2007 AIME II Problems/Problem 10
Problem
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 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 )
Solution 1
Use casework:
- has 6 elements:
- Probability:
- must have either 0 or 6 elements, probability: .
- has 5 elements:
- Probability:
- must have either 0, 6, or 1, 5 elements. The total probability is .
- has 4 elements:
- Probability:
- 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 .
Solution 2
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 .
Solution 3
must be in or must be in . This is equivalent to saying that must be in or is disjoint from . The probability of this is the sum of the probabilities of each event individually minus the probability of each event occurring simultaneously. There are ways to choose , where is the number of elements in . From those elements, there are ways to choose . Thus, the probability that is in is the sum of all the values for values of ranging from to . For the second probability, the ways to choose stays the same but the ways to choose is now . We see that these two summations are simply from the Binomial Theorem and that each of them is . We subtract the case where both of them are true. This only happens when is the null set. can be any subset of , so there are possibilities. Our final sum of possibilities is . We have total possibilities for both and , so there are total possibilities. This reduces down to . The answer is thus .
Solution 4
Let denote the number of elements in a general set . We use complementary counting.
There is a total of elements in , so the total number of ways to choose and is .
Note that the number of -element subset of is . In general, for , in order for to be in neither nor , must have at least one element from both and . In other words, must contain any subset of and except for the empty set . This can be done in ways. As ranges from to , we can calculate the total number of unsuccessful outcomes to be So our desired answer is
-MP8148
Solution 5
To begin with, we note that there are subsets of (which we can assume is ), including the null set. This gives a total of total possibilities for A and B.
Case 1: B is contained in A only. If B has elements, which occurs in ways, A can be anything, giving us . If B has element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us . Summing, we end up with the binomial expansion of .
Case 2: B is contained in S-A only. By symmetry, this case is the same as Case 1, once again giving us possibilities.
Case 3: B is contained in both. We claim here that B can only be the null set. For contradiction, assume that there exists some element in B which satisfies this restriction. Then, A must contain as well, but we also know that contains , contradiction. Hence, B is the null set, whereas A can be anything. This gives us possibilities.
Since we have overcounted Case 3 in both of the other two cases, our final count is . This gives us the probability . Upon simplifying, we end up with , giving the desired answer of . - Spacesam
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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