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Warm up

Warm up. A Ferris wheel holds 12 riders. If there are 20 people waiting in line, how many different groups of people can ride it? You may write your answer in terms of factorials/permutations/combinations. Solution.

peter-perkins
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Warm up

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  1. Warm up • A Ferris wheel holds 12 riders. If there are 20 people waiting in line, how many different groups of people can ride it? • You may write your answer in terms of factorials/permutations/combinations.

  2. Solution • Since only 12 of the 20 people can ride the ferris wheel at a time, there are C(20,12) or 125 970 different groups of riders. • Each group can be placed on the ferris wheel 11! or 39 916 800 ways since it is a circular permutation. • So the total number of ways is: • C(20, 12) x 11! • = 125 970 x 39 916 800 • = 5 028 319 296 000 or 5.03 x 1012 • I hope they purchased the season pass!

  3. Chapter 4 ReviewMDM 4U Mr. Lieff

  4. Test Format • 20 Multiple Choice • 6 long answer • Simulation / Experimental Probability • Theoretical Probability / Additive Principle • Diagram (Venn or Tree) • Conditional Probability • Permutations • Combinations • 1 hour time limit

  5. 4.1 Intro to Simulations and Experimental Probability • Design a simulation to model the probability of an event • Ex: design a simulation to determine the experimental probability that more than one of 5 keyboards chosen in a class will be defective if we know that 25% are defective 1. Get a well-shuffled deck of cards, choosing clubs to represent a defective keyboard 2. Choose 5 cards with replacement and record how many are clubs 3. Repeat a large number of times (e.g., 4 outcomes x 10 = 40) and calculate the experimental probability

  6. 4.2 Theoretical Probability • Work with Venn diagrams • ex: Create a Venn diagram illustrating the sets of face cards and red cards S = 52 red & face = 6 face = 6 red = 20

  7. 4.2 Theoretical Probability • Calculate the probability of an event or its complement • Ex: What is the probability of randomly choosing a male from a class of 30 students if 10 are female? • P(A) = n(A) = 20 = 0.67 n(S) 30

  8. 4.2 Theoretical Probability • Ex: Calculate the probability of not throwing a total of four with 3 dice • There are 63 = 216 possible outcomes with three dice • Only 3 outcomes produce a 4 • P(sum = 4) = 3 216 • probability of not throwing a sum of 4 is: • 1 – 3_ = 213 = 0.986 216 216

  9. 4.3 Finding Probability Using Sets • Recognize the different types of sets (union, intersection, complement) • Utilize the additive principle for unions of sets • The Additive Principle for the Union of Two Sets: • n(A U B) = n(A) + n(B) – n(A ∩ B) • P(A U B) = P(A) + P(B) – P(A ∩ B) • Calculate probabilities using the additive principle

  10. 4.3 Finding Probability Using Sets • Ex: What is the probability of drawing a red card or a face card • Ans: P(A U B) = P(A) + P(B) – P(A ∩ B) • P(red or face) = P(red) + P(face) – P(red and face) • = 26/52 + 12/52 – 6/52 • = 32/52 • = 0.615

  11. 4.3 Finding Probability Using Sets • What is n(B υ C) • 2+8+3+3+6+2+1+8+1 = 34 • What is P(A∩B∩C)? • n(A∩B∩C) = 3 = 0.07 • n(S) 43

  12. 4.4 Conditional Probability • 100 Students surveyed • What is the probability that a student takes Mathematics given that he or she also takes English?

  13. E 4.4 Conditional Probability 17 45 1 5 2 5 25

  14. 4.4 Conditional Probability • To answer the question in (b), we need to find P(Math|English). • We know... • P(Math|English) = P(Math ∩ English) P(English) • Therefore… • P(Math|English) = 30 / 100 = 30 x 100 = 3 80 / 100 100 80 8

  15. 4.4 Conditional Probability • Calculate the probability of event B occurring, given that A has occurred • Need P(A∩B) and P(A) • Use the multiplicative law for conditional probability • Ex: What is the probability of drawing a jack and a queen in sequence, given no replacement? • P(1J ∩ 2Q) = P(2Q | 1J) x P(1J) • = 4 x 4 = 16 = 0.006 51 52 2 652

  16. R R B G R B B G R G B G 4.5 Tree Diagrams and Outcome Tables • A sock drawer has a red, a green and a blue sock • You pull out one sock, replace it and pull another out • Draw a tree diagram representing the possible outcomes • What is the probability of drawing 2 red socks? • These are independent events

  17. 4.5 Tree Diagrams and Outcome Tables • Mr. Lieff is going fishing • He finds that he catches fish 70% of the time when the wind is out of the east • He also finds that he catches fish 50% of the time when the wind is out of the west • If there is a 60% chance of a west wind today, what are his chances of having fish for dinner? • We will start by creating a tree diagram

  18. 4.5 Tree Diagrams and Outcome Tables P=0.3 fish dinner 0.5 west 0.6 0.5 P=0.3 bean dinner fish dinner 0.7 P=0.28 0.4 east 0.3 bean dinner P=0.12

  19. 4.5 Tree Diagrams and Outcome Tables • P(east, catch) = P(east) x P(catch | east) • = 0.4 x 0.7 = 0.28 • P(west, catch) = P(west) x P(catch | west) • = 0.6 x 0.5 = 0.30 • Probability of a fish dinner: 0.28 + 0.3 = 0.58 • So Mr. Lieff has a 58% chance of catching a fish for dinner

  20. Combinatorics (§4.6 & 4.7) • Permutations – order matters • e.g., President • Combinations – order does not matter • e.g., Committee

  21. 4.6 Permutations • Find the number of outcomes given a situation where order matters • Calculate the probability of an outcome or outcomes in situations where order matters • Recognizing how to restrict the calculations when some elements are the same

  22. 4.6 Permutations • Ex: How many ways can 5 students be arranged in a line? • Ans: 5! = 120 • Ex: How many ways are there if Jake must be first? • Ans: (5-1)! = 4! = 60 • Ex: in a class of 10 people, a teacher must pick 3 for an experiment (students are tested in a particular order) • How many ways are there to do this? • Ans: P(10,3) = 10! = 720 (10 – 3)!

  23. Permutations cont’d • How many ways are there to rearrange the letters in the word TOOLTIME? • 8! = 10 080 (2!2!)

  24. 4.6 Permutations • Ex: What is the probability of opening one of the school combination locks by chance? • Second digit must be different from the first • Ans: 1 in 60 x 59 x 59 = 1 in 208 860 • Circular Permutations: There are (n-1)! ways to arrange n objects in a circle

  25. 4.7 Combinations • Find the number of outcomes given a situation where order does not matter • Calculate the probability of an outcome or outcomes in situations where order does not matter • Ex: How many ways are there to choose a 3 person committee from a class of 20? • Ans: C(20,3) = 20!___ = 1140 (20-3)! 3!

  26. 4.7 Combinations • Ex: From a group of 5 men and 4 women, how many committees of 5 can be formed with • a. exactly 3 women • b. at least 3 women • ans a: • ans b:

  27. Review • Read your notes! • pp. 268-269 #1, 4, (5-6) ace…, 8, 9, 11 • p. 270 #1, 2

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