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Events Influencing Each Other Conditional Probability Bayes Theorem Conditional Probability Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Conditional Probability Events

  1. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B � � � � � � � � � � � � � � � � � � logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  2. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B � � � � � � � � � � � � � � � � � � A ∩ B logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  3. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B � � � � � � � � � � � � � � � � � � A ∩ B logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  4. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B B � � � � � � � � � � � � � � � � � � A ∩ B logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  5. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B B � � � � � � � � � � � � � � � � � � A ∩ B logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  6. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B B � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � A ∩ B logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  7. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B B � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � A ∩ B A ∩ B logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  8. Events Influencing Each Other Conditional Probability Bayes’ Theorem Definition. For events A and B with P ( B ) > 0 , the conditional probability of A given that B has occurred is P ( A | B ) : = P ( A ∩ B ) . P ( B ) S A B B � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � A ∩ B A ∩ B For P ( A | B ) , the set B becomes the sample space. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  9. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  10. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  11. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  12. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” P ( B ) = 4 52 = 1 13 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  13. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” P ( B ) = 4 52 = 1 13, P ( A ∩ B ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  14. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 13, P ( A ∩ B ) = 52 · 51 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  15. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  16. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 P ( A | B ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  17. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 P ( A | B ) = P ( A ∩ B ) P ( B ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  18. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) 13 · 17 = 1 P ( B ) 13 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  19. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  20. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 P ( B ′ ) = 48 52 = 12 13 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  21. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 P ( B ′ ) = 48 52 = 12 13, P ( A ∩ B ′ ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  22. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 52 · 51 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  23. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  24. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  25. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) = P ( A ∩ B ′ ) P ( B ′ ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  26. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) = P ( A ∩ B ′ ) 16 13 · 17 = P ( B ′ ) 12 13 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  27. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) = P ( A ∩ B ′ ) 16 = 4 13 · 17 = P ( B ′ ) 12 51 13 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  28. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) = P ( A ∩ B ′ ) 16 = 4 13 · 17 = P ( B ′ ) 12 51 13 Proposition. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  29. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) = P ( A ∩ B ′ ) 16 = 4 13 · 17 = P ( B ′ ) 12 51 13 Proposition. Multiplication rule: P ( A ∩ B ) = P ( A | B ) · P ( B ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  30. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A = “the second card drawn from a deck is a queen” B = “the first card drawn from a deck is a queen” 4 · 3 P ( B ) = 4 52 = 1 1 13, P ( A ∩ B ) = 52 · 51 = 13 · 17 1 P ( A | B ) = P ( A ∩ B ) = 1 13 · 17 = 1 P ( B ) 17 13 13, P ( A ∩ B ′ ) = 48 · 4 P ( B ′ ) = 48 52 = 12 16 52 · 51 = 13 · 17 P ( A | B ′ ) = P ( A ∩ B ′ ) 16 = 4 13 · 17 = P ( B ′ ) 12 51 13 Proposition. Multiplication rule: P ( A ∩ B ) = P ( A | B ) · P ( B ) (Can be used to double check and to compute probabilities of intersections.) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  31. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  32. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  33. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  34. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  35. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  36. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . P ( A 1 ∩ A 2 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  37. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . P ( A 1 ∩ A 2 ) = P ( A 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  38. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . P ( A 1 ∩ A 2 ) = P ( A 1 ) P ( A 2 | A 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  39. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . P ( A 1 ∩ A 2 ) = P ( A 1 ) P ( A 2 | A 1 ) 2 = 3 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  40. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . P ( A 1 ∩ A 2 ) = P ( A 1 ) P ( A 2 | A 1 ) 2 3 · 1 = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  41. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Three cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all three cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 . P ( A 1 ∩ A 2 ) = P ( A 1 ) P ( A 2 | A 1 ) 2 3 · 1 = 2 1 = 3 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  42. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  43. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  44. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  45. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  46. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  47. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  48. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  49. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  50. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 1 ∩ A 2 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  51. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 1 ∩ A 2 ) = P ( A 3 | A 1 ∩ A 2 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  52. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 1 ∩ A 2 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 2 | A 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  53. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 1 ∩ A 2 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 2 | A 1 ) P ( A 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  54. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 1 ∩ A 2 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 2 | A 1 ) P ( A 1 ) 1 2 · 2 3 · 3 = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  55. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. Four cards are placed face down on a table. One of them is the queen of hearts. What is the probability that all four cards must be turned over to find the queen of hearts? A 1 : = first card not the queen of hearts. A 2 : = second card not the queen of hearts. A 3 : = third card not the queen of hearts. The event we are interested in is A 1 ∩ A 2 ∩ A 3 . P ( A 1 ∩ A 2 ∩ A 3 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 1 ∩ A 2 ) = P ( A 3 | A 1 ∩ A 2 ) P ( A 2 | A 1 ) P ( A 1 ) 1 2 · 2 3 · 3 = 4 1 = 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  56. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  57. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  58. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  59. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  60. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  61. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  62. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  63. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  64. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  65. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  66. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  67. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  68. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  69. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. P ( A 2 ) = 0 . 35 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  70. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. P ( A 2 ) = 0 . 35, P ( B | A 2 ) = 0 . 05. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  71. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. P ( A 2 ) = 0 . 35, P ( B | A 2 ) = 0 . 05. A 3 = purchased model 3. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  72. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. P ( A 2 ) = 0 . 35, P ( B | A 2 ) = 0 . 05. A 3 = purchased model 3. P ( A 3 ) = 0 . 25 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  73. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. P ( A 2 ) = 0 . 35, P ( B | A 2 ) = 0 . 05. A 3 = purchased model 3. P ( A 3 ) = 0 . 25, P ( B | A 3 ) = 0 . 15. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

  74. Events Influencing Each Other Conditional Probability Bayes’ Theorem Example. A car company sells three different models of compact cars. The probabilities that one of these models is selected (by a compact car buyer) are 0 . 40 , 0 . 35 and 0 . 25 , respectively. The probabilities that a buyer of any of these cars will buy the same model again are 0 . 30 , 0 . 05 and 0 . 15 , respectively. Given that a buyer has just purchased a compact car of the same model that was previously owned, what are the chances that the model was the first, second or third of the above models? Name the models 1, 2 and 3, respectively. B = purchased previously owned model again. A 1 = purchased model 1. P ( A 1 ) = 0 . 40, P ( B | A 1 ) = 0 . 30. A 2 = purchased model 2. P ( A 2 ) = 0 . 35, P ( B | A 2 ) = 0 . 05. A 3 = purchased model 3. P ( A 3 ) = 0 . 25, P ( B | A 3 ) = 0 . 15. What is P ( A i | B ) ? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Conditional Probability

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