Dependent Events

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(2nd card is K | 1st card is K ) = 3/51 = 1/17 P(2nd card is K | 1st 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 5/10 = 1/2 2/10 = 0.2 P( A and B) = (0.2)/(0.5) = 0.4 P(A) = 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} =1/3 P(A|B) = 0 P(A|C) P(D|C) = 2/3

General events In a sample space S have: P(A) = 1/3 , P(B)= ¼ , P(A and B) =1/6

Seenext slide

P(B) = P(Ac and B) + P( A and B) Sincethey are mutuallyexclusive S A B AC and B A Then P(Ac and B) = P(B) – P(A and B)

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

200 people are interviewed on their breakfast preference Joint distribution P(Male) = 93/200 = 0.465 = 46.5% P(Male AND Cereal) = 47/200 = 0.235 = 23.5%

Joint and Marginaldistributions P(Cereal) = 45% Shorternotation for AND P(Female, Cereal) = 21.5%

Conditional Probabilities P(BB|Female) = P(Female, BB) / P(Female)

Conditioning on the choice

General AND Rule

General AND Rule • New Multiplication Rule • Also • P(Drawing 2 K’s) = (4/52)(3/51) = 0.0045

Drawingcards Draw two cards (in sequence) from a deck of 52. 1. Determine the probability that the first card is hearts (A1) and the second is a flower (A2) We look for P(A1 and A2) P(A1) = 13/52, P(A2 |A1) = 13/51, P(A1and A2) = P(A2 |A1) P(A1) = 132/(52∙51)

Determine the probability that the second is a flower (A2) Seenext slide

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

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 ?

Consider the events: E = pass the exam, A = go to party Whatweknow: P( E | AC)= 0.99, P( E | A) = 0.50, P(A) = P(AC)= 0.5 Look for

Note that P(E) = P(E and A) + P(E and AC) = P(E|A)∙P(A) + P(E| AC)∙P(AC) 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 + ) = ???

An event and itscomplement 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 +)

P(Test + AND Pat doesn’t use) =P( Pat doesn’t use ) P(Test + | Pat doesn’t use) = 0.99(0.01) = 0.0099 • P(Test + AND Pat does use) = =P( Pat does use ) P(Test + | Pat does use) =(0.01)(0.99) = 0.0099 Finally • P( Pat does use | Test +) = 0.0099/0.0198 = 0.5 P(Test + )= P(Test + AND Pat does use) + P(Test + AND Pat doesn’t use) = 0.0099 + 0.0099 = 0.0198

Patuses Patdoesn’t use Test + S

Dependent Events