Question 378733: A positive interger ,n, is picked at random. Find the probability that the GCF of n and 60 is 2. Find the probability that that the GCF of n and 60 is not 1.
Found 2 solutions by robertb, richard1234:
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! So n is not from 1 to 500 inclusive now?
The problem is ill-posed. You have to limit the population set to a finite set, just like what you did in your first problem post 378532, from 1 to 500. Let me demonstrate: If n is any positive integer of the form n = 2p, where p is a prime not equal to 2,3,or 5, then exactly GCF(60,n) = 2. The probability would then be equal to (number of all positive primes not equal to 2,3,5)/(number of all positive integers). This ratio will not have an answer, and we have already restricted ourselves to the domain of the positive integers!
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! For the GCF to equal 2, n must divide 2, and n must not divide 3 nor 5. First, note that there is a 1/2 probability of n being even. Now we will consider the set of even integers 2, 4, 6, 8, ... and compute the probability that a chosen number *will* be divisible by 3 or 5 (then subtract from 1).
By inclusion-exclusion principle, the probability that a number n is divisible by 3 or 5 is 1/3 + 1/5 - 1/15, or 7/15. 1 - 7/15 = 8/15, times 1/2 gives us a probability of 4/15.
For the GCF of n and 60 to not equal 1, n must divide any of 2, 3, or 5. This requires some inclusion-exclusion bashing.
We can compute this by P(2|n) + P(3|n) + P(5|n) - P(2,3|n) - P(2,5|n) - P(3,5|n) + P(2,3,5|n)
= 1/2 + 1/3 + 1/5 - 1/6 - 1/10 - 1/15 + 1/60
= 43/60
I hope that's the right answer :)
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