Math 408, Actuarial Statistics I A.J. Hildebrand Combinatorial Probabilities Key concepts • Permutation: arrangement in some order. • Ordered versus unordered samples: In ordered samples, the order of the elements in the sample matters; e.g., digits in a phone number, or the letters in a word. In unordered samples the order of the elements is irrelevant; e.g., elements in a subset, or lottery numbers. • Samples with replacement versus samples without replacement: In the first case, repetition of the same element is allowed (e.g., numbers in a license plate); in the second, repetition not allowed (as in a lottery drawing—once a number has been drawn, it cannot be drawn again). Formulas • Number of permutations of n objects: n! • Number of ordered samples of size r, with replacement, from n objects: nr • Number of ordered samples of size r, without replacement, from n objects: n(n − 1) · · · (n − r + 1) = n! = n Pr . (n − r)! • Number of unordered samples of size r, without replacement, from a set of n objects (= number of subsets of size r from a set of n elements) (combinations): n n! n(n − 1) . . . (n − r + 1) n Pr = = = . r r! r!(n − r)! r! (See back of page for properties of these binomial coefficients.) • Number of subsets of a set of n elements: 2n 1 Math 408, Actuarial Statistics I A.J. Hildebrand Binomial coefficients n n! . • Definition: For n = 1, 2, . . . and k = 0, 1, . . . , n, = k!(n − k)! k (Note that, by definition, 0! = 1.) • Alternate notations: n Ck or C(n, k) n n(n − 1) . . . (n − k + 1) • Alternate definition: = . k k! (This version is convenient for hand-calculating binomial coefficients.) n n • Symmetry property: = k n−k n n n n • Special cases: = = 1, = =n 0 n 1 n−1 n X n k n−k n • Binomial Theorem: (x + y) = x y k k=0 • Binomial Theorem, special case: n X n k=0 k pk (1 − p)n−k = 1 n • Combinatorial Interpretations: represents k 1. the number of ways to select k objects out of n given objects (in the sense of unordered samples without replacement); 2. the number of k-element subsets of an n-element set; 3. the number of n-letter HT sequences with exactly k H’s and n − k T’s. • Binomial distribution: Given a positive integer n and a number p with 0 <p < 1, the n k binomial distribution b(n, p) is the distribution with density (p.m.f.) f (k) = p (1 − k p)n−k , for k = 0, 1, . . . , n. 2 Probability Examples ● ● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles What is the probability that you draw and replace marbles 3 times and you get NO red marbles? ● There are 55 marbles, 25 of which are not red ● P(getting a color other than red) = P(25/55) ≈ .455 ● Probability of this happening 3 times in a row is found by .455*.455*.455 ≈ .094 Example 2: At least 1 Red ● ● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles What is the probability that you draw and replace marbles 3 times and you get at least 1 Red? ● ● It's easier to calculate the probability of getting NO red marbles, and subtract that from 1 (we use the complement rule : P(AC) = 1 – P(C) From previous example, it is 1 - .094 = .906 Example 3: The First Red ● ● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles You draw and replace marbles 3 times. What is the probability the third marble is the first red marble? ● ● This means the first two are not red. We calculated P(drawing a non-red) = .455. Therefore, P(red)=.545 P(non-red & non-red & Red) = P(non-red) * P(nonred) * P(red) = .455 * .455 * .545 = .113 Example 4: Red, Yellow and Blue ● ● A jar contains 30 red marbles, 12 yellow marbles, 8 green marbles and 5 blue marbles You draw and replace marbles 3 times. What is the probability you draw 1 Red, 1 Yellow, and 1 Blue? ● ● ● This is harder, because we are drawing marbles in an order, but we don't care about which order we get Red, Yellow and Blue, just that there is 1 of each. But we can do it! Example 4: Continued ● ● ● ● Let RBY = “Draw a Red, then Blue, then Yellow” So all disjoint events we want to consider are: RBY, RYB, YRB, YBR, BYR, BRY – there are 6 of them. P(RBY) = P(R)*P(B)*P(Y) = (30/55)*(5/55)*(12/55) = .0108 But we have 6 disjoint cases. Because each one is calculated as a product of the three, and each disjoint case has the same probability (each order is equally likely), our answer is 6*.0108 = .0649