1. LESSON 12.1 Use lists, tables and tree diagrams to represent sample spaces I can use the Fundamental counting Principle to count outcomes

2. experiment outcomes outcome trial event experiment sample space tree diagram

3. H H T T T H HT H T H H T T H T H T H T H T H H T T H T

4. H 1 H 2 H 3 H 4 H 5 H 6 T 1 T 2 T 3 T 4 T 5 T 6 H T 1 2 3 4 5 6 H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6

5. K R D C NC C C P NP P P P P P kids with cheese and pickles 12 different orders possible 3 ⦁ 2 ⦁ 2 = 12

6. N1 ⦁ N2 ⦁ N3 ⦁ …..

7. 10 ⦁ 2 ⦁ 12 ⦁ 5 ⦁ 20 ⦁ 20 ⦁ 2 = 960,000

8. 11 ⦁ 7 ⦁ 5 ⦁ 3 ⦁ 6 ⦁ 4 ⦁ 3 = 83,160

9. 12-1 worksheet ASSIGNMENT

10. LESSON 12.2 I can use permutations with probability I can use combinations with probability

11. n! 4 ⦁ 3 ⦁ 2 ⦁ 1 = 24 10! = 10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 = 3,628,800 5! = 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 = 120

12. order n P r n! (n – r)! 5 P 2 = 5! (5 – 2)! = 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 3 ⦁ 2 ⦁ 1 = 20 6 P 4 = 6! (6 – 4)! = 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 2 ⦁ 1 = 360

13. 10 P 4 =5040 8 P 3 =336

14. order n C r n! r!(n – r)! 10 C 4 = 10! 4! (10 – 4)! 10 ⦁ 9 ⦁ 8 ⦁ 7 ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 4 ⦁ 3 ⦁ 2 ⦁ 1 = ⦁ 6 ⦁ 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 5040 24 = 210

15. 5 C 2 = 5! 2! (5 – 2)! = 5 ⦁ 4 ⦁ 3 ⦁ 2 ⦁ 1 2 ⦁ 1 ⦁ 3 ⦁ 2 ⦁ 1 20 2 = 10

16. 8 P 3 = 336 5 C 3 = 10 60 C 5 =5,461,512 20 C 4 = 4845 4 P 4 = 24

17. 1 20 C 6 = 1 38,760 1 20 P 2 = 1 380

18. 1 26 C 5 = 1 65,780 1 30 P 2 = 1 870

19. 12-2 worksheet ASSIGNMENT

20. LESSON 12.3 I can solve problems involving geometric probability I can solve problems involving sectors and segments of circles

21. chance Area of B Total area

22. Area of square Total area = 4 28 = 0.14

23. Area of shaded Total area = 48 100 = 0.48

24. P(red) = 80 360 = 0.222

25. P(blue) = 120 360 = 0.333

26. π (2 2 ) = 4 π – π (1 2 ) = 1 π = 3 π (area of white region) π (3 2 ) = 9 π (area of target) P(white) = 3π3π 9π9π = 0.333

27. π (6 2 ) = 36 π – π (3 2 ) = 9 π = 27 π (area of blue region) π (6 2 ) = 36 π (area of target) P(blue) = 27 π 36 π = 0.75

28. 12-3 worksheet ASSIGNMENT

29. LESSON 12.4 I can find the probabilities of independent and dependent events I can find the probability of events given the occurrence of other events

30. 2 or more does not affects

31. independent dependent independent

32. dependent

33. P(A and B) = P(A) ⦁ P(B) 1 6 1 6 1 6 ⦁ 1 6 = 1 36

34. 1 3 1 3 1 3 ⦁ 1 3 = 1 9

35. P(A and B) = P(A) ⦁ P(B | A)

36. P(regular 1 st and 2 nd ) 8 13 ⦁ 7 12 = 56 156

37. TV VVVC 3 10 ⦁ 1 9 = 3 90 6 10 ⦁ 5 9 = 30 90

38. 12-4 worksheet ASSIGNMENT

39. LESSON 12.5 I can find the probabilities of events using two-way frequency tables

40. 1218 20 10 30 P(male) =20/30 P(female) =10/30 P(11th) = 12/30 P(12th) =18/30

41. 1218 20 10 30 4/12 6/10 12/18 0

42. 160 44 32 25 59 76 84 69 91 69 76

43. 20% 15.6% 36.9% 47.5% 52.5% 43.1% 56.9% 100% 47.5% of 285 0.475(285) ≈ 135 students 44/76 ≈ 0.579

44. 12-5 worksheet ASSIGNMENT

45. LESSON 12.6 I can find probabilities of events that are mutually exclusive and events that are not mutually exclusive

46. mutually exclusive yes no

47. yes

48. P(A or B) = P(A) + P(B) 35 10 35 + 12 35 = 22 35

49. 6 36 + 4 = 10 36 15 80 + 16 80 = 41 80 + 10 80

50. P(A or B) = P(A) + P(B) – P(A and B) P(H or K) = P(H)+ P(K) –P(H and K) 13 52 + 4 = 16 52 – 1

52. P(~A) = 1 – P(A) ~A: rolling an even number ~B: picking heart, diamond or spade ~C: losing or tying a game P(A) = 3/6 P(~A) = 3/6 P(B) = 13/52 P(~B) = 39/52 P(C) = 0.7 P(~C) = 0.3

53. 2 6 2 6 4 6 2 6 + 2 6 = 4 6

54. 3 6 2 6 4 6 3 6 + 2 6 = 4 6 – 1 6

55. 0.24 0.84 0.54 0.73

56. 12-6 worksheet ASSIGNMENT