# asst4 - York University MATH 2030 3.0AF(Elementary...

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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 , 0 . 7) so P ( X = 4) = ( 8 4 ) (0 . 7) 4 (0 . 3) 4 = 70 × 0 . 2401 × 0 . 0081 = 0 . 13614 (b) P ( X = 4 | X 2) = P ( X = 4 and X 2) /P ( X 2) = P ( X = 4) /P ( X 2). We’ve 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 - 0 . 0000657 - 0 . 001225 = 0 . 99871; Thus P ( X = 4 | X 2) = 0 . 13614 / 0 . 99871 = 0 . 13631 (c) Let Y Bin(6 , 0 . 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 × 0 . 49 × 0 . 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 μ = np = 100 and σ = 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. P ( X = 100) = P (99 . 5 X 100 . 5) P (99 . 5 Y 100 . 5) = P 99 . 5 - 100 9 . 1827 Y - 100 9 . 1827 100 . 5 - 100 9 . 1827 = P ( - 0 . 05445 Z 0 . 05445) = Φ(0 . 05445) - Φ( - 0 . 05445) = Φ(0 . 05445) - (1 - Φ(0 . 05445)) = 2Φ(0 . 05445) - 1 .

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