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Unformatted text preview: York University MATH 2030 3.0AF (Elementary Probability) Assignment 4 Solutions October 2008 Salisbury 2.1 No. 6 (a) Let X be the number of shots that hit the bullseye. Then X Bin(8 , . 7) so P ( X = 4) = ( 8 4 ) (0 . 7) 4 (0 . 3) 4 = 70 . 2401 . 0081 = 0 . 13614 (b) P ( X = 4  X 2) = P ( X = 4 and X 2) /P ( X 2) = P ( X = 4) /P ( X 2). Weve already calculated the numerator, and P ( X 2) = 1 P ( X = 0) P ( X = 1) = 1 (0 . 3) 8 8(0 . 7)(0 . 3 7 ) = 1 . 0000657 . 001225 = 0 . 99871; Thus P ( X = 4  X 2) = 0 . 13614 / . 99871 = . 13631 (c) Let Y Bin(6 , . 7) be the number of bullseyes in the last 6 shots. Then P ( X = 4  1st two shots are bullseyes) = P ( Y = 2) = ( 6 2 ) (0 . 7) 2 (0 . 3) 4 = 15 . 49 . 0081 = 0 . 05954 2.2 No. 8 Let X be the number of sixes in 600 rolls of a die, so X Bin( n,p ) where n = 600 and p = 1 / 6. Then X Y where Y is normal, with = np = 100 and = p np (1 p ) = 9 . 1827; The crude approximation P ( X = 100) P ( Y = 100) = 0 is not very interesting. But we get a better answer using the continuity correction....
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This note was uploaded on 03/12/2009 for the course MATH 2030 taught by Professor Salisbury during the Fall '08 term at York University.
 Fall '08
 Salisbury
 Probability

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