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Unformatted text preview: 2-142-3810! = 3,628,8002-48 a) The total number of possible samples is There are 220 ways to sample 3 parts from 12.The number of samples that result in 1 nonconforming part (and 2 conforming parts) is Therefore, the requested probability is 90/220 = 0.409.b) The number of samples with no nonconforming part is The probability of at least one nonconforming part is 1.2-62a) P(A) = 86/100 = 0.86b) P(B) = 79/100 = 0.79c) P(A') = 14/100 = 0.14d) P(A∩B) = 70/100 = 0.70e) P(A∪B) = (70+9+16)/100 = 0.95f) P(A’∪B) = (70+9+5)/100 = 0.842-96N = event heart failure due to natural causes P(N) = 0.87O = event heart failure due to outside causes P(O) = 0.13I = induced substance P(I|O) = 0.73F = foreign objects P(F|O) = 0.27A = arterial blockage P(A|N) = 0.56D = disease P(D|N) = 0.27E = infection P(E|N) = 0.17a)Find P(I) = P(I|O)P(O) = (0.73)(0.13) = 0.0949b)Find P(DUE) = P(D) + P(E) = P(D|N)P(N) + P(E|N)P(N) = (0.27)(0.87) + (0.17)(0.87) = 0.38282-110(a) Let X = number of opponents defeated in a game. (a) Let X = number of opponents defeated in a game....
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This note was uploaded on 05/02/2008 for the course IE 360 taught by Professor Chen during the Spring '08 term at University of Louisville.
- Spring '08