Conditional Probability and Independent Events

1. Conditional Probabilityand Independent Events

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

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. • That's 40 out of 115.

5. For conditional probability, we define In general... • That is,

6. sedan mini-van truck totals male .16 .10 .20 .46 female .24 .22 .08 .54 .40 .32 .28 1.00 In general... • In terms of the probabilities, we define • P( sedan owner | male ) = _______?

7. sedan mini-van truck totals male .16 .10 .20 .46 female .24 .22 .08 .54 .40 .32 .28 1.00 Compute the probability

8. Compare • NOT conditional: • Are Conditional:

9. 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 calleddependent events.

10. Independent Events • Two events E and F , are calledindependentifor simply the probability of E is unaffected by event F

11. Check Independence • a single card is drawn from a deck... • are the events "a spade is drawn" and "an ace is drawn" independent events? • Check if P( spade and ace ) equalsP(spade)P(ace) ? "drawing a spade doesn't affect the probability that an ace was drawn, an vice versa"

12. “Unaffected” • These events are independent • the given condition had no effect. • That is, P(ace| spade ) = 1/13 = 4/52 = P(ace) • And similarly,P(spade| ace ) = 1/4 = 13/52 = P(spade). • equality is the result of the events being independent

13. 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?

14. Not Independent • Does P(sum = 6 and a 4 was rolled) equalP(sum = 6)P(a 4 was rolled) ? • Equivalently, P(sum = 6| 4 is rolled ) = 2/11 = 0.1818P(sum = 6) = 5/36 = 0.1389 • These are dependent events.

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

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

17. Multiplication Rule P(2nd is spade | 1st is spade) (spade, spade) P(1st card is spade)

18. Multiplicative Law for Probability • For two events A and B, • And when A and B are independent events, this identity simplifies to

19. Additive and Multiplicative Laws and if events A and B are mutually exclusive events, this simplifies to and if events A and B are independent events, this simplifies to

20. Additive law extended …

21. Multiplicative law extended …