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12X1 T08 06 binomial probability (2011)

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12X1 T08 06 binomial probability (2011)

1. 1. Binomial Probability
2. 2. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows;
3. 3. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; A B
4. 4. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; A p  Bq 
5. 5. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AA A p  AB Bq 
6. 6. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AA A p  AB BA Bq  BB
7. 7. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AA p 2  A p  AB  pq  BA  pq  Bq  BB q 2 
8. 8. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AAA AA p 2  AAB A p  ABA AB  pq  ABB BAA BA  pq  BAB Bq  BBA BB q 2  BBB
9. 9. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AAA p 3  AA p 2  AAB  p q  2 A p  ABA  p 2 q  AB  pq  ABB  pq 2  BAA p 2 q  BA  pq  BAB pq  2 Bq  BBA pq  2 BB q 2  BBB p  3
10. 10. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AAA p 3  AAAA AAAB AA p  2 AABA AAB  p q  2 AABB A p  ABA  p q  2 ABAA AB  pq  ABAB ABBA ABB  pq 2  ABBB BAAA BAA p 2 q  BAAB BA  pq  BABA BAB pq  2 BABB Bq  BBA pq 2 BBAA BB q 2  BBAB BBB  p  3 BBBA BBBB
11. 11. Binomial ProbabilityIf an event has only two possibilities and this event is repeated, then theprobability distribution is as follows; AAA p 3  AAAA  p 4  AAAB  p 3q  AA p 2  AABA p 3q  AAB  p q  AABB  p 2 q 2  2 A p  ABA  p 2 q  ABAA  p 3q  AB  pq  ABAB  p 2 q 2  ABB  pq 2  p q  ABBA 2 2 ABBB  pq 3  BAA p q  BAAA 3 p q BAAB  p 2 q 2  2 BA  pq  BABA  p 2 q 2  BAB pq  BABB  pq  2 3 Bq  BBA pq 2 BBAA p q  2 2 BB q 2  BBAB  pq  3 BBB  p   pqq  3 BBBA 3 4 BBBB
12. 12. 1 Event
13. 13. 1 EventP  A  pP B   q
14. 14. 1 Event 2 EventsP  A  pP B   q
15. 15. 1 Event 2 EventsP  A  p P AA  p 2P B   q P AandB   2 pq PBB   q 2
16. 16. 1 Event 2 Events 3 EventsP  A  p P AA  p 2P B   q P AandB   2 pq PBB   q 2
17. 17. 1 Event 2 Events 3 EventsP  A  p P AA  p 2 P AAA   p 3P B   q P AandB   2 pq P2 AandB   3 p 2 q PBB   q 2 P Aand 2 B   3 pq 2 PBBB  q 3
18. 18. 1 Event 2 Events 3 EventsP  A  p P AA  p 2 P AAA   p 3P B   q P AandB   2 pq P2 AandB   3 p 2 q PBB   q 2 P Aand 2 B   3 pq 2 PBBB  q 34 Events
19. 19. 1 Event 2 Events 3 Events P  A  p P AA  p 2 P AAA   p 3 P B   q P AandB   2 pq P2 AandB   3 p 2 q PBB   q 2 P Aand 2 B   3 pq 2 PBBB  q 34 EventsP AAAA   p 4P3 AandB   4 p 3qP2 Aand 2 B   6 p 2 q 2P Aand 3B   4 pq 3PBBBB  q 4
20. 20. 1 Event 2 Events 3 Events P  A  p P AA  p 2 P AAA   p 3 P B   q P AandB   2 pq P2 AandB   3 p 2 q PBB   q 2 P Aand 2 B   3 pq 2 PBBB  q 34 Events If an event is repeated n times and P X   pP AAAA   p 4 and P X   q then the probabilit y that X willP3 AandB   4 p 3q occur exactly k times is;P2 Aand 2 B   6 p 2 q 2P Aand 3B   4 pq 3PBBBB  q 4
21. 21. 1 Event 2 Events 3 Events P  A  p P AA  p 2 P AAA   p 3 P B   q P AandB   2 pq P2 AandB   3 p 2 q PBB   q 2 P Aand 2 B   3 pq 2 PBBB  q 34 Events If an event is repeated n times and P X   pP AAAA   p 4 and P X   q then the probabilit y that X willP3 AandB   4 p 3q occur exactly k times is;P2 Aand 2 B   6 p 2 q 2 P X  k  nCk q nk p kP Aand 3B   4 pq 3PBBBB  q 4 Note: X = k, means X will occur exactly k times
22. 22. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing;a) All black balls?
