Chapter 15 - The Binomial Formula PART IV : PROBABILITY Dr. Joseph Brennan Math 148, BU Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 1 / 19
Pascal s Triangle In this chapter we explore a mathematical concept which gives rise to Pascal s Triangle: Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 2 / 19
Pascal s Triangle Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 3 / 19
Pascal s Triangle Pascal s Triangle is connected to simple combinations. What is a combination? Assume that we have n distinct objects (say rocks painted with numbers) in a bucket. We are going to remove k objects from the bucket, without replacement. Can we count the number of possible ways such a choice can occur? Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 4 / 19
Permutations and Combinations ORDER In order to count the number of possible ways to choose, without replacement, k objects from a collection of n distinct we must be specific as to we acknowledge order. A permutation is a choice where order matters. The choice 1, 3, 4 is different than the choice 4, 1, 3. A combination is a choice where order does not matter. Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 5 / 19
Multiplication Principle The Multiplication Principle of Counting If there are A ways to complete a first action and B ways to complete a second action, then the number of ways to complete both actions is A B. How many complete outfits (fashion aside...) can be made when there are 4 shirts and 3 pants from which to choose? Factorials A mathematical shorthand notation for the multiplication of descending integers: n! = n (n 1) (n 2)... 3 2 1 For example: 1! = 1 3! = 3 2 1 = 6 5! = 5 4 3 2 1 = 120 Mathematical convention: 7! = 7 6 5 4 3 2 1 = 5040 0! = 1 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 6 / 19
Permutations The number of permutations that can be constructed with k objects taken without replacement from n distinct objects is denoted n P k. np k = n! (n k)! Example: Consider the number of permutations from choosing 3 objects from 10 objects. 10P 3 = 10! (10 3)! = 10 9 8 7 6... 1 = 10 9 8 = 720 7 6... 1 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 7 / 19
Example (Permutations) A social club has 50 members. The club would like to have more organization and members will be selected to the office of president, vice president, treasurer, secretary, and webmaster. If no member is able to hold multiple offices, in how many ways can the club organize itself? Solution: As no member can hold multiple offices, we are assigning offices without replacement. Further, as each office has distinct duties order matters. We are counting permutations! Specifically 50 P 5. 50P 5 = 50! 50 49 48 47 46 45... 1 = (50 5)! 45... 1 = 50 49 48 47 46 = 2, 542, 512, 200! Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 8 / 19
Combinations The number of combinations that can be constructed with k objects taken without replacement from n distinct objects is denoted ( n k). ( ) n n! = k k! (n k)! We read ( n k) as n choose k. The only difference between a permutation and a combination is order. This leads to very similar counting formulas: np k = n! (n k)! ( n k ) = n! k! (n k)! Notice that they differ by dividing by k!. Recall that 3! = 6: Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 9 / 19
Example (Combinations) A 12 member jury is to be chosen from a jury pool of 100 individuals. In how many ways can this jury be formed? Solution: We are choosing without replacement from our jury pool, otherwise we won t have 12 jurors. Further, unless we specify a leader for the jury, each member holds an indistinguishable office and order does not matter. The number of ways to form this jury is denoted ( ) 100 12 ; which is read 100 choose 12 : ( ) 100 100! = 12 12! (100 12)! = 100! = 1.050421 1015 12! 88! Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 10 / 19
The Binomial Coefficient ( n ) k is referred to as the binomial coefficient. k 0; you don t choose a negative number of objects. k n; you can t choose more objects than exists in the bin. n 0; there cannot be a negative number of items in the bin. ( 0 ) ( 0 = 1 and n ) ( 0 = 1 and n ) n = 1. Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 11 / 19
