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Chapter 5 Possibilities and Probability

Chapter 5 Possibilities and Probability. Counting Permutations Combinations Probability. 5.1 Counting. Example 5.1: Ice cream cones Flavor: chocolate , vanilla, strawberry Cones: sugar, regular How many different varieties?. Tree diagram. sugar

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Chapter 5 Possibilities and Probability

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  1. Chapter 5 Possibilities and Probability • Counting • Permutations • Combinations • Probability

  2. 5.1 Counting Example 5.1: Ice cream cones Flavor: chocolate , vanilla, strawberry Cones: sugar, regular How many different varieties?

  3. Tree diagram sugar choc reg sugar van reg sugar str reg 3*2=6 choices

  4. Rule 1: Multiplication Rule • A choice consists of 2 distinct steps • 1st step can be made in m different ways • For each of these, 2nd step can be made in n different ways Then the whole choice can be made in m*n ways

  5. Example 5.2 • 4 horses in a race • How many ways can we pick a first and second place finisher? • Answer: 43=12 • You can get it from a tree diagram as well

  6. B A C D A Choice of 1st placechoices of 2nd place B C D 4*3=12 ways A C B D A D B C

  7. Rule 2: Generalized Multiplication Rule • A choice consists of k steps; • Step 1 can be made in n1ways; • Step 2 can be made in n2 ways; … … • Step k can be made in nk ways ; Then whole choice can be made inn1n2…nkways

  8. Example 5.3 • 5 horses in a race • How many ways can we pick a 1st , 2nd, and 3rd place finisher?

  9. Example 5.4 • A multiple choice exam has 5 questions • Each question has 4 possible answers • What is the number of ways to answer the exam?

  10. Example 5.5 • How many license plates can be formed with 3 letters followed by 3 numbers?

  11. Example 5.6 • How many ways can we write the letters O W L ?

  12. 5.2 Permutations • The number of ways to order r of n objects nPr=n(n-1)(n-2)…(n-r+1) (application of multiplication rule)

  13. Example 5.7 • How many ways can we put 3 cards from a deck of cards in order? 52 51 50 (52)(51)(50)=132,600

  14. Notation: n!=(1)(2)…(n-1)(n) • The number of ways to order (permute) n of n objects is nPn = n(n-1)(n-2)…(1) = n! • n! is called n-factorial 3!=(1)(2)(3)=6 4!=(1)(2)(3)(4)=3!(4)=24 5!=4!(5)=120 (0!=1)

  15. Example 5.8 • How many ways can we order 5 horses? 5!=120

  16. Example 5.9 • How many ways can we arrange the letters OLWS?

  17. The Number of Permutations

  18. Exercise How many different ways can we inject 3 of 15 mice with 3 different doses of a serum?

  19. Exercise 5.21 In optics kits there are 5 concave lenses, 5 convex lenses, 2 prisms, and 3 mirrors. how many different ways can a person choose 1 of each kind?

  20. Exercise 5.27 In how many different ways can a television director schedule a sponsor’s six different commercials during a telecast?

  21. 5.3 Combinations • How many ways can we choose 3 of 5 candidates to be in the final election? Here the order of choosing the 3 finalists doesn’t matter!

  22. Think this way …… • When we did worry about the order, there are (5)(4)(3)=60 ways ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA BCD BDC ... … …

  23. and • Each choice of 3 candidates has 3!=6 ordered versions ABC  ABC ACB BAC BCA CAB CBA

  24. Therefore ABC ACB BAC BCA CAB CBAABC ABD ADB BAD BDA DAB DBAABD BCD BDC ... BCD … … The options are decreased by a factor of 6 compared to the ordered options

  25. The number of ways to pick 3 out of 5

  26. The Number of Combinations • The number of unordered ways to choose r of n objects is (n choose r)

  27. Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

  28. Example 5.10 • Choose 5 cards from a deck

  29. Example 5.11 • 5 flavors of ice cream. Choose 2. Order doesn’t matter.

  30. Example 5.12 • 6 candidates in a primary election. Choose 2 for a final election

  31. Exercise • Calculate the number of ways in which a chain of ice cream stores can choose 2 of 12 locations for new franchises.

  32. Exercise • A computer store carries 15 kinds of monitors. Calculate the number of ways in which we can purchase 3 different ones.

  33. Exercise • In planning a garden we have 5 kinds of bushes to choose from and 10 kinds of flowers. How many ways can we choose 2 kinds of bushes and 4 kinds of flowers?

  34. 5.4 Probability • In a deck of cards what is the probability of picking an ace? • What is meant by “Probability”? • Frequency interpretation: The probability of an event happening is the proportion of times that event would occur in the long run.

  35. Facts • Each card is equally likely to be chosen for a shuffled deck of cards • If there are “n” equally likely possibilities and “s” of these are a “success” , then the probability of a success is s/n P(ace)= 4/52=1/13 P(red)=26/52=1/2

  36. Example 5.13 • Draw 2 cards from a deck. What is the probability that we get 2 aces?

  37. Ideas • Each card is equally likely to be selected if one card is selected • All pairs of cards are equally likely to be selected if only two cards are selected • Any three cards are equally likely to be selected if only three cards are selected … … (we call these randomness)

  38. Solution to Example 5.13 • # of possible ways to pick 2 cards • # of ways to pick 2 aces • Probability P(2 aces)

  39. Example 5.14 • Pick up 3 cards, what is the probability of getting 2 aces and 1 king? Let’s work it out!

  40. Step 1 Step 2 Step 3 Probability P(2 aces and 1 king)=s/n= 24/22100 # of ways to pick 2 aces out of 4 # of ways to pick 1 king out of 4

  41. Example 5.15 • Pick up 5 cards. What is the probability of getting 3 aces?

  42. # of ways to pick 3 aces out of 4 Step 1 Step 2 Step 3 Probability P(3 aces in 5 cards )=s/n= # of ways to pick other 2 cards

  43. Example 5.16 • Roll a red die and a white die. • Find the probability that sum=3. red die: 1, 2, 3, 4, 5, 6 white die: 1, 2, 3, 4, 5, 6 n = # of outcomes = (ways for red to land)*(ways for white to land) = 6×6=36 pairs s = # of ways sum=3 (R=1, W=2), (R=2, W=1) =2 Probability=2/36=1/18

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