2007 AIME II Problems/Problem 10
Let be a set with six elements. Let be the set of all subsets of Subsets and of , not necessarily distinct, are chosen independently and at random from . The probability that is contained in at least one of or is where , , and are positive integers, is prime, and and are relatively prime. Find (The set is the set of all elements of which are not in )
- has 6 elements:
- must have either 0 or 6 elements, probability: .
- has 5 elements:
- must have either 0, 6, or 1, 5 elements. The total probability is .
- has 4 elements:
- must have either 0, 6; 1, 5; or 2,4 elements. If there are 1 or 5 elements, the set which contains 5 elements must have four emcompassing and a fifth element out of the remaining numbers. The total probability is .
We could just continue our casework. In general, the probability of picking B with elements is . Since the sum of the elements in the th row of Pascal's Triangle is , the probability of obtaining or which encompasses is . In addition, we must count for when is the empty set (probability: ), of which all sets of will work (probability: ).
Thus, the solution we are looking for is .
The answer is .
we need to be a subset of or we can divide each element of into 4 categories:
- it is in and
- it is in but not in
- it is not in but is in
- or it is not in and not in
these can be denoted as , ,, and
we note that if all of the elements are in , or we have that is a subset of which can happen in ways
similarly if the elements are in ,, or we have that is a subset of which can happen in ways as well
but we need to make sure we don't over-count ways that are in both sets these are when or which can happen in ways so our probability is .
so the final answer is .
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