Stat 321 Lecture 8 Law of Total Probability

Stat 321 – Lecture 8 Law of Total Probability (2. 4)

Last Time - Conditional Probability n n P(A|B) = probability of event A occurring given that event B has occurred To calculate: q q n restrict sample space to outcomes in B (if equally likely) P(A|B) = P(A B)/P(B) Properties q q q P(A|B) vs. P(B|A) P(A'|B) = 1 -P(A|B) Multiplication rule: P(A B)=P(B)P(A|B)=P(A)P(B|A)

Last Time - Independence n Events A and B are independent if q q n A Multiplication Rule for independent events q n P(A|B) = P(A) P(B|A) = P(B) P(A B) = P(A|B)P(B) = P(A)P(B) = P(B|A)P(A) = P(A)P(B) To “check” independence q See if P(A|B) = P(A) or if P(B|A) = P(B) or if P(A B) = P(A)P(B) B

Example 3: ELISA Test n Given q q q n If has AIDS, test says AIDS 97. 7% of time If doesn’t have AIDS, test says no AIDS 92. 6% of time About. 5% of people have AIDS Given that someone has tested positive, what is the probability they have AIDS?

Table Solution Positive test Carries AIDS virus Does not carry AIDS Total 4885 Negative test Total 115 5000 73630 921370 995, 000 78515 921485 1, 000 Of those with positive test, proportion with AIDS: 4885/78515 =. 0622 Probability is only. 062 that a person who tests positive actually has AIDS! Most physicians think the probability is about 60 -70% - confusion of the inverse P(AIDS | positive test) ≠ P(positive test | AIDS)

Example 1: Governator Votes n CNN. com q q q 52% of the white votes 17% of the black votes 31% of the Hispanic votes 37% of the votes from other races. So overall (. 52+. 17+. 31+. 37)/4 =. 3425 => 34% of votes?

Example 1: Governator Votes n Given: q . 52 = P(A|W). 17 = P(A|B). 31 = P(A|H). 37 = P(A|O) q . 70 = P(W)… q q q

Example 1: Governator Votes White Arnold . 52(700)=364 P(A W) =. 364 A' Total Black . 0102 Hisp. . 0558 Other . 0222 Total . 4522 =P(A) P(A|W)P(W) + P(A|B)P(B) + P(A|H)P(H) + P(A|O)P(O) = P(A) . 70 . 06 . 18 . 06 700 60 180 60 1. 00 1000

Law of Total Probability n P(B)=SP(B|Ai)P(Ai) where {Ai}=partition A B P(A|B)P(B) Arnold 3 A H 2 P(A|H)P(H) P(AA |W)P(W) W 4 A W P(A|W)P(W) 1. 70 White . 18 . 06 H B O

Law of Total Probability. 52 A|W P(A W)=(. 52)(. 70)=. 364 Not A|W P(A' W)=(. 48)(. 70)=. 336. 70 W . 18. 06 H B . 31 A|H Not A|H . 17 . 06 A|B P(A B)=(. 17)(. 06)=. 0102 Not A|B O. 37 A|O Not A|O P(A)? P(A H)=(. 31)(. 18)=. 0558 . 364+. 0558+. 0102+. 0222 =. 4522 P(A O)=(. 37)(. 06)=. 0222

General Result n Law of Total Probability P(B)=P(B|A 1)P(A 1)+P(B|A 2)P(A 2)+… = (. 7)(. 52)+(. 18)(. 31)+(. 17)(. 06)+. 37(. 06)

Example 2: Randomized Response n n n Technique for asking sensitive questions Randomly decide which question respondents will answer: sensitive or boring Work backwards with probability rules to estimate proportions for sensitive question

Example 2: Randomized Response n Flip fair coin q q n n Heads: answer sensitive question Tails: answer boring question=“does your home phone number end in even digit? ” Determine proportion of “yeses” Define events q q Y=“response is yes” S=“respondent answered sensitive question”

Example 2: Randomized Response n n Respondents are ensured confidentiality Can still obtain estimate for P(Y|S)

For Friday n n Lab 3 due Finish reading Ch. 2 (Section 2. 4!) Will not begin a new lab but have a regular class session (in lab) Review handout available Friday