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# HW1 - HW1 Solutions Dilip Raghavan September 6 2008 1...

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HW1 Solutions Dilip Raghavan September 6, 2008 1 Section 5.1 Problem 16. We model the outcomes as 4–tuples ( R 1 , W 1 , R 2 , W 2), where R 1 represents the first roll of the red die and R 2 the second, and similarly for W 1 and W 2. a. There are 6 possible outcomes for each of R 1, R 2, W 1 and W 2. So there are a total of 6 4 outcomes. b. We need R1 = R2 and W1 = W2. So if R1 and W1 are choosen, R2 and W2 are determined. There are 6 possible outcomes for each of R1 and W1. So there are 6 2 favorable outcomes. So the probability is no. of favorable outcomes no. of total outcomes = 6 2 6 4 . c. The sum can be anything from 2 to 12. It is best to break up the problem into cases depending on what the sum is. Notice that given a fixed value for the sum, if the value of R 1 is choosen, then the value of W 1 is automatically determined. So for any given sum, it is enough to figure out how many outcomes there are for R 1 and for R 2. (a) The sum is 2. R 1 can only be 1. Same for R 2. So 1 2 outcomes. (b) The sum is 3. R 1 can be either 1 or 2. Same for R 2. So 2 2 outcomes. (c) The sum is 4. R 1 can be anything from 1 to 3. Same for R 2. So 3 2 outcomes. (d) The sum is 5. R 1 can be anything from 1 to 4. Same for R 2. So 4 2 outcomes. (e) The sum is 6. R 1 can be anything from 1 to 5. Same for R 2. So 5 2 outcomes. (f) The sum is 7. R 1 can be anything from 1 to 6. Same for R 2. So 6 2 outcomes. (g) The sum is 8. R 1 can be anything from 2 to 6. Same for R 2. So 5 2 outcomes. (h) The sum is 9. R 1 can be anything from 3 to 6. Same for R 2. So 4 2 outcomes. (i) The sum is 10. R 1 can be anything from 4 to 6. Same for R 2. So 3 2 outcomes. (j) The sum is 11. R 1 can be anything from 5 to 6. Same for R 2. So 2 2 outcomes. (k) The sum is 12. R 1 has to be 6. Same for R 2. So 1 2 outcomes. So the number of favorable outcomes is: 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 + 5 2 + 4 2 + 3 2 + 2 2 + 1 2 . 1

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d. The number of outcome where R 1 + W 1 R 2 + W 2 can be gotten by subtracting the answer to part c. from the total number of outcomes. Since the situation is symmetrical, the number of outcomes where R 1 + W 1 < R 2 + W 2 is half of the number of outcomes where R 1 + W 1 R 2 + W 2. So the number of favorable outcomes is 1 2 6 4 - 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 + 5 2 + 4 2 + 3 2 + 2 2 + 1 2 . Problem 22. The first slot can have any of 0, 1 or 2. So there are 3 possibilities for the first slot. Every slot after the first one can have anything except what is in the slot right before it. So there are 2 possibilities for every slot after the first one. So there are a total of 3 × 2 9 outcomes. Problem 23. In all three problems it is best to first count the possibilities between 1,000 and 9,999 and then to deal with 10,000 separately.
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HW1 - HW1 Solutions Dilip Raghavan September 6 2008 1...

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