hw2sol - Engineering 0020: Probability and Statistics for...

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Unformatted text preview: Engineering 0020: Probability and Statistics for Engineers 1 Spring Term-2007; Solutions: Homework #2 6. a S = { (1,2,3), (1,2,4), (1,2,5), (2,1,3), (2,1,4), (2,1,5), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3), (4), (5) } b A = { (3), (4), (5) } c B = { (1,2,5), (2,1,5), (1,5), (2,5), (5) } d C = { (3), (4), (5), (2,3),(2,4), (2,5) } 8. a A 1 A 2 A 3 b A 1 A 2 A 3 c A 1 A 2 A 3 d (A 1 A 2 A 3 ) (A 1 A 2 A 3 ) (A 1 A 2 A 3 ) e A 1 (A 2 A 3 ) Note : Read as OR and as AND to reinforce your understanding! 22. Define A 1 Stops at first signal and A 2 Stops at second signal Then P(A 1 )=0.4, P(A 2 )=0.5 and P(A 1 A 2 )=0.6 a. Since P(A 1 A 2 )=P(A 1 )+P(A 2 )- P(A 1 A 2 ), we have P ( A 1 A 2 ) = P(A 1 )+P(A 2 )- P(A 1 A 2 ) = 0.4+0.5-0.6 = 0.3 b. The event A 1 = (A 1 A 2 ) (A 1 A 2 ) since the events on the right are mutually exclusive, P(A 1 ) = P (A 1 A 2 ) + P (A 1 A 2 ), i.e., P ( A 1 A 2 ) = P(A 1 ) - P(A 1 A 2 ) = 0.4-0.3 = 0.1 c. P(A 1 A 2 )+P(A 1 A 2 ) = {P(A 1 ) - P(A 1 A 2 )} +{ P(A 2 ) - P(A 1 A 2 )} = (0.4-0.3) + (0.5-0.3) = 0.3 Alternatively P(exactly one of A 1 or A 2 ) = P(A 1 A 2 ) - P(A 1 A 2 ) = 0.6-0.3 = 0.3 A 1 A 2 0.1 0.3 0.2 A 1 A 2 A 3 Let us denote 25. A-automatic transmission, B-sunroof, C-stereo-CD To find the various probabilities for a Venn diagram, let us work backwards from what is given: (i) 0.70 (ii) 0.80 (iii) 0.75 (iv) 0.85 (v) 0.90 (vi) 0.95 (vii) 0.98 From (vii) and (vi): From (vii) and (v): From (vii) and (vi): Pr(only A)= 0.98-0.95= 0.03 Pr(only B)= 0.98-0.90= 0.08 Pr(only C)= 0.98-0.85= 0.13 (viii) (ix) (x) From (vii) and the above three statements it follows that (Pr(more than one event)) Pr(A&B OR A&C OR B&C OR A&B&C)= 0.98-(0.03+0.08+0.13) = 0.74 (xi) Also, From (viii) and (i): Pr(A&B OR A&C OR A&B&C) = 0.70-0.03= 0.67 (xii) From (ix) and (ii): Pr(A&B OR B&C OR A&B&C) = 0.80-0.08= 0.72 (xiii) From (x) and (iii): Pr(A&C OR B&C OR A&B&C) = 0.75-0.13= 0.62 (xiv) So, From (xi) and (xii): Pr(B&C ONLY) = 0.74-0.67 =0.07 (xv) From (xi) and (xiii): Pr(A&C ONLY) = 0.74-0.72 =0.02 (xvi) From (xi) and (xiv): Pr(A&B ONLY) = 0.74-0.62 =0.12 (xvii) Finally, from (xv), (xvi), (xvii) and (xi), it follows that A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C Pr (A&B&C) = 0.74-(0.07+0.02+0.12)=0.53%. Pr (A&B&C) = 0....
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hw2sol - Engineering 0020: Probability and Statistics for...

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