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08062007  #1 
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A lil math (probability) help!?
A graphing calculator is programmed to generate a number between 1 and 20. a) What is the probability that three consecutive prime numbers (1,2,3,7,11,13,17) will be produced?b) What is the probability that atleast one of the first three numbers will be prime?For a) P= 8c3 / 20c3 = 0.049and for b)P = 1 P(2 numbers aren't prime) 1 (2/12) = 0.833Is that somewhat right?

08062007  #2 
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a) What is the probability that three consecutive prime numbers (1,2,3,7,11,13,17) will be produced?Did you mean in three trials, or n>3 trials?(You omitted 5, and let's accept your definition of 1 as prime.)Then the prime numbers are {1,2,3,5,7,11,13,17} i.e. the total sequence has length 8.For three trials, there are 20^3=8000 total possibilitiesFor inorder subsequences of length 3 of primes there are (83+1) = 6 possibilities (i.e. you could pick the starting number from 1>ending number=3 up to starting number =11>ending number=17)So, probability of three inorder primes from 3 trials is 6/8000 = 0.00075b) What is the probability that at least one of the first three numbers will be prime?p = Probability that any given trial is a prime = 8/20 = 0.4=> Probability that any given trial is composite = (1p) = 0.6Probability that three trials are composite = (1p)^3Probability that at least one of first three trials is composite = 1  (1p)^3= 0.784

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