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Unformatted text preview: ECEN 646 Homework 4 Solutions 1 C ¸inlar Exercise (5.11), Chapter 3 The rate of crossings is 4 vehicles per minute. So, the probability of a vehicle passing a point in a given second is 4 60 = 0 . 067. (a) p = P [ X n = 1] = P [a vehicle passing in a given second] = 1 15 = 0 . 067 (b) P [ T 4 T 3 = 12] = P [It takes 12 trials after the 3rd success for the 4th success to occur] = P [3rd success is followed by 11 failures and a success] = q 11 p = 0 . 0312 (c) E [ T 4 T 3 ] = E [time between the 3rd and the 4th successes] = E [time for the 1st success] = 1 p = 15 E [ T 13 T 3 ] = E [time between the 3rd and the 13th successes] = E [time for the first 10 successes] = 10 p = 150 (d) Var ( T 2 + 5 T 3 ) = Var (6 T 2 + 5( T 3 T 2 )) = Var (6 T 2 ) + Var (5( T 3 T 2 )) This follows from the independent increments property that makes T 2 and T 3 T 2 independent. We then use the fact that the variance of a sum of independent r.v’s equals the sum of the individual variances. Var (6 T 2 ) = 36Var ( T 2 ) = 36 2 q p 2 = 15120 Var (5( T 3 T 2 )) = 25Var ( T 3 T 2 ) = 25 q p 2 = 5250 Therefore Var ( T 2 + 5 T 3 ) = 20370. 1 2 C ¸inlar Exercise (5.12), Chapter 3 Let f ( a ,a 1 ,... a 7 ) be defined as P [ T 8 = 17  T = a ,T 1 = a 1 ...T 7 = a 7 ]. For the 8 th success to happen at the 17 th trial, we need the 7 th success to have happened earlier. So, the conditional probability above will be non zero only if the event T 7 ≤ 16 occurs. Assuming this event occurs, we have f ( a ,... a 7 ) = P [ X a 7 +1 = 0 ,... ,X 16 = 0 ,X 17 = 1] = pq 16 − a 7 Therefore P [ T 8 = 17  T ,T 1 ... T 7 ] = f ( T ,T 1 ,... T 7 ) = pq 16 − T 7 if the event T 7 ≤ 16 occurs. 3 C ¸inlar Exercise (5.18), Chapter 3 (a) For the duel to last exactly 13 rounds, both duelists should stay alive for the first 12 rounds and at least one of them should die in the 13 th round....
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This note was uploaded on 01/19/2012 for the course ECEN 646 taught by Professor Serapsavari during the Fall '11 term at Texas A&M.
 Fall '11
 SerapSavari

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