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Probability

Probability. Assignment #41 Review Paper #1 day7. A#41 Review Paper # 1 day7. Probability. 5. [ Maximum mark: 6] Let A and B be independent events, where P(A) = 0.6 and P(B) = x. Write down an expression for P(A  B). (b) Given that P(A  B) = 0.8 (i) Find x;

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Probability

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  1. Probability Assignment #41 Review Paper #1 day7

  2. A#41 Review Paper #1 day7 Probability

  3. 5. [ Maximum mark: 6] Let A and B be independent events, where P(A) = 0.6 and P(B) = x. Write down an expression for P(A  B). (b) Given that P(A  B) = 0.8 (i) Find x; (ii) Find P(A  B) (c) Hence, explain why A and B are not mutually exclusive. 2008/SP1/TZ0 Complement Combined events Mutually exclusive events Conditional Independent events

  4. Independent 2008/SP1/TZ0 Tree diagram P(B) = (a) P(AB) = P(A)  P(B) P(AB) = (0.6)  (x) = 0.6x x P(A) = 0.6 1 – x 0.4 • Defn of Independent events: P(A  B) = P(A)  P(B)

  5. Independent 2008/SP1/TZ0 (b) (i) 0.8 = 0.6 + x – (0.6)(x) 0.2 = 0.4x x = 0.5 (b)(ii) P(A B) = P(A)  P(B) =(0.6)(0.5) = 0.3 (c) A and B are not mutually exclusive because P(A B)  0. • P(AB) = P(A) + P(B) – P(A B) • Independent: • P(AB) = P(A) + P(B) – P(A)  P(B) • Defn of Independent events: P(A  B) = P(A)  P(B)

  6. 07/5/MATME/SP1/ENG/TZ1/ Complement Combined events Mutually exclusive events Conditional Independent events • P(AB) = P(A) + P(B) – P(A B)

  7. 07/5/MATME/SP1/ENG/TZ1/ Conditional probability (c) (a) (b) P(B) = 1 – P’(B) 3SF Conditional probability: P(A  B) = P(A)  P(B|A)

  8. Two standard six-sided dice are tossed. A diagram representing the sample space is shown below. 2008/SP1/TZ0/Section B Let X be the sum of the scores on the two dice. (a) Find (b) Find the value of k for which Elena’s expected number of points is zero

  9. 2008/SP1/TZ0/ Section B Probability (a) (i) number of ways of getting X = 6 is 5 (ii) number of ways of getting X > 6 is 21 (iii) Conditional probability: P(A  B) = P(A)  P(B|A)

  10. 2008/SP1/TZ0/ Section B Probability (b) number of ways of getting X < 6 is 10 • Probability

  11. 2008/SP1/TZ0/ Section B Probability Probability distribution table Sum = 1 Expectation or mean value or expected mean

  12. Probability Assignment #42 Review Paper #1 day8

  13. 08/5/MATME/SP1/ENG/TZ2 ans ans ans

  14. 08/5/MATME/SP1/ENG/TZ2 Probability ? • (i) P(male  T) = P(male) + P(T) – P(Male T) • P(male or Tennis) • (a) (ii) P(not football | female) Conditional probability: P(A  B) = P(A)  P(B|A)

  15. 08/5/MATME/SP1/ENG/TZ2 Probability • (b) ? P(2nd Fb’|1stFb’) = P(1stFb’) = 0.4 Conditional probability

  16. answers 2010/SP1/TZ1/ Consider the events A and B, where P(A) = 0.5, P(B) = 0.7 and P(AB) =0.3. The Venn diagram below shows the events A and B, and the probabilities p, q, and r.

  17. 2010/SP1/TZ1/ Probability • (a) • (i) p = 0.2 • (ii) q = 0.4 • (iii) r = 0.1 • (b) P(A|B ’) = • (c) Independent  P(AB) = P(A)P(B) • P(AB) = 0.3 • P(A)P(B) = (0.5)(0.7) = 0.35 • 0.3  0.35 so A and B are not independent. ? .5 - .3 = .2 .7 - .3 = .4 .3 1 – (.2 + .3 + .4) = .1 • Defn of Independent events: P(A  B) = P(A)  P(B)

