1. B1B1 B 2 (= B 1 c ) A AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities

2. B1B1 B 2 (= B 1 c ) A AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

3. B1B1 B 2 (= B 1 c ) A(.2)(.9)(.6)(.1) AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

4. B1B1 B 2 (= B 1 c ) A.18.06 AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

5. B1B1 B 2 (= B 1 c ) A.18.06.24 AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities P(A) “LAW OF TOTAL PROBABILITY” posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

6. B1B1 B 2 (= B 1 c ) A.18.06.24 AcAc.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 In the highlands… B 1 decreases from a prior of 90% to a posterior of 75% B 2 increases from a prior of 10% to a posterior of 25%

7. B1B1 B 2 (= B 1 c ) A.18.06.24 AcAc.72.04.76.90.10 1 Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 In the lowlands… B 1 increases from a prior of 90% to a posterior of 94.7% B 2 decreases from a prior of 10% to a posterior of 5.3%

8. Exercise: ??? Species B 1 – with a prior prob of ??? – ?????? ??? to a posterior of ??? in the highlands ?????? ??? to a posterior of ??? in the lowlands. ??? Species B 2 – with a prior prob of ??? – ?????? ??? to a posterior of ??? in the highlands ?????? ??? to a posterior of ??? in the lowlands. Example: Two similar bat species, B 1 and B 2, occupy both highland (A) and lowland (A c ) areas. Species B 1 makes up 90% of the population; species B 2, 10%. 20% of spec B 1 live in the highlands, 80% in the lowlands. 60% of spec B 2 live in the highlands, 40% in the lowlands. What is the probability that a randomly caught bat belongs to each species, if it is caught in the highlands? In the lowlands? Bayes’ Law (full form, n = 2) P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4 prior probabilities posterior probabilities P(B 1 ) =.9, P(B 2 ) =.1 P(A|B 1 ) =.2, P(A c |B 1 ) =.8 P(A|B 2 ) =.6, P(A c |B 2 ) =.4

9. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. “10% of pop is B 1 -deficient (only), 20% is B 2 -deficient (only), and 30% is B 3 -deficient (only). The remaining 40% is not B-deficient.” Given:

10. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 A = Alcoholic A c = Not Alcoholic Given: P(A ∩ B 1 ) To find these intersection probabilities, we need more information! Prior probs 1.00 P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 )

11. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 Also given… “Alcoholics comprise 35%, 30%, 25%, and 20% of the B 1, B 2, B 3, B 4 groups, respectively.” Given: A = Alcoholic A c = Not Alcoholic Prior probs 1.00 P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 )

12. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 Also given… P(A | B 1 ) =.35P(A | B 2 ) =.30P(A | B 3 ) =.25P(A | B 4 ) =.20 Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 )

13. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. Also given… Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic P(A ∩ B 1 )P(A ∩ B 2 )P(A ∩ B 3 ) P(A ∩ B 4 ) P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 ) P(A | B 1 ) =.35P(A | B 2 ) =.30P(A | B 3 ) =.25P(A | B 4 ) =.20 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 P(A ∩ B) = P(A | B) P(B) Recall:

14. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. Also given… Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic P(A c ∩ B 1 ) P(A c ∩ B 2 )P(A c ∩ B 3 )P(A c ∩ B 4 ) P(A ∩ B) = P(A | B) P(B) Recall: P(A | B 1 ) =.35P(A | B 2 ) =.30P(A | B 3 ) =.25P(A | B 4 ) =.20.10 .35.20 .30.30 .25.40 .20.035.060.075.080 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

15. Example: Vitamin B-complex deficiency among general population B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. Prior probs 1.00 Given: A = Alcoholic A c = Not Alcoholic.10 .35.20 .30.30 .25.40 .20.035.060.075.080.065.140.225.320 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 P(B 1 | A) = ?P(B 2 | A) = ?P(B 3 | A) = ? P(B 4 | A) = ? Posterior probabilities

16. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40 P(B 1 | A) = ?P(B 2 | A) = ?P(B 3 | A) = ? P(B 4 | A) = ?.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75.035 1.00 P(B 1 ∩ A) P(A) Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities

17. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 1 | A) =P(B 2 | A) =P(B 3 | A) = P(B 4 | A) =.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75.035.060.075.080 Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities 1.00 P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

18. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 1 | A) =.14P(B 2 | A) =.24P(B 3 | A) =.30 P(B 4 | A) =.32.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75 1.00 INCREASE DECREASE NO CHANGE; A and B 3 are independent! Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

19. B1B1 Thiamine B2B2 Riboflavin B3B3 Niacin B4B4 No B deficiency Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. P(B 1 | A c ) = ??P(B 2 | A c ) = ??P(B 3 | A c ) = ?? P(B 4 | A c ) = ??.035.060.075.080.065.140.225.320 P(A) =.25 P(A c ) =.75 1.00 Exercise: Prior probs Given: A = Alcoholic A c = Not Alcoholic Posterior probabilities P(B 2 ) =.20P(B 3 ) =.30P(B 1 ) =.10 P(B 4 ) =.40

20. Example: Vitamin B-complex deficiency among general population Assume B 1, B 2, B 3, B 4 “partition” the population, i.e., they are disjoint and exhaustive. A (Yes) A c (No) Alcoholic etc. Non- deficient Thiamine- deficient Riboflavin- deficient Niacin- deficient C1C1 C2C2 C5C5 C6C6 C4C4 C3C3 C7C7 C8C8 C 1 C 2 C 3 C 4 C 5 C 6 C 7 C 8