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Chapter 6, Counting and Probability CMSC 250 1 Counting Counting elements in a list: – how many integers in the list from 1 to 10? – how many integers in the list from m to n? (assuming m <= n) CMSC 250 2 How many in a list divisible by something How many positive three-digit integers are there? – (this means only the ones that require 3 digits) – 999 – 100 + 1 = 900 How many three-digit integers are divisible by 5? – think about the definition of divisible by x | y ( k Z)[y = kx] then count the k’s that work 100, 101, 102, 103, 104, 105, 106,…,994, 995, 996, 997, 998, 999 20 5 21 5 … 199 5 – count the integers between 20 and 199 – 199 – 20 + 1 = 180 CMSC 250 3 Probability The likelihood of a specific event. Sample space = set of all possible outcomes Event = subset of sample space Equal probability formula: – given a finite sample space S where all outcomes are equally likely – select an event E from the sample space S – the probability of event E from sample space S: n( E ) P( E ) n( S ) CMSC 250 4 Coins and cards and dice Two coins – sample space = {(H,H), (H,T), (T,H), (T,T)} Cards – values: 2,3,4,5,6,7,8,9,10,J,Q,K,A – suits: D(), H(), S(), C() Dice – sample space {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), … (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} CMSC 250 5 Multi-level probability If a coin is tossed once, the probability of head = ½ If it’s tossed 5 times 1 1 1 1 1 1 * * * * – the probability of all heads: 2 2 2 2 2 25 5 – the probability of exactly 4 heads: 25 This is because the coin tosses are all independent events CMSC 250 6 The breakfast problem CMSC 250 Suppose your cereal can be Rice Krispies, cornflakes, Raisin Bran, or Cheerios. Suppose your drink can be coffee, orange juice, or milk. How many ways can you have breakfast? 7 The multiplication rule If the 1st step of an operation can be performed n1 ways And the 2nd step can be performed n2 ways … And the kth step can be performed nk ways Then the operation can be performed n1n2 ∙ ∙ ∙ nk ways Cartesian product n(A)=3, n(B)=2, n(C)=4 – n(A B C) = 24 – n(A B) = 6 – n((A B) C) = 24 CMSC 250 8 Discrete Structures CMSC 250 Lecture 34 April 21, 2008 CMSC 250 9 Using the multiplication rule for selecting a PIN Number of 4 digit PINs of (0,1,2,.) – with repetition allowed = 4 4 4 4 = 256 – with no repetition allowed = 4 3 2 1 = 24 Extra rules : – . (the period) can’t be first or last – 0 can’t be first • with repetition allowed = 2 4 4 3 • without repetition allowed = 2 2 2 1 CMSC 250 10 Probabilities with PINs Number of 4 digit PINs of {0,1,2,.} – with repetition allowed = 4 4 4 4 = 256 – with no repetition allowed = 4 3 2 1 = 24 What is the probability that your 4 digit PIN has no repeated digits? What is the probability that your 4 digit PIN does have repeated digits? Probability of the complement of an event P(E’) = P(Ec) = 1 P(E) CMSC 250 11 The difference rule formally If A is a finite set and B A, then n(A B) = n(A) – n(B) One application: probability of the complement of an event: P(E’) = P(Ec) = 1 P(E) CMSC 250 12 PINs with less specified lengthaddition rule Assume a PIN can be of length 2, 3, or 4, using {0,1,2,.} Partition the problem: – number of length-2 PINs w/rep allowed: 4 4 = 16 – number of length-3 PINs w/rep allowed: 4 4 4 = 64 – number of length-4 PINs w/rep allowed: 4 4 4 4 = 256 Number of PINs if allowing length 2, 3, or 4 = 336 CMSC 250 13 The addition rule formally If A1 A2 A3 … Ak = A and A1, A2 , A3,…,Ak are pairwise disjoint in other words, if these subsets form a partition of A, then n(A) = n(A1) + n(A2) + n(A3) + ∙∙∙ + n(Ak) CMSC 250 14 The inclusion/exclusion rule If there are two sets: n(A B) = n(A) + n(B) – n(A B) If there are three sets: n(A B C) = n(A) + n(B) + n(C) – n(A B) – n(A C) – n(B C) + n(A B C) CMSC 250 15 Discrete Structures CMSC 250 Lecture 35 April 23, 2008 CMSC 250 16 Permutations Different ways of arranging all n of n objects – in either a line or a circle – without duplication/all items distinguishable – note: order is taken into account Number of linear permutations of N objects = N! N possible for 1st position (N – 1) for 2nd ∙ ∙ ∙ 1 for last Number of circular permutations of N objects = (N 1)! Fix one