1. 7.4 Basic Concepts of Probability This presentation is copyright by Paul Hendrick © 2003-2005, Paul Hendrick All rights reserved

2. 7.4 Basic Concepts of Probability Union rule for probability –Union rule (general case – always true) –IF P(E F) = 0, it can be omitted! –Union rule for mutually exclusive events, only

3. 7.4 Basic Concepts of Probability Complement rule –back-door approach (because it doesnt calculate directly)

4. –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

5. –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)

6. –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)

7. –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

8. 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; were 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!

9. 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. Whats 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.

10. 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: –Probability distribution for sum of 2 fair dice: 1/6 4 2 Pr 6531E 1/36 12 5/36 6 1/6 7 5/36 8 1/9 9 1/12 10 1/12 4 1/36 2 1/181/91/18Pr 1153E 7.4 Further probability notions

11. Properties of probability –Let S be a sample space consisting of n distinct (i.e., mutually exclusive) outcomes, s 1, s 2, …, s n. An acceptable probability assignment consists of assigning to each outcome s i a number p i (the probability of s i ) according to these rules: –1. The probability of each outcome is a number between 0 and 1. (PINGTO! and PINN!) 0 <= p 1 <= 1, 0 <= p 2 <= 1, …, 0 <= p n <= 1, –2. The sum of the probabilities of all possible outcomes is 1. p 1 + p 2 + p 3 + … + p n =1 (or p i for short) – Dont forget: 7.4 Further probability notions