ch04 e SOLUTION - 7.doc - Chapter 4 Probability 1 Chapter 4...

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Chapter 4: Probability 1 Chapter 4 Probability Ans Not Included is 4.30 4.1 Enumeration of the six parts: D 1 , D 2 , D 3 , A 4 , A 5 , A 6 D = Defective part A = Acceptable part Sample Space: D 1 D 2 , D 2 D 3 , D 3 A 5 D 1 D 3 , D 2 A 4 , D 3 A 6 D 1 A 4 , D 2 A 5 , A 4 A 5 D 1 A 5 , D 2 A 6 , A 4 A 6 D 1 A 6 , D 3 A 4 , A 5 A 6 There are 15 members of the sample space The probability of selecting exactly one defect out of two is: 9/15 = .60 4.2 X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9} and Z = {1, 2, 3, 4, 7,} a) X Z = {1, 2, 3, 4, 5, 7, 8, 9} b) X Y = {7, 9} c) X Z = {1, 3, 7} d) X Y Z = {1, 2, 3, 4, 5, 7, 8, 9} e) X Y Z = {7} f) (X Y) Z = {1, 2, 3, 4, 5, 7, 8, 9} {1, 2, 3, 4, 7} = {1, 2, 3, 4, 7} g) (Y Z) (X Y) = {2, 4, 7} {7, 9} = {2, 4, 7, 9} h) X or Y = X Y = {1, 2, 3, 4, 5, 7, 8, 9} i) Y and Z = Y Z = {2, 4, 7} 4.3 If A = {2, 6, 12, 24} and the population is the positive even numbers through 30, A’ = {4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30} 4.4 6(4)(3)(3) = 216 4.5 Enumeration of the six parts: D 1 , D 2 , A 1 , A 2 , A 3 , A 4 D = Defective part A = Acceptable part Sample Space: D 1 D 2 A 1 , D 1 D 2 A 2 , D 1 D 2 A 3 , D 1 D 2 A 4 , D 1 A 1 A 2 , D 1 A 1 A 3 , D 1 A 1 A 4 , D 1 A 2 A 3 , D 1 A 2 A 4 , D 1 A 3 A 4 , D 2 A 1 A 2 , D 2 A 1 A 3 , D 2 A 1 A 4 , D 2 A 2 A 3 , D 2 A 2 A 4 , D 2 A 3 A 4 , A 1 A 2 A 3 , A 1 A 2 A 4 ,A 1 A 3 A 4 , A 2 A 3 A 4 Combinations are used to counting the sample space because sampling is done without replacement. 6 C 3 = ! 3 ! 3 ! 6 = 20 Probability that one of three is defective is: 12/20 = 3/5 .60 There are 20 members of the sample space and 12 of them have 1 defective part. 4.6 10 7 = 10,000,000 different numbers 4.7 20 C 6 = ! 14 ! 6 ! 20 = 38,760 It is assumed here that 6 different (without replacement) employees are to be selected. 4.8 P(A) = .10, P(B) = .12, P(C) = .21 P(A C) = .05 P(B C) = .03 a) P(A C) = P(A) + P(C) - P(A C) = .10 + .21 - .05 = .26 b) P(B C) = P(B) + P(C) - P(B C) = .12 + .21 - .03 = .30 c) If A, B mutually exclusive, P(A B) = P(A) + P(B) = .10 + .12 = .22 4.9 D E F A 5 8 12 25 B 10 6 4 20 C 8 2 5 15 23 16 21 60 a) P(A D) = P(A) + P(D) - P(A D) = 25/60 + 23/60 - 5/60 = 43/60 = .7167 b) P(E B) = P(E) + P(B) - P(E B) = 16/60 + 20/60 - 6/60 = 30/60 = .5000 c) P(D E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500 d) P(C F) = P(C) + P(F) - P(C F) = 15/60 + 21/60 - 5/60 = 31/60 = .5167
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Chapter 4: Probability 2 4.10 E F A .10 .03 .13 B .04 .12 .16 C .27 .06 .33 D .31 .07 .38 .72 .28 1.00 a) P(A F) = P(A) + P(F) - P(A F) = .13 + .28 - .03 = .38 b) P(E B) = P(E) + P(B) - P(E B) = .72 + .16 - .04 = .84 c) P(B C) = P(B) + P(C) =.16 + .33 = .49 d) P(E F) = P(E) + P(F) = .72 + .28 = 1.000 4.11 A = event - flown in an airplane at least onceT = event - ridden in a train at least once P(A) = .47 P(T) = .28 P (ridden either a train or an airplane) = P(A T) = P(A) + P(T) - P(A T) = .47 + .28 - P(A T) Cannot solve this problem without knowing the probability of the intersection. We need to know the
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  • Summer '18
  • Keyur Popat
  • Probability theory, Zagreb, Zagreb bypass, Initialisms, = P

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