mont4e_sm_ch02_sec03 - (b There are 36 experiments that use...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (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 c) P( A ∩ B ) = 0 , because A ∩ B = ∅ d) P( ( A ∪ B ) ∩ C ) = 0, because ( A ∪ B ) ∩ C = ( A ∩ C ) ∪ ( B ∩ C ) e) P( A′ ∩ B′ ∩ C ′ ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1 =∅ 2-68. (a) P(Caused by sports) = P(Caused by contact sports or by noncontact sports) = P(Caused by contact sports) + P(Caused by noncontact sports) =0.46+0.44 =0.9 (b) 1- P(Caused by sports)=0.1. 2-69.a) 70/100 = 0.70 b) (79+86-70)/100 = 0.95 c) No, P( A ∩ B ) ≠ 0 2-70. (a) P(High temperature and high conductivity)= 74/100 =0.74 (b) P(Low temperature or low conductivity) = P(Low temperature) + P(Low conductivity) – P(Low temperature and low conductivity) =(8+3)/100 + (15+3)/100 – 3/100 =0.26 (c) No, they are not mutually exclusive. Because P(Low temperature) + P(Low conductivity) =(8+3)/100 + (15+3)/100 =0.29, which is not equal to P(Low temperature or low conductivity). a) 350/370 345 + 5 + 12 362 b) = 370 370 345 + 5 + 8 358 c) = 370 370 d) 345/370 2-71. 2-72.a) 170/190 = 17/19 b) 7/190 2-73.a) P(unsatisfactory) = (5+10-2)/130 = 13/130 b) P(both criteria satisfactory) = 117/130 = 0.90, No 2-74. (a) 5/36 (b) 5/36 (c) P ( A ∩ B ) = P( A) P( B) = 0.01929 (d) P ( A ∪ B ) = P( A) + P ( B ) = 0.2585 Section 2-4 2-75. a) P(A) = 86/100 b) P(B) = 79/100 2-14 ...
View Full Document

This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.

Ask a homework question - tutors are online