1. Conditional Probability and Independent Events

2. Conditional Probability if we have some information about the result…use it to adjust the probability likelihood an event E occurs under the condition that some event F occurs notation: P(E | F ) "the probability of E, given F ". called a “conditional probability”

3. Given They’re Male If an individual is selected at random, what is the probability a sedan owner is selected, given that the owner is male? P( sedan owner | male ) = _______?

4. Smaller Sample Space Given the owner is male reduces the total possible outcomes to 115.

5. In general... In terms of the probabilities, we define sedan mini-van truck totals male.16.10.20.46 female.24.22.08.54.40.32.28 1.00 P( sedan owner | male ) = _______?

6. Compute the probability sedan mini-van truck totals male.16.10.20.46 female.24.22.08.54.40.32.28 1.00

7. Compare NOT conditional: P( truck ) = Are Conditional: P( truck | male ) =

8. Dependent Events? probability of owning a truck…...was affected by the knowledge the owner is male events "owns a truck" and "owner is male" are called dependent events.

9. Independent Events Two events E and F, are called independent if or simply the probability of E is unaffected by event F

10. Roll the Dice Using the elements of the sample space: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Compute the conditional probability: P( sum = 6 | a “4 was rolled” ) = ? are the events “sum = 6" and “a 4 was rolled" independent events?

11. “Affected” The events are NOT independent the given condition does have an effect. That is, P(sum = 6 | 4 is rolled ) = 2/11 = 0.1818 but P(sum = 6) = 5/36 = 0.1389 These are dependent events.

12. Not Independent These are dependent events. As a result, P(sum = 6 and a 4 was rolled) does not equal P(sum = 6)P(a 4 was rolled) ?

13. Probability of “A and B” Draw two cards in succession, without replacing the first card. P(drawing two spades) = ________? may be written equivalently as

14. Multiplication Rule P(1 st card is spade) P(2 nd is spade | 1 st is spade) (spade, spade) Compare with “combinations approach”, ( 13 C 2 )( 52 C 2 ).

15. Multiplicative Law for Probability For two events A and B, But when A and B are independent events, this identity simplifies to

16. Example In a factory, 40% of items produced come from Line 1 and others from Line 2. Line 1 has a defect rate of 8%. Line 2 has a defect rate of 10%. For randomly selected item, find probability the item is not defective. D: the selected item is defective (i.e., ~D means not defective)

17. The Decision Tree Line 1 Line 2 defective not defective

18. The Two Lines ~D: the selected item is not defective. S L1L1 L2L2 ~D

19. Total Probability S B1B1 B2B2 BkBk … A

20. B1B1 B2B2 B3B3 P(A|B1)P(B1)P(A|B1)P(B1) P(A|B2)P(B2)P(A|B2)P(B2) P(A|B3)P(B3)P(A|B3)P(B3)

21. Bayes’ Theorem follows…

22. Bayes’ B1B1 B2B2 B3B3 P(A|B1)P(B1)P(A|B1)P(B1) P(A|B2)P(B2)P(A|B2)P(B2) P(A|B3)P(B3)P(A|B3)P(B3)

23. Back at the factory… For randomly selected item, find probability it came from Line 1, given the item is not defective. P( L 1 | W ) = Line 1 Line 2 defective not defective

24. The 3 Urns Three urns contain colored balls. UrnRedWhite Blue 1 3 4 1 2 1 2 3 3 4 3 2 An urn is selected at random and one ball is randomly selected from the urn. Given that the ball is red, what is the probability it came from urn #2 ?