Review Counting Principles Theorems Examples. John Venn. Arthur Berg Counting Rules 2/ 21

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1 Counting Rules

2 John Venn Arthur Berg Counting Rules 2/ 21

3 Augustus De Morgan Arthur Berg Counting Rules 3/ 21

4 Algebraic Laws Let S be a sample space and A, B, C be three events in S. Commutative Laws: A B = B A AB = BA Associative Laws: (A B) C = A (B C) (AB)C = A(BC) Distributive Laws: A(B C) = AB AC A (BC) = (A B)(A C) De Morgan s Laws: n i=1 A i = n i=1 A n i i=1 A i = n i=1 A i Arthur Berg Counting Rules 4/ 21

5 Definition of a Probability Definition (probability) A probability is a mapping of events to the real numbers such that the following three conditions hold: (i) P(A) 0 for any event A; (ii) P(S) = 1 where S is the sample space; (iii) If A 1, A 2,... are mutually exclusive (meaning A i A j = for any i j) then ( ) P A i = i=1 P(A i ). i=1 Arthur Berg Counting Rules 5/ 21

6 Implications of conditions (i)-(iii) (a) P( ) = 0. (b) If AB =, then P(A B) = P(A) + P(B). (c) For any events A and B, (d) If A B, then P(A) P(B). (e) For any event A, 0 P(A) 1. (f) P(Ā) = 1 P(A) (g) Principle of Inclusion-Exclusion P(A B) = P(A) + P(B) P(AB). Theorem (Principle of Inclusion-Exclusion) Given events E 1,..., E n, n P ( n i=1 E i) = P(E i ) P(E i1 E i2 ) + i=1 i 1 <i 2 + ( 1) r+1 P(E i1 E i2 E ir ) + + ( 1) n+1 P(E 1 E n ) i 1 <i 2 < i r Arthur Berg Counting Rules 6/ 21

7 PIE with three events Arthur Berg Counting Rules 7/ 21

8 PIE with four events Arthur Berg Counting Rules 8/ 21

9 Shopping at two stores Arthur Berg Counting Rules 9/ 21

10 Fundamental Principle of Counting Theorem (Fundamental Principle of Counting) If the first task has n 1 possible outcomes and the second task has n 2 possible outcomes, then there are n 1 n 2 possible outcomes for the two tasks together. Example Suppose the first task is to roll a five-sided blue-colored die and the second task is to role a six-sided orange-colored die. How many different outcomes are there? Arthur Berg Counting Rules 10/ 21

11 Five-Sided Die Arthur Berg Counting Rules 11/ 21

12 Fundamental Principle of Counting Theorem (Fundamental Principle of Counting) If the first task has n 1 possible outcomes and the second task has n 2 possible outcomes, then there are n 1 n 2 possible outcomes for the two tasks together. Example Suppose the first task is to roll a five-sided blue-colored die and the second task is to role a six-sided orange-colored die. How many different outcomes are there? Arthur Berg Counting Rules 12/ 21

13 Fundamental Principle of Counting Theorem (Fundamental Principle of Counting) If the first task has n 1 possible outcomes and the second task has n 2 possible outcomes, then there are n 1 n 2 possible outcomes for the two tasks together. Example Suppose the first task is to roll a five-sided blue-colored die and the second task is to role a six-sided orange-colored die. How many different outcomes are there? Ans: 6 5 = 30. Arthur Berg Counting Rules 12/ 21

14 Fundamental Principle of Counting Theorem (Fundamental Principle of Counting) If the first task has n 1 possible outcomes and the second task has n 2 possible outcomes, then there are n 1 n 2 possible outcomes for the two tasks together. Example Suppose the first task is to roll a five-sided blue-colored die and the second task is to role a six-sided orange-colored die. How many different outcomes are there? Ans: 6 5 = ,12,13,14,15, 16,21,22,23,24, 25,26,31,32,33, 34,35,36,41,42, 43,44,45,46,51, 52,53,54,55,56 Arthur Berg Counting Rules 12/ 21

15 Example 2.6 (p.35) Example (Example 2.6) Five cans of paint (numbered 1 through 5) were delivered to a professional painter. Unknown to her, some of the cans (1 and 2) are satin finish and the remaining cans (3, 4, and 5) are glossy finish. Suppose she selects two cans at random for a particular job. Let A denote the event that the painter selects the two cans of satin-finish paint, and let B denote the even that the two cans have different finishes (one of satin and one of glossy). Find P(A) and P(B). Arthur Berg Counting Rules 13/ 21