23. 23. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing;a) All black balls? 0 7P X  7  7C7     2 3      5 5
24. 24. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? 0 7P X  7  7C7     2  3   5 5
25. 25. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? 0 7P X  7  7C7     2  3   5 5 7 C7 37  7 5
26. 26. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? 0 7P X  7  7C7     2  3   5 5 7 C7 37  7 5 2187  78125
27. 27. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? b) 4 black balls? 0 7P X  7  7C7     2  3   5 5 7 C7 37  7 5 2187  78125
28. 28. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? b) 4 black balls? 0 7 3 4P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37  7 5 2187  78125
29. 29. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? b) 4 black balls? 0 7 3 4P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187  78125
30. 30. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawna) All black balls? b) 4 black balls? 0 7 3 4P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187 4536   78125 15625
31. 31. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawn a) All black balls? b) 4 black balls? 0 7 3 4 P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187 4536   78125 15625 (ii) At an election 30% of voters favoured Party A. If at random an interviewer selects 5 voters, what is the probability that;a) 3 favoured Party A?
32. 32. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawn a) All black balls? b) 4 black balls? 0 7 3 4 P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187 4536   78125 15625 (ii) At an election 30% of voters favoured Party A. If at random an interviewer selects 5 voters, what is the probability that;a) 3 favoured Party A? Let X be the number favouring Party A
33. 33. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawn a) All black balls? b) 4 black balls? 0 7 3 4 P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187 4536   78125 15625 (ii) At an election 30% of voters favoured Party A. If at random an interviewer selects 5 voters, what is the probability that; 2 3 P3 A5C3     7 3a) 3 favoured Party A?      10   10  Let X be the number favouring Party A
34. 34. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawn a) All black balls? b) 4 black balls? 0 7 3 4 P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187 4536   78125 15625 (ii) At an election 30% of voters favoured Party A. If at random an interviewer selects 5 voters, what is the probability that; 2 3 P3 A5C3     7 3a) 3 favoured Party A?      10   10  Let X be the number 5 C3 7 233 favouring Party A  105
35. 35. e.g.(i) A bag contains 30 black balls and 20 white balls. Seven drawings are made (with replacement), what is the probability of drawing; Let X be the number of black balls drawn a) All black balls? b) 4 black balls? 0 7 3 4 P X  7  7C7     2  3 P X  4 7C4     2  3    5 5  5 5 7 C7 37 7 C4 2334  7  5 57 2187 4536   78125 15625 (ii) At an election 30% of voters favoured Party A. If at random an interviewer selects 5 voters, what is the probability that; 2 3 P3 A5C3     7 3a) 3 favoured Party A?      10   10  Let X be the number 5 C3 7 233 1323 favouring Party A  5  10 10000
36. 36. b) majority favour A?
37. 37. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10 
38. 38. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10  5 C3 7 233  5C4 7  34  5C5 35  105
39. 39. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10  5 C3 7 233  5C4 7  34  5C5 35  105 4077  25000
40. 40. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10  5 C3 7 233  5C4 7  34  5C5 35  105 4077  25000c) at most 2 favoured A?
41. 41. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10  5 C3 7 233  5C4 7  34  5C5 35  105 4077  25000c) at most 2 favoured A? P X  2  1  P X  3
42. 42. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10  5 C3 7 233  5C4 7  34  5C5 35  105 4077  25000c) at most 2 favoured A? P X  2  1  P X  3 4077  1 25000
43. 43. b) majority favour A? 2 3 1 4 0 5 P X  35C3      5C4      5C5     7 3 7 3 7 3             10   10   10   10   10   10  5 C3 7 233  5C4 7  34  5C5 35  105 4077  25000c) at most 2 favoured A? P X  2  1  P X  3 4077  1 25000 20923  25000
44. 44. 2005 Extension 1 HSC Q6a)There are five matches on each weekend of a football season.Megan takes part in a competition in which she earns 1 point ifshe picks more than half of the winning teams for a weekend, andzero points otherwise.The probability that Megan correctly picks the team that wins inany given match is 2 3(i) Show that the probability that Megan earns one point for a given weekend is 0 7901, correct to four decimal places.
45. 45. 2005 Extension 1 HSC Q6a)There are five matches on each weekend of a football season.Megan takes part in a competition in which she earns 1 point ifshe picks more than half of the winning teams for a weekend, andzero points otherwise.The probability that Megan correctly picks the team that wins inany given match is 2 3(i) Show that the probability that Megan earns one point for a given weekend is 0 7901, correct to four decimal places. Let X be the number of matches picked correctly
46. 46. 2005 Extension 1 HSC Q6a)There are five matches on each weekend of a football season.Megan takes part in a competition in which she earns 1 point ifshe picks more than half of the winning teams for a weekend, andzero points otherwise.The probability that Megan correctly picks the team that wins inany given match is 2 3(i) Show that the probability that Megan earns one point for a given weekend is 0 7901, correct to four decimal places. Let X be the number of matches picked correctly 2 3 1 4 0 5 P X  35C3      5C4      5C5     1 2 1 2 1 2             3  3   3  3   3  3 
47. 47. 2005 Extension 1 HSC Q6a)There are five matches on each weekend of a football season.Megan takes part in a competition in which she earns 1 point ifshe picks more than half of the winning teams for a weekend, andzero points otherwise.The probability that Megan correctly picks the team that wins inany given match is 2 3(i) Show that the probability that Megan earns one point for a given weekend is 0 7901, correct to four decimal places. Let X be the number of matches picked correctly 2 3 1 4 0 5 P X  35C3      5C4      5C5     1 2 1 2 1 2             3  3   3  3   3  3  5 C3 23  5C4 2 4  5C5 25  35
48. 48. 2005 Extension 1 HSC Q6a)There are five matches on each weekend of a football season.Megan takes part in a competition in which she earns 1 point ifshe picks more than half of the winning teams for a weekend, andzero points otherwise.The probability that Megan correctly picks the team that wins inany given match is 2 3(i) Show that the probability that Megan earns one point for a given weekend is 0 7901, correct to four decimal places. Let X be the number of matches picked correctly 2 3 1 4 0 5 P X  35C3      5C4      5C5     1 2 1 2 1 2             3  3   3  3   3  3  5 C3 23  5C4 2 4  5C5 25  35  0 7901
49. 49. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places.
