Introductory Probability - PDF Free Download

Introductory Probability Combinations Nicholas Nguyen nicholas.nguyen@uky.edu Department of Mathematics UK

Agenda Assigning Objects to Identical Positions Denitions Committee Card Hands Coin Toss Counts Adding binomial coecients Announcement: Quiz on Friday, HW due Saturday

A Common Product (Note) We have (n) 0 = 1, and 0! = 1. There is only one way to make an ordered list of no objects: an empty list.

Subsets We have (n) 0 = 1, and 0! = 1. There is only one way to make an ordered list of no objects: an empty list. Given a set A of n objects, the act of choosing a random subset of A is a sequence of n stages where we decide if each object is in the subset or not in it. Hence there are a total of 2 n subsets of A.

Subsets Given a set A of n objects, the act of choosing a random subset of A is a sequence of n stages where we decide if each object is in the subset or not in it. Hence there are a total of 2 n subsets of A. The number of subsets with k objects will be denoted by the binomial coecient ( n k ), "n choose k."

Permutations and Subsets Let A be a set {a,b,c}. Then there are (3) 2 = 6 2-permutations of A: However, there are 3 subsets of A that have two elements each: a,b a,c b,c b,a c,a c,b {a,b} {a,c} {b,c}

Let A be a set {a,b,c}. Then there are (3) 2 = 6 2-permutations of A: However, there are 3 subsets of A that have two elements each: a,b a,c b,c b,a c,a c,b {a,b} {a,c} {b,c} A 2-permutation is formed by picking a subset of A and ordering the 2 objects in the subset. Hence ( ) ( ) 3 3 2! = (3) 2 2, so = (3) 2 = 3 2 2 2! 2 1 = 3.

Binomial Coecients In general, for non-negative integers n and k with 0 k n, ( ) n = (n) k n! = k k! k!(n k)!. The binomial coecient counts the number of ways to choose k objects from a set of n while ignoring the order. As such, ) ) ( n 0 = ( n n = 1 (one way to choose all or none of the objects).

Combinations When picking from a collection of objects without replacement (repeats), use combinations when the order in which the objects are chosen does not matter.

Combinations When picking from a collection of objects without replacement (repeats), use combinations when the order in which the objects are chosen does not matter. This means it only matters if an object is chosen or not. All of the chosen objects are placed into indistinguishable positions.

Committee Four people must be chosen from a pool of nine to form a committee. Each person on the committee will have identical roles. How many committees are possible? Use combinations: ( 9 4 ) = (9) 4 4! = 9 8 7 6 4 3 2 1 = 3,024 = 126 24

Poker Hands A poker hand is a subset of 5 cards from a deck of 52 cards. There are ( ) 52 = (52) 5 5 5! 52 51 50 49 48 = 5 4 3 2 1 = 2, 598, 960 poker hands.

Poker Hands A poker hand is a subset of 5 cards from a deck of 52 cards. There are ( ) 52 = (52) 5 5 5! 52 51 50 49 48 = 5 4 3 2 1 = 2, 598, 960 poker hands. The deck has 4 suits, and 13 cards of each suit. How many hands have 5 cards of the same suit 1? 1 Note that this includes royal ushes, straight ushes, and ushes.

Poker Hands A poker hand is a subset of 5 cards from a deck of 52 cards. There are ( ) 52 = (52) 5 5 5! 52 51 50 49 48 = 5 4 3 2 1 = 2, 598, 960 poker hands. The deck has 4 suits, and 13 cards of each suit. How many hands have 5 cards of the same suit? 1. Pick the suit: 4 choices 2. From the 13 cards of that suit, pick 5 of them in a subset:

Poker Hands A poker hand is a subset of 5 cards from a deck of 52 cards. There are ( ) 52 = (52) 5 5 5! 52 51 50 49 48 = 5 4 3 2 1 = 2, 598, 960 poker hands. The deck has 4 suits, and 13 cards of each suit. How many hands have 5 cards of the same suit? 1. Pick the suit: 4 choices 2. From the 13 cards of that suit, pick 5 of them in a subset: ( 13 5 ) = (13) 5 5!

