~HW3_Sol - ECEN 646 Homework 3 Solutions 1 C ¸inlar...

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ECEN 646 Homework 3 Solutions 1 C ¸inlar Exercise (5.1), Chapter 3 (a) ( X 1 ,...,X 8 ) = (1 , 0 , 0 , 0 , 1 , 1 , 0 , 1) (b) ( N 0 ,...,N 8 ) = (0 , 1 , 1 , 1 , 1 , 2 , 3 , 3 , 4) (c) ( T 0 ,...,T 4 ) = (0 , 1 , 5 , 6 , 8) 2 C ¸inlar Exercise (5.3), Chapter 3 (a) This is the probability that, out of the first four vehicles, only the 3rd vehicle turns left. It has probability (1 - 0 . 62)(1 - 0 . 62)(0 . 62)(1 - 0 . 62) = 0 . 0340. (b) This is the probability that the first vehicle turned left, exactly one of the 2nd or 3rd turned left, and the 4th turned right. There are two ways for this to happen, so the probability is 2(0 . 62) 2 (1 - 0 . 62) 2 = 0 . 1110. (c) This is the probability that 6 of the first 8 vehicles turned left, and then 12 - 6 = 6 of the next 7 turned left. Since these two are independent (the drivers act independently), the probability is, P { N 8 = 6 ,N 15 = 12 } = P { N 8 = 6 ,N 15 - N 8 = 6 } = P { N 8 = 6 ,N 7 = 6 } = parenleftbigg 8 6 parenrightbigg (0 . 62) 6 (1 - 0 . 62) 2 × parenleftbigg 7 6 parenrightbigg (0 . 62) 6 (1 - 0 . 62) 1 = 0 . 0347 3 C ¸inlar Exercise (5.5), Chapter 3 (a) E [ N 3 ] = E [ X 1 + X 2 + X 3 ] = 3 p = 3(0 . 8) = 2 . 4 E [ N 7 ] = 7(0 . 8) = 5 . 6 E [ N 3 + 4 N 7 ] = E [ N 3 ] + 4 E [ N 7 ] = 2 . 4 + 4(5 . 6) = 24 . 8 (b) Var ( N 3 ) = 3 pq = 3(0 . 8)(0 . 2) = 0 . 48 1 https://www.coursehero.com/file/6709625/HW3-Sol/
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