a posterior of ??? in the lowlands. P(B1 | A) = .75 P(B1 | Ac ) = .947 P(B2 | A) = .25 P(B2 | Ac ) = .053 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. Thiamine Riboflavin B1 B2 Niacin B3 No B deficiency B4 “10% of pop is B1-deficient (only), 20% is B2-deficient (only), and 30% is B3-deficient (only). The remaining 40% is not B-deficient.” Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Thiamine Riboflavin Niacin B1 B2 P(A ∩ B1) P(A ∩ B2) P(A ∩ B3) P(A ∩ B4) P(Ac ∩ B1) P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4) P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 B3 No B deficiency B4 To find these intersection probabilities, we need more information! 1.00 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Also given… Thiamine Riboflavin Niacin B1 B2 P(A ∩ B1) P(A ∩ B2) P(A ∩ B3) P(A ∩ B4) P(Ac ∩ B1) P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4) P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 B3 No B deficiency B4 “Alcoholics comprise 35%, 30%, 25%, and 20% of the B1, B2, B3, B4 groups, respectively.” 1.00 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Also given… Thiamine Riboflavin B1 B2 P(A ∩ B1) P(A ∩ B2) P(A ∩ B3) P(A ∩ B4) P(Ac ∩ B1) P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4) P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 P(A | B3) = .25 P(A | B4) = .20 P(A | B1) = .35 P(A | B2) = .30 Niacin B3 No B deficiency B4 1.00 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Also given… Recall: Thiamine Riboflavin B1 B2 P(A ∩ B1) P(A ∩ B2) P(A ∩ B3) P(A ∩ B4) P(Ac ∩ B1) P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4) P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 P(A | B3) = .25 P(A | B4) = .20 P(A | B1) = .35 P(A | B2) = .30 Niacin No B deficiency B3 P(A ∩ B) = P(A | B) P(B) B4 1.00 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Also given… Recall: Thiamine Riboflavin B1 B2 .10.035 .35 .20.060 .30 .30.075 .25 .40.080 .20 P(Ac ∩ B1) P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4) P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 P(A | B3) = .25 P(A | B4) = .20 P(A | B1) = .35 P(A | B2) = .30 Niacin No B deficiency B3 P(A ∩ B) = P(A | B) P(B) B4 1.00 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Thiamine Riboflavin Niacin B1 B2 .10.035 .35 .20.060 .30 .30.075 .25 .40.080 .20 .065 .140 .225 .320 P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 B3 No B deficiency B4 1.00 Posterior probabilities P(B1 | A) = ? P(B2 | A) = ? P(B3 | A) = ? P(B4 | A) = ? Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Thiamine Riboflavin Niacin B1 B2 .035 .060 .075 .080 P(A) = .25 .065 .140 .225 .320 P(Ac) = .75 P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 1.00 B3 No B deficiency B4 Posterior probabilities P(B1 ∩ A) P(B1 | A) = ? P(B2 | A) = ? P(A) P(B3 | A) = ? P(B4 | A) = ? Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Thiamine Riboflavin Niacin B1 B2 .035 .060 .075 .080 P(A) = .25 .065 .140 .225 .320 P(Ac) = .75 P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 1.00 No B deficiency B3 B4 Posterior probabilities .035 P(B1 | A) = .25 .060 P(B2 | A) = .25 .075 P(B3 | A) = .25 P(B4 | A) = .080 .25 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Thiamine Riboflavin B1 B2 .035 .060 .075 .080 P(A) = .25 .065 .140 .225 .320 P(Ac) = .75 P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 1.00 INCREASE P(B1 | A) = .14 INCREASE P(B2 | A) = .24 Niacin No B deficiency B3 NO CHANGE; A and B3 are independent! P(B3 | A) = .30 B4 DECREASE Posterior probabilities P(B4 | A) = .32 Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. A= Alcoholic Ac = Not Alcoholic Prior probs Thiamine Riboflavin Niacin B1 B2 .035 .060 .075 .080 P(A) = .25 .065 .140 .225 .320 P(Ac) = .75 P(B1) = .10 P(B2) = .20 P(B3) = .30 P(B4) = .40 1.00 No B deficiency B3 Posterior probabilities Exercise: P(B1 | Ac) = ?? B4 P(B2 | Ac) = ?? P(B3 | Ac) = ?? P(B4 | Ac) = ?? Example: Vitamin B-complex deficiency among general population Assume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive. Thiaminedeficient C2 C5 C4 Alcoholic C2 C3 C6 C3 Nondeficient C7 Niacindeficient C1 C1 Riboflavindeficient C4 C5 C6 C7 C8 C8 A (Yes) etc. Ac (No)