Stats 100A Notes

Lecture 1 (10/2/23) Probability and Set Operations (1.2~1.4) Lecture Notes Summary We will use set theory for the mathematical model of events. Outcomes of an experiment are elements of some sample space S, and each event is a subset of S. Two events both occur if the outcome is in the intersection of the two sets. At least one of a collection of events occurs if the outcome is in the union of the sets. Two events cannot both occur if the sets are disjoint. An event fails to occur if the outcome is in the complement of the set. The empty set is every event that cannot possibly occur. Video Notes Lecture 2 (10/4/23) Properties of Probability (1.5) Lecture Notes Summary IfAi,..., aredisjoint,Pr(Uf^jA,)= Pr(A/). ■Pr(A^)= 1- Pr(A). ■ A C B implies that Pr(A) < Pr(5). ■Pr(A U5) = Pr(A)+ Pr(5)-Pr(A5). Simple Space (1.6) A simple sample space is a finite sample space S such that every outcome in S has the same probability. If there are n outcomes in a simple sample space S, then each one must have probability 1/n. The probability of an event £ in a simple sample space is the number of outcomes in E divided by n. In the next three sections, we will present some useful methods for counting numbers of outcomes in various events.
Counting Methods (1.7) Suppose that the following conditions are met: ■ Each element of a set consists of k distinguishable parts. ■ There are n i possibilities for the first part. ■Foreachi=2,...,kandeachcombinationofthefirsti—1parts,therearen, possibilities for the /th part. Under these conditions, there are tii •■■ elements of the set. The third condition requires only that the number of possibilities for part i be n,- no matter what the earlier partsare.Forexample,for/=2,itdoesnotrequirethatthesame ^2 possibilitiesbe available for part 2 regardless of what part 1 is. It only requires that the number of possibilities be n 2 no matter what part 1 is. In this way, the general rule includes the multiplication rule, the calculation of permutations, and sampling with replacement as special cases. For permutations of m items k at a time, we have n,- = m —/ + 1 for / = 1,. . ., and the n, possibilities for part i are just the n, items that have not yet appeared in the first i — 1parts. For sampling with replacement from m items, we have n,- = m for all /, and the m possibilities are the same for every part. In the next section, we shall consider how to count elements of sets in which the parts of each element are not distinguishable.
Multiplication method- sampling with replacement Factorial method- sampling without replacement (where binomial theorem comes from- not a simple sample space) Complex example- screen shot above Combinatorial Methods (1.8)
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