person, then (N – 1) possible for next position * (N – 2) for 2nd ∙ ∙ ∙ 1 for last CMSC 250 17 r-permutations If there are n things in a set, and you want to line up only r of them n! P(n, r ) n P r (n r )! Example: Class = {Alice, Bob, Carol, Dan} – select a president and a vice president to represent the class – select a president, vice president, and webmaster CMSC 250 18 Combinations Different ways of selecting objects – counting subsets – without duplication/all items distinguishable – note: order is not taken into account 0 (n, r Z where n r ) C (n, r ) n r P(n, r ) n! r! (n r )!r! Examples: suppose {Alice, Bob, Carol, Dan} are on the ballot – select two superdelegates – select three superdelegates CMSC 250 19 Permutations but of indistinguishable items Examples: – Arrangements of the word “baboons” – Assume you have a set of 15 beads: • 6 green • 4 orange • 3 red • 2 black select positions of the green ones, then the orange ones, then the red ones, then the black ones (or a different order of selecting their positions would work as well) 15 6 CMSC 250 15! * * * 6!4!3!2! 9 4 5 3 2 2 20 Discrete Structures CMSC 250 Lecture 36 April 25, 2008 CMSC 250 21 Combinations with repetition {a,b,c,d,e} How many 3-combinations can be made without repetition? How many 3-combinations can be made with unlimited repetition allowed? These are multisets [a,b,c] – not sets {a,b,c} – and not tuples (a,b,c) How many combinations of 20 a's, b's, and c's can e made with unlimited repetition allowed? CMSC 250 22 Notice similarities The number of nonnegative integer solutions of the equation x1 x2 xn r xi 0, i Z 1i n The number of selections, with repetition, of size r from a collection of size n. The number of ways r identical objects can be distributed among n distinct containers. CMSC 250 23 Choosing r elements out of n elements order doesn’t matter order matters repetition allowed repetition not allowed CMSC 250 n n n r times r n r 1 r n! n n! P(n, r ) (n r )! r (n r )!r! 24 Probability with combinations Assume: – there are 32 people in a class – seven will be chosen to get extra homework What is the probability that you get extra homework? Number of ways to select the lucky seven Number of ways to select if you get homework P(you get homework) CMSC 250 25 Tournament play Team A and Team B compete in a “best of 3” tournament They each have an equal likelihood of winning each game – – – – CMSC 250 Do the leaves add up to 1? Do they always have to play 3 games? What's the probability the tournament finishes in 2 games? Do A and B have an equal chance of winning? 26 What if A wins 2 of every 3 games? Each line for A must have a 2/3 Each line for B must have a 1/3 – How likely is A to win the tournament? – How likely is B to win the tournament? – What is the probability the tournament finishes in two games? CMSC 250 27 Where the multiplication rule doesn’t work People= {Alice, Bob, Carolyn, Dan} Need to be appointed as president, vice-president, and treasurer, and nobody can hold more than one office – how many ways can it be done with no restrictions? – how many ways can it be done if Alice doesn’t want to be president? – how many ways can it be done if Alice doesn’t want to be president, and only Bob and Dan are willing to be vicepresident? CMSC 250 28 Discrete Structures CMSC 250 Lecture 37 April 28, 2008 CMSC 250 29 Harder examples of selecting representatives 1. 2. 3. 4. 5. CMSC 250 Candidates= {Azar, Barack, Clinton, Dan, Erin, Fred} select two, with no restrictions select two, assuming that Azar and Dan must stay together select three, with no restrictions select three, assuming that Azar and Dan must stay together select three, assuming that Barack and Clinton refuse to serve together 30 Properties of combinations and their proofs n 2 1 n n 0 n 1 n( n 1) 2 n n 1 n n 1 n n2 n n( n 1) 2 n r CMSC 250 n nr 31 The binomial theorem ( x y) x y ( x y)2 ( x y)(x y) x 2 xy xy y 2 x 2 2xy y 2 ( x y)3 ( x y) 2 ( x y) ( x 2 2 xy y 2 )(x y) x 3 2 x 2 y xy 2 x 2 y 2 xy 2 y 3 x 3 3x 2 y 3xy 2 y 3 ( x y ) 4 ( x y )3 ( x y ) ( x 3 3x 2 y 3xy 2 y 3 )(x y ) x 4 3x 3 y 3x 2 y 2 xy3 x 3 y 3x 2 y 2 3xy3 y 4 x 4 4 x 3 y 6 x 2 y 2 4 xy3 y 4 4 4 4 3 4 2 2 4 3 4 4 ( x y) x x y x y xy y 0 1 2 3 4 n n i n i n ( x y) x y i 0 i CMSC 250 4 32