Difference between revisions of "2007 AIME II Problems/Problem 10"
Nathantang (talk | contribs) (→Solution 3) |
Nathantang (talk | contribs) (→Solution 3) |
||
Line 48: | Line 48: | ||
== Solution 3 == | == Solution 3 == | ||
− | <math>B</math> must be in <math>A</math> or <math>B</math> must be in <math>S-A</math>. This is equivalent to saying that <math>B</math> must be in <math>A</math> or <math>B</math> is disjoint from <math>A</math>. The probability of this is the sum of the probabilities of each event individually minus the probability of each event occurring simultaneously. There are 6C<math>x</math> ways to choose <math>A</math>, where <math>x</math> is the number of elements in <math>A</math>. From those <math>x</math> elements, there are <math>{2^x}</math> ways to choose <math>B</math>. Thus, the probability that <math>B</math> is in <math>A</math> is the sum of all the values <math>6Cx({2^x})</math> for values of <math>x</math> ranging from <math>0</math> to <math>6</math>. For the second probability, the ways to choose <math>A</math> stays the same but the ways to choose <math>B</math> is now <math>{2^6-x}</math>. We see that these two summations are simply from the Binomial Theorem and that each of them is {(2+1)^6}. We subtract the case where both of them are true. This only happens when <math>B</math> is the null set. <math>A</math> can be any subset of <math>S</math>, so there are {2^6} possibilities. Our final sum of possibilities is <math>2\cdot 3^6-2^6</math>. We have <math>{2^6}</math> total possibilities for both <math>A</math> and <math>B</math>, so there are <math>{2^12}</math> total possibilities. <math>\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}</math>. | + | <math>B</math> must be in <math>A</math> or <math>B</math> must be in <math>S-A</math>. This is equivalent to saying that <math>B</math> must be in <math>A</math> or <math>B</math> is disjoint from <math>A</math>. The probability of this is the sum of the probabilities of each event individually minus the probability of each event occurring simultaneously. There are 6C<math>x</math> ways to choose <math>A</math>, where <math>x</math> is the number of elements in <math>A</math>. From those <math>x</math> elements, there are <math>{2^x}</math> ways to choose <math>B</math>. Thus, the probability that <math>B</math> is in <math>A</math> is the sum of all the values <math>6Cx({2^x})</math> for values of <math>x</math> ranging from <math>0</math> to <math>6</math>. For the second probability, the ways to choose <math>A</math> stays the same but the ways to choose <math>B</math> is now <math>{2^(6-x)}</math>. We see that these two summations are simply from the Binomial Theorem and that each of them is <math>{(2+1)^6}</math>. We subtract the case where both of them are true. This only happens when <math>B</math> is the null set. <math>A</math> can be any subset of <math>S</math>, so there are <math>{2^6}</math> possibilities. Our final sum of possibilities is <math>2\cdot 3^6-2^6</math>. We have <math>{2^6}</math> total possibilities for both <math>A</math> and <math>B</math>, so there are <math>{2^12}</math> total possibilities. <math>\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}</math>. |
This reduces down to <math>\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}</math>. | This reduces down to <math>\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}</math>. | ||
The answer is thus <math>697 + 2 + 11 = 710</math>. | The answer is thus <math>697 + 2 + 11 = 710</math>. |
Revision as of 13:10, 18 May 2017
Contents
[hide]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 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 )
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 6C 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.