  18. 08/SP1/TZ1/ Section B ans ans ans ans ans ans

  19. 08/SP1/TZ1/ Section B ans Find the value of p, of s and of t. ans www.curiousminds.com.au

  20. 08/SP1/TZ1/ Section B Probability ? (a) (c) (i) Score = 2 (b) Score = 3 Score = 2 Intersection

  21. 08/SP1/TZ1/ Section B ans ans ans ans

  22. 08/SP1/TZ1/ Section B Probability ? (c) (ii) Score = 2 Score = 3 Score = 2 Intersection & union

  23. 08/SP1/TZ1/ Section B Probability ? (d) (i) Sum = 1 (d) (ii) Expectation or mean value or expected mean

  24. 08/SP1/TZ1/ Section B Probability ? (e) List all possible outcomes

  25. 2010/SP1/TZ2 2010/SP1/TZ2 ans ans ans ans

  26. 2010/SP1/TZ2 Probability ? (a) (i) P(EF) P(E’F) (a) (ii) P(F) = P(EF) + P(E’F) Intersection & union

  27. 2010/SP1/TZ2 Probability ? (b) (i) P(EF’) (b) (ii) P(E | F) = Conditional probability: P(A  B) = P(A)  P(B|A)

  28. 2010/SP1/TZ2 Probability ? (c) List all possible outcomes

  29. 2010/SP1/TZ2 Probability ? (c) Sum = 1 (d) Expectation or mean value or expected mean

  30. 09/SP1/TZ0 Section B ans ans ans ans ans

  31. 09/SP1/TZ0 Section B 09/SP1/TZ0 Section B

  32. Probability 09/SP1/TZ0 Section B (a) (i) n(AB) = n(A) + n(B) – n(A  B) 100 = 130 – n(A  B) n(A  B) = 130 – 100 = 30 boys play both (a) (ii) 45 boys play only rugby n(B) – n(A  B) = 75 – 30 = 45 ? Universe: 100 boys A: Football 55 B : Rugby 75 – n(A  B) + 75 100 = 55 A  B both ? 45 ? 30 ? Neither: 0 Venn diagram: Union & intersection

  33. Probability 09/SP1/TZ0 Section B (b) (i) 25 boys play only football n(A) – n(A  B) = 55 – 30 = 25 n(Boys who play only 1 sport) = 100 – (boys who play both) = 100 – 30 = 70 P(he plays only 1 sport) = = 0.70 ? Universe: 100 boys A: Football 55 B : Rugby 75 AB ? 25 45 30 Neither: 0 Venn diagram

  34. Probability 09/SP1/TZ0 Section B 70 (b) (ii) P(Rugby | he plays only 1 sport) “given” ? Universe: 100 boys who play only 1 sport Universe: 100 boys A: Football 55 B : Rugby 75 25 45 AB 25 45 30 Neither: 0 Conditional probability: P(A  B) = P(A)  P(B|A)

  35. Probability 09/SP1/TZ0 Section B (c) Events A and B are mutually exclusive iffP(A  B) is zero. So events A and B are not mutually exclusive. ? Universe: 100 boys A: Football 55 B : Rugby 75 P(A  B) = 0.30  0 AB 25 45 30 Neither: 0 • Mutually exclusive events: P(A  B) = 0

  36. Probability 09/SP1/TZ0 Section B (d) Events A and B are independent iff P(A  B) = P(A) P(B) 0.30 (0.55)(0.75) 0.30  0.4125 So events A and B are not independent. ? Universe: 100 boys A: Football 55 B : Rugby 75 AB 25 45 30 Neither: 0 • Defn of Independent events: P(A  B) = P(A)  P(B)

  37. End of Probability slides

  38. 09/SP1/TZ0 Section B 09/SP1/TZ0 Section B

  39. 09/SP1/TZ0 Section B 09/SP1/TZ0 Section B

  40. Statistics

  41. N09/SP1/TZ0/ N09/SP1/TZ0/

  42. M08/5/MATME/SP1/ENG/TZ2/ M08/5/MATME/SP1/ENG/TZ2/ STATISTICS

  43. 07/5/MATME/SP1/ENG/TZ1 07/5/MATME/SP1/ENG/TZ1 Statistics

  44. 2007/SP1/ENG/TZ1/ 2007/SP1/ENG/TZ1/ Statistics – Calculator

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