Dependent Events Conditional Probability General rule for “AND” events

Conditional Probability Roll a die What is the probability of a 6? What is the probability of a 6 if I know, before looking at the result, that an even number showed up?

Dependence A says something about B P(B|A) = conditional probability of B given A P(2 nd card is K | 1 st card is K ) = 3/51 = 1/17 P(2 nd card is K | 1 st card is Q ) = 4/51 If A and B are independent, P(B|A) = P(B)

Conditions Change the Sample Space Probability of Event B given Event A TexPoint Display

B given A B A ∩ B A

A given B A A ∩ B B

Choose a number between 1 and 10 A={your number is even} B={your number is ≤ 5} P(A) = P(B) = P(A and B) = P(A|B) = 5/10 = 1/2 2/10 = 0.2 P( A and B) P(A) = (0.2)/(0.5) = 0.4 = 2/5

Cast a die S = {1,2,3,4,5,6} A={6}, B={2,4,6}, C={1,3,5}, D={1,3} P(A|B)=1/3 P(A|C)= 0 P(D|C) = 2/3

General events

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S A A C and B P(A c and B) = P(B) – P(A and B) P(B) = P(A c and B) + P( A and B) Since they are mutually exclusive Then A B

Two children S = { (B,B); (B,G); (G,B); (G,G)} Assuming babies’ sexes are equally probable and independent. P( 2 nd is Girl) = ½ P( 2 nd is Girl | 1 st is a Girl) = ½ P(2 nd is Girl | one is a Girl)=?? P(one is a Girl)= P(2 nd is Girl and one is a Girl)= 3/4 1/2 2/3

200 people are interviewed on their breakfast preference CerealBreakfast Bar Oat Meal Male Female P(Male) = 93/200 = = 46.5% P(Male AND Cereal) = 47/200 = = 23.5% Joint distribution

Joint and Marginal distributions CerealBBOM Male23.5%21%2%46.5% Female21.5%30%2%53.5% 45%51%4% P(Cereal) = 45% P(Female, Cereal) = 21.5% Shorter notation for AND

Conditional Probabilities CerealBB OM Male23.5%21%2%46.5% Female21.5%30%2%53.5% 45%51%4% 40.2%56.1%3.7% 50.5%41.2%4.3% P(BB|Female) = P(Female, BB) / P(Female)

Conditioning on the choice CerealBBOM Male23.5%21%2%46.5% Female21.5%30%2%53.5% 45%51%4% 52.2% 47.8% 41.2% 58.8% 50%

General AND Rule

New Multiplication Rule Also P(Drawing 2 K’s) = (4/52)(3/51) =

Drawing cards Draw two cards (in sequence) from a deck of Determine the probability that the first card is hearts (A1) and the second is a flower (A2) We look for P(A 1 and A 2 ) P(A 1 ) = 13/52, P(A 2 |A 1 ) = 13/51, P(A 1 and A 2 ) = P(A 2 |A 1 ) P(A 1 ) = 13 2 /(52∙51)

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A1A1 A2A2

Sampling without Replacement Jar has 6 black olives and 6 green olives. You pull out 3 P(1 st is black) = 6/12 = 0.5 P(2 nd is black | 1 st is black) = 5/11 = 0.45 P(3 rd is black | 1 st two were black) = 4/10=0.4 P(1 st two are black) = 0.5(.45) =.227

Independence A and B are independent P(A | B) = P(A) P(A | B c ) = P(A) P(B | A) = P(B) P(B | A c ) = P(B) P(A and B) = P(A) P(B |A) = P(A) P(B)

P(A |B) ≠ P(B|A) Think: P(A |B) = probability of A (with information about B) P(2 nd is Girl | one is a Girl)= 2/3 P(one is a Girl | 2 nd is a Girl) = 1

Example A student has to sit an exam. The night before the exam: if he studies passes with probability 99% if he goes to a dance party his chance of promotion is reduced to 50%. He decides to go to the party if heads tossing a fair coin. The day after he passes the exam. What is the probability that he went to dance ?

Note that P(E) = P(E and A) + P(E and A C ) = P(E|A)∙P(A) + P(E| A C )∙P(A C ) From which

Drug Test 1% of students use cocaine. Drug test is 99% accurate. Pat tests positive for cocaine. –What is the probability that Pat uses cocaine? P( Test + | Pat uses) =.99 P( Pat uses | Test + ) = ??? ≠

Available information P(Pat uses) = 0.01 P( Pat doesn’t use ) = 0.99 P(Test + | Pat doesn’t use ) = 0.01 P(Test + | Pat does use) = 0.99 Looking for P( Pat does use | Test +) P(Test + AND Pat doesn’t use) = P(Test +) An event and its complement

P(Test + AND Pat doesn’t use) = P( Pat doesn’t use ) P(Test + | Pat doesn’t use) = 0.99(0.01) = P(Test + AND Pat does use) = = P( Pat does use ) P(Test + | Pat does use) =(0.01)(0.99) = Finally P( Pat does use | Test +) = / = 0.5 P(Test + )= P(Test + AND Pat does use) + P(Test + AND Pat doesn’t use) = =

Pat doesn’t use S Test + Pat uses