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DS212_HW4_Solutions

# DS212_HW4_Solutions - DS 212 Business Statistics Soorapanth...

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DS 212 Business Statistics Soorapanth Homework #4 Solutions 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} 4.6 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 1

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DS 212 Business Statistics Soorapanth 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 4.11 A = event of having flown in an airplane at least once T = event of having 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.
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DS212_HW4_Solutions - DS 212 Business Statistics Soorapanth...

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