Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
7.4 Basic Concepts of Probability • This presentation is copyright by Paul Hendrick • © 2003-2005, Paul Hendrick • All rights reserved 7.4 Basic Concepts of Probability • Union rule for probability – Union rule (general case – always true) P( E F ) P( E ) P( F ) P( E F ) – IF P(EF) = 0, it can be omitted! – Union rule for mutually exclusive events, only P( E F ) P( E ) P( F ) 7.4 Basic Concepts of Probability Complement rule – “back-door” approach • (because it doesn’t calculate directly) P( E ) P( S ) P( E ) P( E ) 1 P( E ) P( E ) P( S ) P( E ) P( E ) 1 P( E ) 7.4 Odds – Probability uses two numbers -- # ways for a successful outcome & # ways for any outcome • • • • • n(E) and n(S) P(E) = n(E) / n(S) Let S={1,2,3,4,5,6} be the sample space for rolling a single fair die. Let E={1,6} be the event of rolling an “extreme” number. n(E) = 2; n(S) = 6; so P(E) = 2/6 = 1/3. – There is a third number -- # ways for an unsuccessful outcome • n(E’) • n(E’) = n(S) - n(E) • For the above, n(E’) = 6-2 = 4, the number of ways of NOT rolling an “extreme” number. • Note: P(E’) = 4/6 = 2/3 or P(E’) = 1 – P(E) = 1-1/3 = 2/3. 7.4 Odds (cont) – odds uses a different two of the three numbers – odds in favor of an event E • = n(E) / n(E’) (assumes all outcomes are equally-likely) • Odds in favor of an “extreme” roll are 2 / 4 or 1/2 • = P(E) / P(E’) (uniform sample space NOT necessary for this formula) • Odds for “extreme” roll also by 1/3 / 2/3 = 1/2 – odds against an event E • = n(E’) / n(E) (numerator and denominator switched!) • Odds against an “extreme” roll are 4 / 2 or 2/1 – Again, note the 2 & 4 are reversed from the first example above. 7.4 Odds (cont) – Instead of as fractions, odds are commonly shown as ratios with a colon used to show comparison – n(E) : n(E’) instead of n(E) / n(E’) – Odds in favor of an “extreme” roll would then be 2:4 or 1:2 instead of 2/4 or ½ – (read as “two to four”, or “1 to 2”, resp.) – Odds against an “extreme” roll would then be 4:2 or 2:1 instead of 4/2 or 2/1 or even just 2 – Note in the above, that odds are generally reduced, just as fractions are. 7.4 Odds (cont) – A lot of people confuse odds with probability -- they are similar ideas (and sometimes close numbers), but are not the same. – Recapping the previous example of the event E = the “extreme” roll of a die, – P(E) = 2/6 = 1/3 ; odds for E are 2:4 or 1:2 – P(E’) = 4/6 = 2/3 ; odds against E are 4:2 or 2:1 – Different example, consider F = “rolling a sum of 12” on two fair dice: – P(F) = 1/36 ; odds for F are 1:35 – P(F’) = 35/36 ; odds against F are 35:1 7.4 Odds (cont) – You should understand the similarities and also the differences between “odds” and “probability”. – You should be able to calculate both: • Odds in favor of an event E (or simply “for” the event) • Odds against an event E – You should be able to convert from probabilities to odds, or vice versa, on a given problem. • The book gives some formulas for this on page 349, if you want to do it “by formula” – Odds are mainly used by gamblers for handling money; we’re not too concerned with this in class – We will predominantly use probabilities in class – in fact combinations such as union and intersection are much easier to do with probabilities than with odds! 7.4 Further probability notions • Types of probability – theoretical (by counting in a uniform sample space -- the formula P(E) = n(E) / n(S) ) – empirical (by having observed typical outcomes -- an experiment) • In a city study at an intersection, out of the 500 northbound cars, 35 of them turned left. What’s the probability of such a car turning left? • The empirical probability = 35 / 500 = 7 / 100 or 7% • This kind of probability is sometimes referred to as “relative frequency” – intuitive ( a “gut feeling” -- from experience?) • You think you have a “fifty-fifty” chance of “acing” exam 1. • A businessman who has a successful chain of 20 pizza restaurants estimates a new restaurant on Texas at University will have an 85% chance of being successful. 7.4 Further probability notions • Probability distribution for an experiment – Simply a list of all possible outcomes and their associated probabilities – Easiest given as a table – Probability distribution for 1 fair die: E Pr 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 – Probability distribution for sum of 2 fair dice: E 2 3 4 5 6 7 8 9 10 11 12 Pr 1/36 1/18 1/12 1/9 5/36 1/6 5/36 1/9 1/12 1/18 1/36 7.4 Further probability notions • Properties of probability – Let S be a sample space consisting of n distinct (i.e., mutually exclusive) outcomes, s1, s2, … , sn. An acceptable probability assignment consists of assigning to each outcome si a number pi (the probability of si ) according to these rules: – 1. The probability of each outcome is a number between 0 and 1. (PINGTO! and PINN!) • 0 <= p1 <= 1, 0 <= p2 <= 1, … , 0 <= pn <= 1, – 2. The sum of the probabilities of all possible outcomes is 1. • p1+ p2+ p3+ … + pn=1 • (or Spi 1, for short) 3. Don’t forget: P( E F ) P( E ) P( F ) P( E F )