1. 1 Probability Scott Matthews Courses: 12-706 / 19-702/ 73-359 Lecture 14 - 10/17/2005

2. 12-706 and 73-3592 Admin Issues  PS 4, Project 1 due Wednesday  Office hours tomorrow  Final Project Ideas (Brad& Don from FMS)  Lecture

3. 12-706 and 73-3593 Probability  Only reviewing the more advanced concepts (and what we’ll need in course)  Basic concepts: between 0 and 1, additive, total probability must be 1, Venn diagrams, etc.

4. 12-706 and 73-3594 Conditional Probability  “Probability (P) that A occurs conditional on B occurring”  Also referred to as “P of A given B”  Joint Probability: P(A and B) Cool MS equation

5. 5 HIV Test Example: Table of Actual Condition, Test Results HIV-HIV+Total Test Result Positive3.960.94.86 Negative95.040.195.14 Total991100

6. 12-706 and 73-3596 Conditional Probabilities  False Positive Test: P(HIV-|pos)  = 3.96/4.86 =.815  False Negative): P(HIV+|neg)  = 0.1/95.14 =.001

7. 12-706 and 73-3597 Total Probability  Probability of an event occuring alone is combination of all possible joint outcomes with another event  Given n mutually exclusive events (A 1..A n ) whose probabilities sum to 1:

8. 12-706 and 73-3598 Total Probability zSuppose the lightbulbs you can buy at store are manufactured by three factories. What is the total probability that a lightbulb sold at the store is defective? yFactory One produces 60% of the light bulbs sold yFactory Two produces 30% of the light bulbs sold yDefective bulb probabilities: 0.01 for Factory One, 0.02 for products of Factory Two, and 0.05 for products of Factory Three.

9. 12-706 and 73-3599 Answer  P(defect) = P(defect|factory1)*P(factory1) + P(Defect|factory 2)*P(factory 2) +P(defect|factory 3)*P(factory 3)  =0.01*0.6 + 0.02*0.3 + 0.05 *0.1 =.006+.006+.005 =.017

10. 12-706 and 73-35910 Bayes’ Theorem  “Opposite” of old conditional equation is:  But P(A and B) must equal P(B and A)..  So P(B|A)*P(A) = P(A|B)*P(B), thus  Using total probability..  Way of finding P(B|A) knowing only P(A|B)

11. 12-706 and 73-35911 Bayes Example z+ event that drug test is positive for person z- event that drug test is negative for person zA event that person tested uses drug tested for  Assume P[A] .1, P[  |A] .98, P[  |Abar] .1  Bayes Theorem:

12. 12-706 and 73-35912 Discrete Distributions  Values can only take on a set of countable values  Probability mass function (pmf) is map of probabilities of each possible outcome  Aka a histogram.  Cumulative distribution function (cdf) is P(X <= x)

13. 12-706 and 73-35913 Discrete Dist’ns (cont.)  Should look familiar - recall lecture on risk profiles. Those were pmf’s, cdf’s.  http://www.weibull.com/hotwire/issue12/rel basics12.htm

14. 12-706 and 73-35914 Continuous distributions  Analogous to pmf/cdf for discrete case  Except pmf=> probability density function (pdf)

15. 12-706 and 73-35915 Reading pdf/cdf graphs  What information can we see from just looking at a randomly selected pdf or cdf?

16. 12-706 and 73-35916 Subjective Probabilities