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HW1Sol

HW1Sol - r = n x i> m i y i Σ i = 1 r = n – Σ i = 1 r...

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EE/NIS 605-CS505 Probability and Stochastic Processes I Homework 1 Solutions Professor Heffes Textbook: S. Ross, A First Course in Probability, Seventh Edition , Prentice Hall, 2006 Chapter 1 : TE * 20, TE21, Problems: 1, 5, 19a, 21, 22, 32 TE20 : The number of integer solutions of is the same as the number of nonnegative solutions to which, from proposition 6.2 is given by TE21 : There are Choices for the k of the x’s to set equal zero. For any of these choices, the other r-k of the x’s must be positive and sum to n. By proposition 6.1 there are such solutions. Hence the desired result follows. P1 : (a) (26) 2 (10) 5 (b) (26)(25)(10)(9)(8)(7)(6) P5 : There are 8(2)(9) = 144 possible area codes. There are 1(2)(9) = 18 possible area codes starting with 4. P19 a: n – m i Σ n = 1 r + r – 1 r – 1 x i Σ i = 1

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Unformatted text preview: r = n, x i > m i y i Σ i = 1 r = n – Σ i = 1 r m r k 8 3 4 3 + 2 1 4 2 = 896 possible committees n – 1 r–k – 1 = n – 1 n – r + k P21 : Each path is a string of 7 characters, 4R’s(Right) and 3 U’s(Up). The number of paths corresponds to the number of ways of selecting the character positions for either the 4R’s (or equivalently the 3U’s). e.g., RRUURUR. P22 : There are Therefore there are P32: (a) Number of nonnegative solutions to is (b) Professor Heffes 7 4 = 35 4 2 paths from A to the circle and 3 2 paths from the circle to B. 4 2 3 2 = 18 paths from A to B passing through the circlethe circle x i Σ i = 1 6 = 8 13 5 # nonnegative solutions to x i Σ i = 1 6 = 5 # nonnegative solutions to y i Σ i = 1 6 = 3 = 10 5 8 5...
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HW1Sol - r = n x i> m i y i Σ i = 1 r = n – Σ i = 1 r...

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