hw5_soln6 - HW Set 5: Due 5/4 in quiz session. HW12: hw_U,...

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HW Set 5: Due 5/4 in quiz session. HW12: hw_U, hw_V, hw_W (lecture 14), 5.10, 5.11 (warning: answer has typos). HW13: hw_X, hw_Y (lecture 15), 5.14, 5.59. HW14: 5.20, 5.21 (Warning: hard!), 5.24 (Note: Venn diagrams are useful for illustrating events, not probs. So, the hint given in the problem is misleading). HW15: 5.46, 5.51 (a,b,c, warning: typos), 5.54, 5.56. Graded problems (and points): TBA. Hw_V. A Venn diagram of these three events is (a) The event ‘at least one plant is completed by the contract date’ is represented by the shaded area covered by all three circles: (b) The event ‘all plants are completed by the contract date’ is the shaded area where all three circles overlap: (c) The event ‘none of the plants is completed by the contract date’ is the complement of the shaded area in (a):
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(d) The event ‘only the plant at site 1 is completed by the contract date’ is shown shaded: (e) The event ‘exactly one of the three plants is completed by the contract date is: (f) The event ‘either the plant at Site 1 or Site 2 or both plants are completed by the contract date’ is:
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10. For i = 1, 2, 3, . .. , 10, let A i denote the event ‘component i functions correctly’. The problem indicates that P(A i ) = .999 for each component which, by the law of complementary events, means that each P(A i ) = 1-.999 = .001. For a series system built from these 10 components to function correctly, all ten must function, so P(system functions correctly) = P(A 1 and A 2 and A 3 and . .. and A k ). According to the problem, this probability exceeds 1 - [P(A 1 ) + P(A 2 ) + P(A 3 ) + . .. + P(A k )] = 1 - [(.001) + (.001) + . .. + (.001)] = 1 - 10(.001) = 1 - .01 = .99. That is, there is at least a 99% probability that the system will function correctly. 11. Letting A i denote the event that the i th component fails (note that this is different from the definition of A i used in problem 5.10), the probability that the entire series system fails is denoted by P(A 1 or A 2 or A 3 or . .. or A k ). Given that each P(A i ) = .01, the problem states that P(A 1 or A 2 or A 3 or . .. or A k ) P(A 1 ) + P(A 2 ) + P(A 3 ) + . .. + P(A k ) = 5(.01) = .05. That is, there is at most a 5% chance of system failure. Hw_Y (a) Note that this question is equivalent to asking ‘what is the probability A (or B, etc.) is chosen given that we know E is not chosen’. That is, the new probabilities are now conditional probabilities, where the conditioning event is that E is not chosen. Therefore, P(A E ) = P(A and E )/P(E ) = P(A)/P(E ) = .20/(1-.10) = .20/.90 = 20/90. P(B
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hw5_soln6 - HW Set 5: Due 5/4 in quiz session. HW12: hw_U,...

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