text: 320_hw2_ans
  
  # 320_hw2_ans - To make the table we need to find M O P ∩ M...
  
  This preview shows pages 1–2. Sign up to view the full content.
  
  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
  
  This preview has intentionally blurred sections. Sign up to view the full version.
  
  View Full Document
  This is the end of the preview. Sign up to access the rest of the document.
  
  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....
  View Full Document
  
  ## This homework help was uploaded on 04/22/2008 for the course ECON 320 taught by Professor Martin during the Spring '08 term at Allegheny.
  
  ### Page1 / 2
  
  320_hw2_ans - To make the table we need to find M O P ∩ M...
  
  This preview shows document pages 1 - 2. Sign up to view the full document.
  
  View Full Document
  Ask a homework question - tutors are online
explanation: The provided text is a solution set for a problem set, specifically Problem Set 2, which contains answers to several probability questions. Each problem is numbered and includes a detailed solution with calculations and explanations. The text does not contain any standalone questions but rather provides solutions to the problems.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 320_hw2_ans
  
  # 320_hw2_ans - To make the table we need to find M O P ∩ M...
  
  This preview shows pages 1–2. Sign up to view the full content.
  
  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
  
  This preview has intentionally blurred sections. Sign up to view the full version.
  
  View Full Document
  This is the end of the preview. Sign up to access the rest of the document.
  
  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....
  View Full Document
  
  ## This homework help was uploaded on 04/22/2008 for the course ECON 320 taught by Professor Martin during the Spring '08 term at Allegheny.
  
  ### Page1 / 2
  
  320_hw2_ans - To make the table we need to find M O P ∩ M...
  
  This preview shows document pages 1 - 2. Sign up to view the full document.
  
  View Full Document
  Ask a homework question - tutors are online
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a solution set for a problem set, specifically Problem Set 2, which contains answers to several probability questions. Each problem is numbered and includes a detailed solution with calculations and explanations. The text does not contain any standalone questions but rather provides solutions to the problems.
  
  Label: no-question