Conditional Probability and Independent Events

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”

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 ) = _______?

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

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

Compute the probability sedan mini-van truck totals male female

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

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.

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

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?

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

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) ?

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

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 ).

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

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)

The Decision Tree Line 1 Line 2 defective not defective

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

Total Probability S B1B1 B2B2 BkBk … A

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)

Bayes’ Theorem follows…

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)

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

The 3 Urns Three urns contain colored balls. UrnRedWhite Blue 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 ?