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320_hw2_ans

# 320_hw2_ans - To make the table we need to find M O P ∩ M...

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Solutions to Problem Set 2 1. A=MBA; B = Undergraduate; P(A)= 0.29, P(B)= 0.24, ) ( B A P =0.08 a. ) ( B A P = P(A)+P(B)- ) ( B A P =0.29+0.24-0.08=0.45 b. ) ( B A P =1- ) ( B A P =1-0.45=0.55 c. ) ( B A P + ) ( B A P = P(A)+P(B)- 2 ) ( B A P i. Here we subtract ) ( B A P two times to completely eliminate it: = 0 .29+0.24-0.16 =0.37 2. P(A) = 0.025, P(B) = 0.9, P(C) = 0.075. a. Events A and C are mutually exclusive, so the special rule of addition applies. P(AUC) = P(A) + P(C) = 0.10 3. P(C) =0.5; C = Air conditioning; P(P) =0 .49; P = Power steering; ) ( P C P =0.26 a. Venn table: b. ) ( P C P =0 .24 c. ) ( P C P = 0.27 4. P(O) =0.75 where O = on time, P(M | O ) = 0.80 where M= meet the specifications; and P ( M | O ) = .60 a. ) ( M O P = P(M | O ). P(O )=(0.80)(0.75) =0 .60

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Unformatted text preview: To make the table we need to find ) ( M O P ∩ ; ) ( M O P ∩ ; ) ( M O P ∩ = P(M | O ). P( O ) = (0.60)(0.25)=0.15; b. P(M) = 0.75 5. P(J) = 0.4 where J = Jacket requires alteration; P(T) = 0.3 where T= Trousers require alteration; and ) ( T J P ∩ =0.22 a. Note: ) ( T J P ∩ =1-) ( T J P ∪ ; ) ( T J P ∪ = P(J)+P(T)-) ( T J P ∩ =0.4 + 0.3 -0 .22 = 0.48 a. Therefore, ) ( T J P ∩ =1 - 0.48 =0 .52. C C P 0.26 0.23 0.49 P 0.24 0.27 0.51 0.5 0.5 O O M 0.60 0.15 0.75 M 0.15 0.10 0.25 0.75 0.25 b. ) ( T J P ∪-) ( T J P ∩ =0 .48-0.22=0.26....
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