16 Two questions to ask Is the order important? Is it with or without replacement? Arthur Berg Counting Rules 14/ 21

17 Permutations Theorem (Permutations) The number of ordered arrangements (permutations) of r objects selected from n distinct objects (r n) is P n r = n(n 1) (n r + 1) = n! (n r)! Example How many three-letter words can be made from the letters A, B, C, D, and E? (Here, any combination of letters counts as a word.) Arthur Berg Counting Rules 15/ 21

18 Permutations Theorem (Permutations) The number of ordered arrangements (permutations) of r objects selected from n distinct objects (r n) is P n r = n(n 1) (n r + 1) = n! (n r)! Example How many three-letter words can be made from the letters A, B, C, D, and E? (Here, any combination of letters counts as a word.) Ans: P 5 3 = = 60. Arthur Berg Counting Rules 15/ 21

19 Combinations Theorem (Combinations) The number of distinct subsets (combinations) of size r that can be selected from n distinct objects (r n) is given by ( ) n = Pn r r r! = n! ( ) r!(n r)! Any permutation x 1 x 2 x r can be ordered in r! different ways, and so ( ) follows. Example (Example 2.8) In Florida s lottery, balls numbered from 1 to 53 are placed in a hopper and six balls are drawn at random without replacement. What is the probability of winning this lottery? Arthur Berg Counting Rules 16/ 21

20 Combinations Theorem (Combinations) The number of distinct subsets (combinations) of size r that can be selected from n distinct objects (r n) is given by ( ) n = Pn r r r! = n! ( ) r!(n r)! Any permutation x 1 x 2 x r can be ordered in r! different ways, and so ( ) follows. Example (Example 2.8) In Florida s lottery, balls numbered from 1 to 53 are placed in a hopper and six balls are drawn at random without replacement. What is the probability of winning this lottery? Ans: 1/22,957,480 since ( ) 53 53! = 6 6!(53 6)! = 53! 6!(47!) = = 22, 957, 480 Arthur Berg Counting Rules 16/ 21

21 Partitions Theorem (Partitions) The number of ways of partitioning n distinct objects into k groups containing n 1, n 2,..., n k objects, respectively, is where k i=1 n i = n. n! n 1!n 2! n k! ( )( ) ( ) n n n1 n n1 n k 1 n 1 n 2 n k ( ) ( ) ( ) n! (n n 1 )! (n n1 n k 1 )! = n 1!(n n 1 )! n 2!(n n 1 n 2 )! n k!0! Arthur Berg Counting Rules 17/ 21

22 Partitions Example Example (Example 2.11) Suppose ten employees are divided among three job assignments, with three employees going to job A, four to job B, and three to job C. In how many ways can the job assignments be made? Arthur Berg Counting Rules 18/ 21

23 Partitions Example Example (Example 2.11) Suppose ten employees are divided among three job assignments, with three employees going to job A, four to job B, and three to job C. In how many ways can the job assignments be made? Ans: This can be accomplished in ways. 10! 3!4!3! = = 4, 200 (3 2)(3 2) Arthur Berg Counting Rules 18/ 21

24 Birthday Problem Example (Birthday Problem) Suppose n people are in a room. In this problem, you may ignore the possibility of someone s birthday being on February What is the probability that no two of them have the same birthday? 2 Determine the smallest n such that the probability of at least two people have the same birthday is greater than 1/2. Arthur Berg Counting Rules 19/ 21

25 Birthday Problem Example (Birthday Problem) Suppose n people are in a room. In this problem, you may ignore the possibility of someone s birthday being on February What is the probability that no two of them have the same birthday? 2 Determine the smallest n such that the probability of at least two people have the same birthday is greater than 1/2. 1 Ans: There are 365 n different outcomes. The number of possibilities of everyone having a different birthday is 365(364)(363) (365 n + 1). Therefore the probability is 1 for n 365 and 1 if n > ! (365 n)!365 n Arthur Berg Counting Rules 19/ 21

26 Birthday Problem Continued for(n in 1: 100){ cat(n,1-prod(365-(0:(n-1)))/365^n,"\n") } Arthur Berg Counting Rules 20/ 21

27 Poker Probabilities Consider a well-shuffled 52-card deck of cards. After being drawn 5 cards, calculate the following probabilities: 1 royal flush (A,K,Q,J,10 in the same suit) 2 straight flush (straight in one suit) 3 four of a kind 4 full house (pair plus 3 of a kind) 5 straight (sequential cards in different suits) 6 three of a kind 7 two pair 8 one pair 9 none of the above Arthur Berg Counting Rules 21/ 21