Binomial Probabilities Recall: An event E in the sample space S has probability number of outcomes in E P(E) = number of outcomes in S Example: There are 22 men and 8 women in a jury pool. What is the probability of forming a 7 person jury where 4 members are women. Solution: As order doesn t matter, the outcomes MMMMWWWW and WWWWMMM are the same, we are dealing with combinations! Let E be the event where 4 women are chosen. The number of combinations with 4 women and 3 men from the pool of 22 men and 8 women: ( ) ( ) 8 22 Size of E = 4 3 The sample space S is ( ) 30 7. ) ( 22 ) 3 P(E) = ( 8 4 ( 30 7 ) = 70 1, 540 2, 035, 800 = 0.053 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 12 / 19
Example (Poker) A deck of cards consists of 52 cards divided into: Two Colors (26 of each): Red and Black. Four Suits (13 of each): Clubs, Diamonds, Hearts, and Spades. Cards (4 of each, one per suit): Numbers and Face Cards. The standard hand in poker consists of 5 cards, dealt to a player without replacement. As the order of a hand is irrelevant, we focus on a hand s composition, the act of being dealt a hand is a combination. How many five card hands are possible? ( ) 52 = 52! = 2, 598, 960 5 5! 47! Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 13 / 19
Example (Poker) Using combinations what is the probability of being dealt 5 cards and obtaining: A pair of 2 s ( 4 ) ( 2 48 ) 3 ( 52 ) = 0.04 5 A full house; 5 s full and 2 s ( 4 ) ( 3 4 ( 2) 52 ) = 0.000009 5 3 Kings ( 4 ) ( 3 48 ) 2 ( 52 ) = 0.0017 5 A flush of diamonds ( 13 5 ) ( 52 5 ) = 0.0005 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 14 / 19
The Binomial Formula Let S be the sample space for a single occurrence and let E be an event in S. Let p = P(E). The probability that E will occur k times in n trials is given by the Binomial Formula: ( ) n P(n occurences of E) = p k (1 p) n k k Be careful to note: n is predetermined. p is P(E) while (1-p) is P(not E). Trials are independent; p is unchanged trial to trial. Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 15 / 19
Example (Binomial Formula) Example: A box contains 3 red, 4 green, and 2 blue marbles. What is the probability of choosing with replacement in 5 trials: (a) Two red marbles. (b) Three green marbles. Solution: As we are choosing with replacement, the trials are independent. We may use the Binomial Formula with n = 5. (a) Here p = 3 9 and k = 2. The probability can be found: ( ) ( ) 5 3 2 ( ) 6 3 P(2 red marbles) = = 0.33 2 9 9 (b) Here p = 4 9 and k = 3. The probability can be found: ( ) ( ) 5 4 3 ( ) 5 2 P(3 green marbles) = = 0.27 3 9 9 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 16 / 19
Example (Blood Disorder) The probability to recover from a certain blood disorder is 0.3. Five patients are observed. Compute the probabilities of the following events. (a) Two out of 5 patients will recover. (b) Four or more patients will recover. (c) No one will recover. (d) All 5 patients will recover. (e) Not exactly 2 patients will recover. Note that trials are independent and we may use the Binomial Formula with n = 5. The probability p is universal for patients and p = 0.3. Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 17 / 19
Example (Blood disorder) (a) We proceed with k = 2. P(2 patients recover) = ( ) 5 (0.3) 2 (1 0.3) 5 2 = 0.3087 2 (b) We need to sum two calculations involving the Binomial Formula; one with k=4 the other with k=5. P(4 + patients recover) = P(4 patients recover) + P(5 patients recover) ( ) ( ) 5 5 = (0.3) 4 (1 0.3) 5 4 + (0.3) 5 (1 0.3) 5 5 4 5 = 0.03078 (c) We proceed with k = 0. P(no patients recover) = ( ) 5 (0.3) 0 (1 0.3) 5 0 = 0.16807 0 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 18 / 19
Example (Blood disorder) (d) We proceed with k = 5. P(5 patients recover) = ( ) 5 (0.3) 5 (1 0.3) 5 5 = 0.3 5 = 0.00243 5 Notice that the chance that nobody will recover is much greater than the chance that everyone will recover... (e) The probability that exactly two patients will recover is calculated using the Binomial Formula with k = 2; by part (a): P(2 patients recover) = 0.3087 Recall that P(A) = 1 P(not A), therefore: P(not exactly 2 patients recover) = 1 P(2 patients recover) = 0.6913 Dr. Joseph Brennan (Math 148, BU) Chapter 15 - The Binomial Formula 19 / 19