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Old 08-06-2007   #2
smci
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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 in-order subsequences of length 3 of primes there are (8-3+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 in-order 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 = (1-p) = 0.6Probability that three trials are composite = (1-p)^3Probability that at least one of first three trials is composite = 1 - (1-p)^3= 0.784