( 52 5 ) = (52) 5 5! 52 51 50 49 48 = 5 4 3 2 1 = 2, 598, 960 poker hands. The deck has 4 suits, and 13 cards of each suit. How many hands have 5 cards of the same suit? 1. Pick the suit: 4 choices 2. From the 13 cards of that suit, pick 5 of them in a subset: ( ) 13 = (13) 5 5 5! 13 12 11 10 9 = 5 4 3 2 1 = 154,440 120 = 1,287

52 51 50 49 48 = 5 4 3 2 1 = 2, 598, 960 poker hands. The deck has 4 suits, and 13 cards of each suit. How many hands have 5 cards of the same suit? 1. Pick the suit: 4 choices 2. From the 13 cards of that suit, pick 5 of them in a subset: ( ) 13 = (13) 5 5 5! 13 12 11 10 9 = Multiply: 5 4 3 2 1 = 154,440 120 4 1,287 = 5,148 hands. = 1,287

= 2, 598, 960 poker hands. The deck has 4 suits, and 13 cards of each suit. How many hands have 5 cards of the same suit? 1. Pick the suit: 4 choices 2. From the 13 cards of that suit, pick 5 of them in a subset: ( ) 13 = (13) 5 5 5! 13 12 11 10 9 = Multiply: 5 4 3 2 1 = 154,440 120 4 1,287 = 5,148 hands. = 1,287 In particular the probability of getting 5 cards of the same suit in a hand is 5,148 2,598,960 0.198%

Coin Tosses A coin is tossed 5 times. There are 2 5 = 32 possible sequences of tosses (5 stages with 2 outcomes each). How many sequences have 3 heads? Use combinations on the 5 tosses: we pick 3 of them to be heads. For example, picking tosses 1, 2, and 3 to be heads should be the same as picking tosses 3, 2, and 1 to be heads. Hence there are ( 5 3 ) = 5 4 3 3 2 1 = 60 6 = 10 sequences where 3 tosses landed heads.

A coin is tossed 5 times. There are 2 5 = 32 possible sequences of tosses (5 stages with 2 outcomes each). How many sequences have 3 heads? Use combinations on the 5 tosses: we pick 3 of them to be heads. For example, picking tosses 1, 2, and 3 to be heads should be the same as picking tosses 3, 2, and 1 to be heads. Hence there are ( 5 3 ) = 5 4 3 3 2 1 = 60 6 = 10 sequences where 3 tosses landed heads. Thus, the probability of getting 3 heads in 5 tosses is 10 32. We will nd this probability using another method soon.

Adding Binomial Coecients Theorem (3.4) For natural numbers n and k with 0 < k < n, ( ) ( ) ( ) n 1 n 1 n + =. k 1 k k

Theorem 3.4 Proof For natural numbers n and k with 0 < k < n, ( ) ( ) ( ) n 1 n 1 n + =. k 1 k k Let A be a set of n objects, and let a be a specic object in A. First, there are ( ) n k subsets of A with k objects in them. Each subset of A with k objects falls into one of two disjoint cases: either the subset has a in it, or it does not have a in it. We will count the number of subsets for each case, and then add them to get the total number of subsets.

Theorem 3.4 Proof (Case 1) The subset has a in it. Then we need to choose k 1 more objects to ll up the subset, and these k 1 objects are chosen from the remaining n 1 objects in A (n 1 because a cannot go in the subset more than once). There are ( ) n 1 ways to do this. k 1

Theorem 3.4 Proof (Case 2 and Conclusion) The subset does not have a in it. Then we still need to choose k objects to ll up the subset, and these k objects are chosen from the remaining n 1 objects in A (n 1 because a is not in the subset). There are ( ) n 1 ways to do this. k Since a subset either has a in it or not in it, but not both, the total number of subsets with k objects must be the sum of the number of subsets with a in them and the number of subsets that do not have a: ( n 1 k 1 ) ( n 1 + k ) = ( n k ).

@Home: Reading Note: page numbers refer to printed version. Add 8 to get page numbers in a PDF reader. Take a look at Pascal's triangle at the top of page 94. It is a table of values of binomial coecients. There is another relation among binomial coecients on page 95. Compare this to any symmetry in each row of Pascal's triangle. The example at the bottom of page 95 discusses two other hands in poker.

Next Time Please read Section 3.2 (you can skip the historical remarks). We will look at Bernoulli trials and binomial probabilities. The third homework is due this Saturday. There is a quiz this Friday.