1. Welcome to MM207 Unit 3 Seminar Dr. Bob Probability and Excel 1

2. Probability Experiment: Rolling a single die Sample Space: All possible outcomes from experiment S = {1, 2, 3, 4, 5, 6} Event: a collection of one or more outcomes (denoted by capital letter) Event A = {3} Event B = {even number} Probability = (number of favorable outcomes) / (total number of outcomes) –P(A) = 1/6 –P(B) = 3/6 = ½ 2

3. Probability Notation P(A) translates as “the probability of Event A occurring” Recall that we defined A = {3} therefore P(A) = “the probability of 3 being the result” Recall that we defined B = {even number} P(B) = “the probability that an even number will be the result”

4. Probability Probability will always be between 0 and 1. It will never be negative or greater than 1. Complement of an event: Represented as “not A” or A c If A = {3} then “not A” = A c = {1, 2, 4, 5, 6} “not A” or A c is everything but 3 4

5. Complement of an Event, page 140 A = 25 – 34 P(A) = P(25 – 34) 366/1000 = 0.366 B = not 25 – 34 P(B) = P(not 25 – 34) 1 – P(A) = 1 – 0.366 = 0.634 C = not 55 + P(C) = P(not 55 + ) = 1 – P(55 – 64) – P(65 +) = 1 – 125/1000 – 42/1000 = 1 – 167/1000 = 1 – 0.167 = 0.833 Employee Frequency Ages 15 – 24 54 25 – 34 366 35 – 44 233 45 – 54 180 55 – 64 125 65 + 42 total 1000

6. Conditional Probability {#15, page 155} The probability of an event occurring given that another event has already occurred: P(A|B) Vacation Vacation Total Yes No Computer Yes 46 11 57 Computer No 55 34 89 ------------------------------------------------------------------------------------- Total 101 45 146 Find the probability that a randomly selected family is taking a vacation given that they own a computer. Taking a vacation is 46 and owning a computer is 57 P(A|B) = 46/57 ≈ 0.807 Find the probability that a randomly selected family is taking a vacation and owns a computer. Owning a computer and taking a vacation P(A and B) = 46/146 ≈ 0.315

7. The Multiplication Rule, pages 150 - 151 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 ≈ 0.0059 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 ≈ 0.0060 Your chances are a bit lower with replacement because there is one more card available (52 instead of 51) on the second draw.

8. 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 ≈ 0.154 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 * 1 2 4 6

9. Basic Probability using PHStat2 9

10. 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.

11. Permutations and Combinations Formula 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)

12. Permutations and Combinations Examples What are the total number of permutations of 3 things taken from 7 things? Order matters {1 2 3 is not the same 3 2 1} What are the total number of combinations of 3 taken from 7 things? ( Order doesn’t matter {1 2 3 is the same as 3 2 1}

13. Insert Function Click on the f x 13

14. Insert Function Dialog Box 14

15. Excel Function for Permutations Problem 7, Page 178 15

16. Excel Function for Combinations Problem 9, Page 178 16