First Course in Probability, A (7th Edition)

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Math 151 HW 1 Solutions Problems 1. With replacing: S= { (red, red),(red,green),(red,blue),(green,red),(green, green), (green, blue),(blue, red),(blue,green),(blue,blue) } Without replacing: S= { (red,green),(red,blue),(green,red), (green, blue),(blue, red),(blue,green) } 2. S = { 6 , x 6 , xx 6 , xxx 6 , ... } ∪ { no 6 appears } Here xxx 6 means that 6 appears on 4th trial. x is anything other than 6. x s on different positions may represent different numbers. Using this notation, E n = { n - 1 times x followed by 6 } And ( E n ) C ) = { no 6 appears } 3. E F = { (1,2), (1,4), (1,6), (2,1), (4,1), (6,1) } E F = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,5), (3,1), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,1), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5) } F G = { (1,4), (4,1) } E F C = { (2,3), (2,5), (3,2), (3,4), (3,6), (4,3), (4,5), (5,2), (5,4), (5,6), (6,3), (6,5) } E F G = { (1,4), (4,1) } 4. (a) This is similar to the solution of problem 2 above. If 1 comes up at positions 3 k + 1, 3 k + 2, 3 k + 3 then A,B,C wins, respectively. If all positions are zeros, it means nobody wins. (b) (i) A = { 1, 0001, 0000001,... } (ii) B = { 01, 00001, 00000001,... } (iii) ( A B ) C = { 001, 000001, 000000001... } ∪ { all zeros } 8. (a) P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = 0 . 3 + 0 . 5 - 0 = 0 . 8 (b) P ( A B C ) = P ( A ) = 0 . 3 (c) P ( A B ) = P ( ) = 0 25.
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  • Winter '08
  • Liu
  • Math, Probability, Trigraph, 3k, EF C GC E C F GC E C F C G E C F C, EF GC EF

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