hw8-solutions

# So with probability 1 4 k 1 the numerator and

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), so with probability 1 / 4 k +1 the numerator and denominator before reduction were 1

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both multiples of 2 k but nor 2 k +1 . Summing over k , the probability that both the randomly chosen numerator and denominator are odd is 1 4 + 1 4 2 + 1 4 3 + · · · = 1 3 . 4. Dungeons and Dragons players use dice in the shape of each of the regular solids. The faces are always numbered 1 through the number of total faces there are. You shake all five dice. what is the probability of your throwing a total of 6? (B+S 7.2.31) Solution. There are five outcomes that have a total of 6: we must get 2 on one die and 1 on the other four dice. There are 4 · 6 · 8 · 12 · 20 = 46080 possible outcomes. So the probability of throwing a 6 is 5 / 46080 = 1 / 9216. 5. Suppose you flip a fair coin 10 times on two different ocassions. One time you see 10 heads, the other time you see HHTHHHTTHT . Is either one of these outcomes more likely than the other? Which one is random? Explain. (B+S 7.3.31) Solution. Neither outcome is more likely than the other; they both have probability 1 / 2 10 of occuring. Both are random! You might think the second one is “more random” then the first. 6. Suppose you deal three cards from a regular deck of 52 cards. What is the probability that they will all be jacks? (B+S 7.4.11) Solution. The probability that the first card dealt is a jack is 4 / 52; the probability that the second card dealt is a jack, given that the first one was a jack, is 3 / 51 (since there are 51 cards left, of which 3 are jacks); the probability that the third card dealt is a jack, given that the first two are jacks, is 2 / 50. So the answer is 4 · 3 · 2 52
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