Lecture Topic: Probability Experiment: An act where the outcome is subject to uncertainty Sample space: The set of all possible outcomes of an experiment Example: - Flip a coin: $S = \{H, T\}$ - Throw a dice: $S = \{1, 2, 3, 4, 5, 6\}$ Events: An event is a collection (subset) of outcomes contained in the sample space S $$ S = \begin{Bmatrix} 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \end{Bmatrix} $$ A = First and second elements are the same $A = \{(1,1) \ (2,2) \ (3,3) \ (4,4) \ (5,5) \ (6,6)\}$ Union: Union of two events, A and B, ($A \cup B$) Compliment: The compliment of an event A, is the set of all outcomes in S that are not in A Mutually Exclusive Events: When and B have no common outcomes, they are said to be mutually exclusive or disjoint events, i.e. $P=(A \cap B) = 0$ Axiom: 1. For event A, P(A) >= 0 2. P(S) = 1 3. (Need to look at slides to correct) (a) If $A_1, A_2 ... A_k$ is a finite collection of mutually exclusive events then: $$P(A_1 \cup A_2 \cup ... \cup A_k) = \sum^{\infty}_{i=1} P(A_i) $$ (b) If $A_1, A_2 ... A_k$ is a finite collection of mutually exclusive events then: $$P(A_1 \cup A_2 \cup ... \cup A_k) = \sum^{\infty}_{i=1} P(A_i) $$ Proposition: 1. For event A, P(A) = 1 - P(A') 2. If and and B are mutually exclusive then $P(A \cap B) = 0$ Permutation: Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from n objects is denoted by $P_{k,n}$ and calculated by: $$ P_{k,n} = \frac{n!}{(n-k)!} $$ Combination: Given a set of n, any unordered sub-set of k of the objects is called a combination. The set of combinations of size k that can be formed from n distinct objected will be denoted by $C_{k,n}$ and calculated by: $$ C_{k,n} = \frac{n!}{k!(n-k)!} = \begin{pmatrix}n \\ k \end{pmatrix} $$ (NEED TO REVIEW THIS) Example: S = {1, 2, 3, 4, 5} $A_1$ = {1,2,3}, $A_2$ = {2,3,5}, $A_3$ = {2,3,1} Permutation: 5! = 120 (5-3)! = 2! = 2 120 / 2 = 60 Combination: 5!/2!3! = 60/6 = 10 0! = 1 5 out of 5 5!/(5-5)! = 120/1 = 120