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.
262. 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 = 0.84 263. 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 = 0.92 264. (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 ) =
265. 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 23 266. 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( 267. 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 = ∅ 213 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 =∅ 268. (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. 269.a) 70/100 = 0.70 b) (79+8670)/100 = 0.95 c) No, P( A ∩ B ) ≠ 0 270. (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) 345/370 271. 272.a) 170/190 = 17/19 b) 7/190 273.a) P(unsatisfactory) = (5+102)/130 = 13/130 b) P(both criteria satisfactory) = 117/130 = 0.90, No 274. (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 24 275. a) P(A) = 86/100 b) P(B) = 79/100 214 ...
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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.
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