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Unformatted text preview: P[v] =P[vI]P[I] + P[vL]P[L] + P[vC]P[C] = .4862 P[ ∪ i = 1 n E i ] = 1 – P[ ∪ i =1 n E i c ] = 1 – P[ ∩ i =1 n E i c ] = 1 – P[E i c ] = 1 – (1 – P[E i ]) Π i = 1 n Π i = 1 n P20 : P30 : P66a : Let T be the event that both relays 1 and 2 are closed. Let B be the event that relays 3 and 4 are closed. Let C be the event that relay 5 is closed. Then we want Since all events are independent Professor Heffes (a) P[FCS] = P[FandCS] P[CS] = .02 .05 = .4 (b) P[CSF] = P[FandCS] P[F] = .02 .52 = .038 P[black] = 1 2 × 1 2 + 2 3 × 1 2 = 7 12 P[black] = P[blackbox1]P[box1] + P[blackbox2]P[box2] = P[blackbox1] 1 2 + P[blackbox2] 1 2 P[(T ∪ B) ∩ C] P[(T ∪ B) ∩ C] = P[(T ∪ B)]P[C] = P[T]+P[B] – P[T ∩ C] P[C] = P[T]+P[B] – P[T]P[C] P[C] P[(T ∪ B) ∩ C] = p 1 p 2 + p 3 p 4 – p 1 p 2 p 3 p 4 p 5 P[box1white] = P[whitebox1]P[box1] P[white] = 1 2 × 1 2 1 – 7 12 = 3 5...
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 Spring '08
 sss
 Probability, Probability theory, Professor Heffes, Self Test Exercises

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