HW3Sol

# HW3Sol - P[v =P[v|I]P[I P[v|L]P[L P[v|C]P[C =.4862 P ∪ i...

This preview shows pages 1–2. Sign up to view the full content.

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

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: 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 ]) 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[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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online