HW3Sol - P[v] =P[v|I]P[I] + P[v|L]P[L] + P[v|C]P[C] = .4862...

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EE/NIS 605-CS505 Probability and Stochastic Processes I Homework 3 Solutions Professor Heffes Textbook: S. Ross, A First Course in Probability, Seventh Edition , Prentice Hall, 2006 Chapter 3 : Theoretical Exercises: 6; Self Test Exercises: 4; Problems: 16,18,20,308,66a TE6 : Where we have used DeMorgan’s Law and the fact that if the E i ’s are independent then so are the E i c . ST4 : P16 : P[survive delivery]=P[sd]= .98, P[C cection]=P[C] =.15, P[sd|C]=.96. Find P[sd|C c ] P18 : (a) (b) Similarly (c) (d) From law of total probability P[W transferred | W drawn] = P[W drawn|W transferred]P[W transferred] P[W drawn|W transferred]P[W transferred] + P[W drawn|B transferred]P[B transferred] P[W transferred | W drawn] = 2 7 × 2 3 2 7 × 2 3 + 1 7 × 1 3 = 4 5 P[sd] = P[sd|C]P[C] + P[sd|C c ]P[C c ] .98 = .96 × .15 + P[sd|C c ] × .85 P[sd|C c ]= .9835 P[I|v] = P[v|I]P[I] P[v|I]P[I] + P[v|L]P[L] + P[v|C]P[C] P[I|v] = .35 × .46 .35 × .46 + .62 × .30 + .58 × .24 = .161 .4862 = .331 P[L|v] = .62 × .30 .35 × .46 + .62 × .30 + .58 × .24 = .383 P[C|v] = .58 × .24 .35 × .46 + .62 × .30 + .58 × .24 = .286
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Unformatted text preview: P[v] =P[v|I]P[I] + P[v|L]P[L] + P[v|C]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 ]) i = 1 n 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[F|CS] = P[FandCS] P[CS] = .02 .05 = .4 (b) P[CS|F] = P[FandCS] P[F] = .02 .52 = .038 P[black] = 1 2 1 2 + 2 3 1 2 = 7 12 P[black] = P[black|box1]P[box1] + P[black|box2]P[box2] = P[black|box1] 1 2 + P[black|box2] 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[box1|white] = P[white|box1]P[box1] P[white] = 1 2 1 2 1 7 12 = 3 5...
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HW3Sol - P[v] =P[v|I]P[I] + P[v|L]P[L] + P[v|C]P[C] = .4862...

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