mont4e_sm_ch02_sec02

mont4e_sm_ch02_sec02 - Section 2-2 2-50. All outcomes are...

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Unformatted text preview: Section 2-2 2-50. All outcomes are equally likely a) P(A) = 2/5 b) P(B) = 3/5 c) P(A') = 3/5 d) P(A∪B) = 1 e) P(A∩B) = P(∅)= 0 a) P(A) = 0.4 b) P(B) = 0.8 c) P(A') = 0.6 d) P(A∪B) = 1 e) P(A∩B) = 0.2 2-51. 2-52. a) 0.5 + 0.2 = 0.7 b) 0.3 + 0.5 = 0.8 2-53. a) 1/10 b) 5/10 a) S = {1, 2, 3, 4, 5, 6} b) 1/6 c) 2/6 d) 5/6 a) S = {1,2,3,4,5,6,7,8} b) 2/8 c) 6/8 2-54. 2-55. 2-56. The sample space is {95, 96, 97,…, 103, and 104}. (a) Because the replicates are equally likely to indicate from 95 to 104 mL, the probability that equivalence is indicated at 100 mL is 0.1. (b) The event that equivalence is indicated at less than 100 mL is {95, 96, 97, 98, 99}. The probability that the event occurs is 0.5. (c) The event that equivalence is indicated between 98 and 102 mL is {98, 99, 100, 101, 102}. The probability that the event occurs is 0.5. The sample space is {0, +2, +3, and +4}. (a) The event that a cell has at least one of the positive nickel charged options is {+2, +3, and +4}. The probability is 0.35+0.33+0.15= 0.83. (b) The event that a cell is not composed of a positive nickel charge greater than +3 is {0, +2, and +3}. The probability is 0.17+0.35+0.33= 0.85. 2-57. 2-58.Total possible: 1016, Only 108 valid, P(valid) = 108/1016 = 1/108 2-59.3 digits between 0 and 9, so the probability of any three numbers is 1/(10*10*10); 3 letters A to Z, so the probability of any three numbers is 1/(26*26*26); The probability your license plate is chosen is then (1/103)*(1/263) = 5.7 x 10-8 2-60.a) 5*5*4 = 100 b) (5*5)/100 = 25/100=1/4 2-61. (a) The number of possible experiments is 4 + 4 × 3 + 4 × 3 × 3 = 52 2-12 (b) There are 36 experiments that use all three steps. The probability the best result uses all three steps is 36/52 = 0.6923. (a) No, it will not change. With k amounts in the first step the number of experiments is k + 3k + 9k = 13k. The number of experiments that complete all three steps is 9k out of 13k. The probability is 9/13 = 0.6923. 2-62. a) P(A) = 86/100 = 0.86 b) P(B) = 79/100 = 0.79 c) P(A') = 14/100 = 0.14 d) P(A∩B) = 70/100 = 0.70 e) P(A∪B) = (70+9+16)/100 = 0.95 f) P(A’∪B) = (70+9+5)/100 = 0.84 2-63. a) P(A) = 30/100 = 0.30 b) P(B) = 77/100 = 0.77 c) P(A') = 1 – 0.30 = 0.70 d) P(A∩B) = 22/100 = 0.22 e) P(A∪B) = 85/100 = 0.85 f) P(A’∪B) =92/100 = 0.92 2-64. (a) The total number of transactions is 43+44+4+5+4=100 P ( A) = 44 + 4 + 4 = 0.52 100 100 − 5 = 0.95 100 44 + 4 + 4 (c) P ( A ∩ B ) = = 0.52 100 (d) P ( A ∩ B ' ) = 0 100 − 5 = 0.95 (e) P ( A ∪ B ) = 100 (b) P ( B ) = 2-65. a) Because E and E' are mutually exclusive events and E ∪ E ′ = S 1 = P(S) = P( E ∪ E ′ ) = P(E) + P(E'). Therefore, P(E') = 1 - P(E) b) Because S and ∅ are mutually exclusive events with S = S ∪ ∅ P(S) = P(S) + P(∅). Therefore, P(∅) = 0 c) Now, B = A ∪ ( A ′ ∩ B) and the events A and A ′ ∩ B are mutually exclusive. Therefore, P(B) = P(A) + P( A ′ ∩ B ). Because P( A ′ ∩ B ) ≥ 0 , P(B) ≥ P(A). Section 2-3 2-66. a) P(A') = 1- P(A) = 0.7 b) P ( A ∪ B ) = P(A) + P(B) - P( A ∩ B ) = 0.3+0.2 - 0.1 = 0.4 c) P( A ′ ∩ B ) + P( A ∩ B ) = P(B). Therefore, P( A ′ ∩ B ) = 0.2 - 0.1 = 0.1 d) P(A) = P( A ∩ B ) + P( A ∩ B ′ ). Therefore, P( A ∩ B ′ ) = 0.3 - 0.1 = 0.2 e) P(( A ∪ B )') = 1 - P( A ∪ B ) = 1 - 0.4 = 0.6 f) P( A ′ ∪ B ) = P(A') + P(B) - P( A ′ ∩ B ) = 0.7 + 0.2 - 0.1 = 0.8 a) P( 2-67. A ∪ B ∪ C ) = P(A) + P(B) + P(C), because the events are mutually exclusive. Therefore, P( A ∪ B ∪ C ) = 0.2+0.3+0.4 = 0.9 b) P ( A ∩ B ∩ C ) = 0, because A ∩ B ∩ C = ∅ 2-13 ...
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