Stat 11 Lecture 9 Handout & Chapter 3.3-3.4 0. Venn Diagrams (pages 9-15 of the handout) These diagrams are not required for the course, but it may make some of the reading easier for those of you who struggle through this material. There is also a link on the course webpage at: http://web.stat.ucla.edu/~vlew/stat11/lectures/venn/venn.html 1. Postulates of probability (formally) (pages 15-19 of the handout) There are three basic rules of probability. All other rules stem from proofs based on these three. Postulate 1: The probability of any event must be a positive real number or zero; symbolically, P(A) ≥ 0 for any event A. This basically means that you cannot have a negative probability event. Postulate 2: The probability of a sample space is always 1; symbolically, P(S) = 1 for any sample space S So this means that the probability of some event from the sample space happening is certain. This is why when we describe the sample space, we must be exhaustive. If we miss something that is possible, the probability of something happening from the sample space will not be 1 (or 100%). Postulate 3: If the set of events A1, A2, A3 … AN are mutually exclusive events, the probability that one or the other will occur equals the sum of their respective probabilities; symbolically, P(A1∪ A2 ∪… ∪ AN) = P(A1) + P(A2)+… + P(AN) for any N mutually exclusive events. This could be rewritten as P(A1∪ A2 ∪… ∪ AN) =∑=niiAP1)( Therefore suppose A1, A2, A3 are mutually exclusive, then the probability of A1 or A2 or A3 happening is the sum of their respective probabilities Here’s a table of 133 movies that grossed over 100 million dollars by quarter of opening day (release date) between 1996-2003 quarter | Freq. Percent Cum. ------------+----------------------------------- 1 | 10 7.52 7.52 2 | 48 36.09 43.61 3 | 34 25.56 69.17 4 | 41 30.83 100.00 ------------+----------------------------------- Total | 133 100.00 If we could pretend that this was a stable population, we could apply some of the rules listed above and extensions to those rules. Let’s let 1=Jan 1-Mar 31 2=Apr 1- Jun 30 3=Jul 1 – Sep 30 4= Oct 1 – Dec 31 Example 1. Suppose you choose a $100 million dollar movie at random, what is the probability that it was released in the first half of the year? 2. Suppose you choose a movie at random, what is the probability that it was released between April 1 and the end of September?Stat 11 Lecture 9 Handout & Chapter 3.3-3.4 Result 1. If A and B are two mutually exclusive events defined over S, then P(A ∪ B) is just the sum of their individual probabilities. This is a simplification of postulate 3. 3. Suppose you choose a movie at random, what is the probability that it WAS NOT released in the last quarter of the year? Result 2. P(A′) = 1.0 – P(A). The rule of complements. The probability that A does not occur is 1 – probability that A occurs. (Note sometimes P(A′) is written P(Ac) depends on the textbook) Proof: If P(S)=1 = P(A ∪ A’) = P(A) + P(A’) Result 3. P(φ) = 0. Impossible events have zero probability Result 4. If A is contained in B (that is A⊆ B) then P(A) ≤ P(B) Here is the distribution of the day of week for release dates: dow | Freq. Percent Cum. ------------+----------------------------------- Monday| 0 0.00 0.00 Tuesday| 1 0.75 0.75 Wednesday| 26 19.55 20.30 Thursday| 5 3.76 24.06 Friday| 101 75.94 100.00 Saturday| 0 0.00 100.00 Sunday| 0 0.00 100.00 ------------+----------------------------------- Total | 133 100.00 If Event B were defined as a “midweek release” (Tue-Thu) and if event A were defined as “a Wednesday release” then we can see that P(A) ≤ P(B). This leads us to result 5: Result 5. 0 ≤ P(A) ≤ 1. Probabilities can’t be negative and they cannot exceed 1.0 Result 6. P (A ∪ B) = P(A) + P(B) – P (A∩B) The probability of observing event A or B is the sum of the individual probabilities less the probability that they both occur. If A and B are mutually exclusive, result 6 is just result 1. If they are not mutually exclusive, this piece: – P (A∩B) , removing the intersection removes the double counting that occurs when summing the probabilities. Of all UCLA undergraduates, 10% are Economics Majors, 15% are Psychology Majors and 3% are both. What is the probability that a randomly selected undergraduate is in at least one of the two majors? (.10 + .15) - .03 = .22 or 22% chance.Stat 11 Lecture 9 Handout & Chapter 3.3-3.4 2. Conditional probability (informally and formally) (pages 19-21 of handout) A conditional probability is the probability of an event occurring given that some other event has already occurred. We write it P(A | B) and say “the probability of A given B.” This second handout is a “cross-tabulation” or basically the result of a tabulate in Stata with two variables – race/ethnicity & educational attainment. Let us assume that this is the population and it is stable. Now suppose we are doing a study in this low-income area of Los Angeles and we select an adult at random from the population. What is the chance that he or she is Hispanic? (.5180 or about 52%). That’s basic probability. We can find this value in the “row marginal” of the table. Conditional is this: what is the probability that the person has an advanced degree GIVEN that we know he or she is Hispanic? (about 1.65%) How? Conditional probability is as if the sample space (S) shrinks from covering all of the groups represented in the table, to just one group (Hispanic). Then we are interested in some outcome WITHIN the new sample space (just Hispanics). Formally, )()()'()()()|(BPBAPBABAPBAPBAP∩=∩+∩∩= The probability of selecting a Hispanic person is P(B) or .5180, the probability selecting a person with an advanced degree who is also Hispanic is P(A∩B) (0.85% or .0085). So P(A|B) = .0085/.5180 = .0165 So you have about a .0165 probability (chance) that if you have selected a Hispanic person he or she will have an advanced degree.