STAT 400 Lecture AL1 Answers for 1.3 Spring 2015 Dalpiaz The conditional probability of A, given B (the probability of event A, computed on the assumption that event B has happened) is P ( A ∩ B ) P ( A B ) = ( assuming P ( B ) ≠ 0 ). P ( B ) Similarly, the conditional probability of B, given A is P ( A ∩ B ) P ( B A ) = ( assuming P ( A ) ≠ 0 ). P ( A ) 3. (continued) The probability that a randomly selected student at Anytown College owns a bicycle is 0.55, the probability that a student owns a car is 0.30, and the probability that a student owns both is 0.10. C C' B 0.10 0.45 0.55 B' 0.20 0.25 0.45 0.30 0.70 1.00 P(B) = 0.55, P(C) = 0.30, P(B ∩ C) = 0.10. a) What is the probability that a student owns a bicycle, given that he/she owns a car? P( B | C ) = 0.10/0.30 = 1/3. b) Suppose a student does not have a bicycle. What is the probability that he/she has a car? P( C | B' ) = 0.20/0.45 = 4/9.5. (continued) Suppose P ( A ) = 0.22, P ( B ) = 0.25, P ( C ) = 0.28, P ( A ∩ B ) = 0.11, P ( A ∩ C ) = 0.05, P ( B ∩ C ) = 0.07, P ( A ∩ B ∩ C ) = 0.01. Find the following: a) P ( B | A ) P ( B | A ) = 0.11/0.22 = 0.50. b) P ( B | C ) c) P ( B ∩ C | A ) P ( B | C ) = 0.07/0.28 = 0.25. P ( B ∩ C | A ) = 0.01/0.22 = 1/22. d) P ( B ∪ C | A ) e) P ( C | A ∪ B ) P ( B ∪ C | A ) = 0.15/0.22 = 15/22. P ( C | A ∪ B ) = 0.11/0.36 = 11/36. f) P ( C | A ∩ B ) g) P ( A ∩ B ∩ C | A ∪ B ∪ C ) P ( C | A ∩ B ) = 0.01/0.11 = 1/11. P ( A ∩ B ∩ C | A ∪ B ∪ C ) = 0.01/0.53 = 1/53. Multiplication Law of Probability If A and B are any two events, then P ( A ∩ B ) = P ( A ) ⋅ P ( B A ) P ( A ∩ B ) = P ( B ) ⋅ P ( A B )8. It is known that 30% of all the students at Anytown College live off campus. Suppose also that 48% of all the students are females. Of the female students, 25% live off campus. P( Off ) = 0.30, P( F ) = 0.48, P( Off | F ) = 0.25. a) What is the probability that a randomly selected student is a female and lives off campus? P( F ∩ Off ) = P( F ) × P( Off | F ) = 0.48 × 0.25 = 0.12. Off On F 0.12 0.36 0.48 M 0.18 0.34 0.52 0.30 0.70 1.00 b) What is the probability that a randomly selected student either is a female or lives off campus, or both? P( F ∪ Off ) = P( F ) + P ( Off ) – P( F ∩ Off ) = 0.48 + 0.30 – 0.12 = 0.66. OR P( F ∪ Off ) = P( F ∩ Off ) + P( F' ∩ Off ) + P( F ∩ Off' ) = 0.12 + 0.18 + 0.36 = 0.66. OR P( F ∪ Off ) = 1 – P( F' ∩ Off' ) = 1 – 0.34 = 0.66. c) What proportion of the off-campus students are females? P( F | Off ) = 0.12/0.30 = 0.40. d) What proportion of the male students live off campus? P( Off | M ) = 0.18/0.52 = 9/26 ≈ 0.346154.9. Suppose that Joe's Discount Store has received a shipment of 25 television sets, 5 of which are defective. On the following day, 2 television sets are sold. a) Find the probability that both of the television sets are defective. P( both defective ) = P( 1st D ∩ 2nd D ) = P( 1st D ) × P( 2nd D | 1st D ) = 5/25 × 4/24 = 1/30 . b) Find the probability that at least one of the two television sets sold is defective. D D 5/25 × 4/24 = 1/30. D D' 5/25 × 20/24 = 5/30. D' D 20/25 × 5/24 = 5/30. × D' D' P( at least one D ) = 1/30 + 5/30 + 5/30 = 11/30 . OR P( at least one D ) = 1 – P( D' D' ) = 1 – 20/25 × 19/24 = 1 – 19/30 = 11/30 . 10. Cards are drawn one-by-one without replacement from a standard 52-card deck. What is the probability that … a) … both the first and the second card drawn are ♥’s? P( 1st ♥ ∩ 2nd ♥ ) = P( 1st ♥ ) × P( 2nd ♥ | 1st ♥ ) = 13/52 × 12/51 = 1/17 .b) … the first two cards drawn are a ♥ and a ♣ ( or a ♣ and a ♥ ) ? P( 1st ♥ ∩ 2nd ♣ ) + P( 1st ♥ ∩ 2nd ♣ ) = 13/52 × 13/51 + 13/52 × 13/51 . c) … there are at least two ♥’s among the first three cards drawn? . ♥ ♥ ♥ . 13/52 × 12/51 × 39/50 or + ♥ ♥ ♥ 13/52 × 39/51 × 12/50 or + ♥ ♥ ♥ 39/52 × 13/51 × 12/50 or + ♥ ♥ ♥ 13/52 × 12/51 × 11/50 19968/132600 ≈
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