Welcome to MM207 - Statistics! Unit 3 Seminar Good Evening Everyone! To resize your pods: Place your mouse here. Left mouse click and hold. Drag to the right to enlarge the pod. To maximize chat, minimize roster by clicking here

Probability and Roulette Roulette – reference Basics: –38 numbers: 1-36, 0, 00 –3 colors: 18 black numbers 18 red numbers 2 green numbers (0 and 00) Purpose: Spin the roulette wheel, place a bet. If the ball lands on your number, you win!

Probability and Roulette cont. Experiment: Spinning the roulette wheel Sample Space: All possible outcomes from experiment S = {0, 00, 1, 2, ……, 35, 36} Event: a collection of one or more outcomes (denoted by capital letter) Event A = {14} Event B = {black (list the 18 numbers)} Event C = {14, 15, 16, 17} Probability = (number of favorable outcomes) / (total number of outcomes)

Probability and Roulette cont. Event A = {14} Event B = {black (list the 18 numbers)} Event C = {14, 15, 16, 17} Probability = (number of favorable outcomes) / (total number of outcomes) Probability of A occurring = 1 / 38 = 0.03 = 3% Probability of B occurring = 18 / 38 = 9 / 19 = 0.47 = 47% Probability of C occurring = 4 / 38 = 2 / 19 = 0.11 = 11% Note: Probability will always be between 0 and 1. It will never be negative or greater than 1.

Probability Notation P(A) translates as “the probability of Event A occurring” Recall that we defined A = {14} therefore P(A) = “the probability of 14 being the result” Recall that we define B = {black} P(B) = “the probability that black will be the result” Compliment of an event: Represented as “not A” or A c If A = {14} the “not A”= A c = {0, 00, 1, 2,...,12,13, 15, 16, …..35, 36} “not A” or A c is everything but 14

Example 7 page P(25 – 34) 366/1000 = P(25 – 54) P(25 – 34) + P(35 – 44) + P(45 – 54) 366/ / / /100 = P(age ≤ 34) P(15 – 24) + P(25 – 34) 54/ / /1000 = Note: Do not round Employee Frequency Ages 15 – – – – – total 1000

Complement of an Event, page 140 A = 25 – 34 P(A) = P(25 – 34) 366/1000 = B = not 25 – 34 P(B) = P(not 25 – 34) 1 – P(A) = 1 – = C = not 55 + P(C) = P(not 55 + ) = 1 – P(55 – 64) – P(65 +) = 1 – 125/1000 – 42/1000 = 1 – 167/1000 = 1 – = Employee Frequency Ages 15 – – – – – total 1000

Conditional Probability, page 149 The probability of an event occurring given that another event has already occurred: P(A|B) Gene Present Gene Absent Total High IQ Normal IQ Total Find the probability that a child has a high IQ, given that the child has the gene. High IQ, gene present is 33 and Gene Present Total is 72 P(A|B) = 33/72 ≈ Find the probability that a child has a normal IQ, given that the child does not have the gene. Normal IQ, Gene Absent is 11 and Gene Absent Total is 30 P(A|B) = 11/30 ≈ 0.367

The Multiplication Rule, pages P(A and B) = “Probability of event A AND event B occurring at the same time” Independent Events P(A and B) = P(A) P(B) Two cards are drawn with replacement. Find P(Ace and King) P(Ace and King) = P(Ace) * P(King) = 4/52 * 4/52 = 16/2704 ≈ Note: There are 4 Aces and 4 kings Dependent Events P(A and B) = P(A) P(B|A) or P(A and B) = P(B) P(A|B) Two cards are drawn without replacement. Find P(Ace and King) P(Ace and King) = P(Ace) * P(Ace | King) = 4/52 * 4/51 = 16/2652 ≈ Your chances are a bit lower with replacement because there is one more card available (52 instead of 51) on the second draw.

The Addition Rule, pages 160 – 161 P(A or B) = “Probability that either A or B (but not both) occur” Two Mutually Exclusive Events (can’t occur at same time) P(A or B) = P(A) + P(B) One card is drawn, find the probability that it is a 7 or a Queen. A 7 and Queen cannot occur at the same time, the events are mutually exclusive. P(7 or Queen) = P(7) + P(Queen) = 4/52 + 4/52 = 8/52 = 2/13 ≈ Any Two Events P(A or B) = P(A) + P(B) – P(A and B) A die is rolled. What is the probability of getting a 2 or an even number? The events are NOT mutually exclusive because 2 is an even number. P(2 or even number) = P(2) + P(even) – P(2 and even) = 1/6 + 3/6 – 1/6 = 4/6 – 1/6 = 3/6 = 0.5 Note: 2 is an outcome of both events, so P(2 and even) = 1/6 *

Permutations and Combinations Permutation: Arranging “n” distinct objects taken “r” at a time (order matters) Given the letters A, B, and C. What are the total number of permutations of 2 letters? AB, BA, AC, CA, BC, CB 6 outcomes AB and BA are two different outcomes, order matters Combination: Arranging “n” distinct objects taken “r” at a time (order does not matter) What are the total number of combinations of 2 letters? AB, AC, BC 3 outcomes AB and BA, for example, are the same in a combination, order doesn’t matter.

Permutations and Combinations Factorial notation: n! = n *(n-1)*(n-2)*….*3*2*1 Example: 5! = 5*4*3*2*1 = 120 Note: 1! = 1 and 0! = 1 Permutation: Arranging “n” distinct objects taken “r” at a time (order matters) Combination: Arranging “n” distinct objects taken “r” at a time (order does not matter)

Permutations and Combinations What are the total number of permutations of 2 letters? (arrange 2 letters, order matters) AB, BA, AC, CA, BC, CB 6 total What are the total number of combinations of 2 letters? (arrange 2 letters, order doesn’t matter) AB, AC, BC 3 total