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Bruce Hajek - Solutions

# Bruce Hajek - Solutions - Chapter 12 Solutions to Problems...

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Chapter 12 Solutions to Problems 1.2 Independent vs. mutually exclusive (a) If E is an event independent of itself, then P [ E ] = P [ E E ] = P [ E ] P [ E ]. This can happen if P [ E ] = 0 . If P [ E ] 6 = 0 then cancelling a factor of P [ E ] on each side yields P [ E ] = 1. In summary, either P [ E ] = 0 or P [ E ] = 1. (b) In general, we have P [ A B ] = P [ A ] + P [ B ] - P [ AB ]. If the events A and B are independent, then P [ A B ] = P [ A ] + P [ B ] - P [ A ] P [ B ] = 0 . 3 + 0 . 4 - (0 . 3)(0 . 4) = 0 . 58 . On the other hand, if the events A and B are mutually exclusive, then P [ AB ] = 0 and therefore P ] A B ] = 0 . 3 + 0 . 4 = 0 . 7. (c) If P [ A ] = 0 . 6 and P [ B ] = 0 . 8, then the two events could be independent. However, if A and B were mutually exclusive, then P [ A ] + P [ B ] = P [ A B ] 1, so it would not possible for A and B to be mutually exclusive if P [ A ] = 0 . 6 and P [ B ] = 0 . 8. 1.4 Frantic search Let D , T , B , and O denote the events that the glasses are in the drawer, on the table, in the brief- case, or in the office, respectively. These four events partition the probability space. (a) Let E denote the event that the glasses were not found in the first drawer search. P [ T | E ] = P [ TE ] P [ E ] = P [ E | T ] P [ T ] P [ E | D ] P [ D ]+ P [ E | D c ] P [ D c ] = (1)(0 . 06) (0 . 1)(0 . 9)+(1)(0 . 1) = 0 . 06 0 . 19 0 . 315 (b) Let F denote the event that the glasses were not found after the first drawer search and first table search. P [ B | F ] = P [ BF ] P [ F ] = P [ F | B ] P [ B ] P [ F | D ] P [ D ]+ P [ F | T ] P [ T ]+ P [ F | B ] P [ B ]+ P [ F | O ] P [ O ] = (1)(0 . 03) (0 . 1)(0 . 9)+(0 . 1)(0 . 06)+(1)(0 . 03)+(1)(0 . 01) 0 . 22 (c) Let G denote the event that the glasses were not found after the two drawer searches, two table searches, and one briefcase search. P [ O | G ] = P [ OG ] P [ G ] = P [ G | O ] P [ O ] P [ G | D ] P [ D ]+ P [ G | T ] P [ T ]+ P [ G | B ] P [ B ]+ P [ G | O ] P [ O ] = (1)(0 . 01) (0 . 1) 2 (0 . 9)+(0 . 1) 2 (0 . 06)+(0 . 1)(0 . 03)+(1)(0 . 01) 0 . 4225 1.6 Conditional probabilities–basic computations of iterative decoding (a) Here is one of several approaches to this problem. Note that the n pairs ( B 1 , Y 1 ) , . . . , ( B n , Y n ) 345

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346 CHAPTER 12. SOLUTIONS TO PROBLEMS are mutually independent, and λ i ( b i ) def = P [ B i = b i | Y i = y i ] = q i ( y i | b i ) q i ( y i | 0)+ q i ( y i | 1) . Therefore P [ B = 1 | Y 1 = y 1 , . . . , Y n = y n ] = X b 1 ,...,b n : b 1 ⊕···⊕ b n =1 P [ B 1 = b 1 , . . . , B n = b n | Y 1 = y 1 , . . . , Y n = y n ] = X b 1 ,...,b n : b 1 ⊕···⊕ b n =1 n Y i =1 λ i ( b i ) . (b) Using the definitions, P [ B = 1 | Z 1 = z 1 , . . . , Z k = z k ] = p (1 , z 1 , . . . , z k ) p (0 , z 1 , . . . , z k ) + p (1 , z 1 , . . . , z k ) = 1 2 Q k j =1 r j (1 | z j ) 1 2 Q k j =1 r j (0 | z j ) + 1 2 Q k j =1 r j (1 | z j ) = η 1 + η where η = k Y j =1 r j (1 | z j ) r j (0 | z j ) . 1.8 Blue corners (a) There are 24 ways to color 5 corners so that at least one face has four blue corners (there are 6 choices of the face, and for each face there are four choices for which additional corner to color blue.) Since there are ( 8 5 ) = 56 ways to select 5 out of 8 corners, P [ B | exactly 5 corners colored blue] = 24 / 56 = 3 / 7. (b) By counting the number of ways that B can happen for different numbers of blue corners we find P [ B ] = 6 p 4 (1 - p ) 4 + 24 p 5 (1 - p ) 3 + 24 p 6 (1 - p ) 2 + 8 p 7 (1 - p ) + p 8 . 1.10 Recognizing cumulative distribution functions (a) Valid (draw a sketch) P [ X 2 5] = P [ X ≤ - 5]+ P [ X 5] = F 1 ( - 5)+1 - F 1 ( 5) = e - 5 2 .
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