50. 50. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point
51. 51. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point PY  1818C18 0  2099 0  7901 0 18
52. 52. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point PY  1818C18 0  2099 0  7901 0 18  0 01 (to 2 dp)
53. 53. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point PY  1818C18 0  2099 0  7901 0 18  0 01 (to 2 dp)(iii) Find the probability that Megan earns at most 16 points during the eighteen week season. Give your answer correct to two decimal places.
54. 54. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point PY  1818C18 0  2099 0  7901 0 18  0 01 (to 2 dp)(iii) Find the probability that Megan earns at most 16 points during the eighteen week season. Give your answer correct to two decimal places. PY  16  1  PY  17 
55. 55. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point PY  1818C18 0  2099 0  7901 0 18  0 01 (to 2 dp)(iii) Find the probability that Megan earns at most 16 points during the eighteen week season. Give your answer correct to two decimal places. PY  16  1  PY  17   118C17 0  2099 0  7901 18C18 0  2099 0  7901 1 17 0 18
56. 56. (ii) Hence find the probability that Megan earns one point every week of the eighteen week season. Give your answer correct to two decimal places. Let Y be the number of weeks Megan earns a point PY  1818C18 0  2099 0  7901 0 18  0 01 (to 2 dp)(iii) Find the probability that Megan earns at most 16 points during the eighteen week season. Give your answer correct to two decimal places. PY  16  1  PY  17   118C17 0  2099 0  7901 18C18 0  2099 0  7901 1 17 0 18  0 92 (to 2 dp)
57. 57. 2007 Extension 1 HSC Q4a)In a large city, 10% of the population has green eyes.(i) What is the probability that two randomly chosen people have green eyes?
58. 58. 2007 Extension 1 HSC Q4a)In a large city, 10% of the population has green eyes.(i) What is the probability that two randomly chosen people have green eyes? P  2 green   0.1 0.1  0.01
59. 59. 2007 Extension 1 HSC Q4a)In a large city, 10% of the population has green eyes.(i) What is the probability that two randomly chosen people have green eyes? P  2 green   0.1 0.1  0.01(ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal eyes.
60. 60. 2007 Extension 1 HSC Q4a)In a large city, 10% of the population has green eyes.(i) What is the probability that two randomly chosen people have green eyes? P  2 green   0.1 0.1  0.01(ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal eyes. Let X be the number of people with green eyes
61. 61. 2007 Extension 1 HSC Q4a)In a large city, 10% of the population has green eyes.(i) What is the probability that two randomly chosen people have green eyes? P  2 green   0.1 0.1  0.01(ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal eyes. Let X be the number of people with green eyes P  X  2   C2  0.9   0.1 20 18 2
62. 62. 2007 Extension 1 HSC Q4a)In a large city, 10% of the population has green eyes.(i) What is the probability that two randomly chosen people have green eyes? P  2 green   0.1 0.1  0.01(ii) What is the probability that exactly two of a group of 20 randomly chosen people have green eyes? Give your answer correct to three decimal eyes. Let X be the number of people with green eyes P  X  2   C2  0.9   0.1 20 18 2  0.2851  0.285 (to 3 dp)
63. 63. (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.
64. 64. (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.P  X  2  1  P  X  2
65. 65. (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.P  X  2  1  P  X  2  1  C2  0  9   0 1  C1  0  9   0 1  C0  0  9   0 1 20 18 2 20 19 1 20 20 0
66. 66. (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.P  X  2  1  P  X  2  1  C2  0  9   0 1  C1  0  9   0 1  C0  0  9   0 1 20 18 2 20 19 1 20 20 0  0.3230  0  32 (to 2 dp)
67. 67. (iii) What is the probability that more than two of a group of 20 randomly chosen people have green eyes? Give your answer correct to two decimal places.P  X  2  1  P  X  2  1  C2  0  9   0 1  C1  0  9   0 1  C0  0  9   0 1 20 18 2 20 19 1 20 20 0  0.3230  0  32 (to 2 dp) Exercise 10J; 1 to 24 even