hw8-solutions

# Heads so the probability of getting three heads is 1

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heads, so the probability of getting three heads is 1 / 8. Three show exactly two heads, so the probability of getting two heads is 3 / 8. Now, if we’re told a president is showing, then there are seven equally likely outcomes (all but TTT ), of which one shows three heads, so the probability of getting three heads is 1 / 7. If we’re told Lincoln is showing, that means the penny comes up heads. So there are four equally likely outcomes, of which one shows three heads, so the probability of getting three heads is 1 / 4. The two answers differ because we have different information in the two cases, so different sets of outcomes remain possible. 3. Suppose someone has randomly generated two natural numbers and used them to make a fraction. Reduce the fraction to its lowest terms. Is there a 0.5 probability that both the numerator and the denominator are odd numbers? Why or why not? (B+S 7.2.29. Note that part of this question is to come up with a reasonable interpretation of the idea of a randomly generated natural number.) Solution. In order for the numerator and the denominator to both be odd, the same power of 2 must occur in the prime factorization of both the numerator and the denominator. Now, half of all numbers are odd, so with probability 1 / 4 the numerator and denominator before reduction were odd. One-fourth of all numbers are even but not multiples of 4, so with probability 1 / 16 the numerator and denominator before reduction were both even but not multiples of 4. In general, the proportion of numbers which are multiples of 2 k but not 2 k +1 is 1 / (2 k +1

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• Summer '09
• Lugo
• Math, Probability, Elementary arithmetic, Randomness, Dice, Dungeons & Dragons

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