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Unformatted text preview: ) P ( X ≥ 1) = 1 − P( X = 0) P ( X = 0) = ( )( ) = ()
5 0 135 20 140 20 135! 20!115! 140! 20!120! = 135!120! = 0.4571 115!140! P ( X ≥ 1) = 1 − 0.4571 = 0.5429
3103. N=300 (a) K=243, n=3, P(X=1)=0.087 (b) P(X≥1)=0.9934 (c) K=26+13=39 P(X=1)=0.297 (d) K=30018=282 P(X≥1)=0.9998
Let X denote the count of the numbers in the state's sample that match those in the player's sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6. 3104. a) ( )( ) = ⎛ 40! ⎞ = 2.61× 10 ( ) ⎜ 6!34! ⎟ ⎝ ⎠ ( )( ) = 6 × 34 = 5.31× 10 b) P ( X = 5) = () () ( )( ) = 0.00219 c) P ( X = 4) = ()
P( X = 6) =
6 6
6 5 6 4 34 0 40 6 34 1 40 6 34 2 40 6 −1 −7 −5 40 6 d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p = 320 1 3,838,380
3105. and E(Y) = 1/p = 3,838,380 weeks. This is more than 738 centuries! Let X...
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This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
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