Difference between revisions of "2007 AIME II Problems/Problem 10"
Mathgeek2006 (talk | contribs) m (→Solution 1) |
(→Solution 5) |
||
(14 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math>S</math> be a [[set]] with six [[element]]s. Let <math>P</math> be the set of all [[subset]]s of <math>S.</math> Subsets <math>A</math> and <math>B</math> of <math>S</math>, not necessarily distinct, are chosen independently and at random from <math>P</math>. The [[probability]] that <math>B</math> is contained in | + | Let <math>S</math> be a [[set]] with six [[element]]s. Let <math>\mathcal{P}</math> be the set of all [[subset]]s of <math>S.</math> Subsets <math>A</math> and <math>B</math> of <math>S</math>, not necessarily distinct, are chosen independently and at random from <math>\mathcal{P}</math>. The [[probability]] that <math>B</math> is contained in one of <math>A</math> or <math>S-A</math> is <math>\frac{m}{n^{r}},</math> where <math>m</math>, <math>n</math>, and <math>r</math> are [[positive]] [[integer]]s, <math>n</math> is [[prime]], and <math>m</math> and <math>n</math> are [[relatively prime]]. Find <math>m+n+r.</math> (The set <math>S-A</math> is the set of all elements of <math>S</math> which are not in <math>A.</math>) |
== Solution 1 == | == Solution 1 == | ||
Line 46: | Line 46: | ||
so the final answer is <math>697 + 2 + 11 = 710</math>. | so the final answer is <math>697 + 2 + 11 = 710</math>. | ||
+ | |||
+ | == 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 <math>\binom{6}{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>\binom{6}{x}({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. 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>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | Let <math>|S|</math> denote the number of elements in a general set <math>S</math>. We use complementary counting. | ||
+ | |||
+ | There is a total of <math>2^6</math> elements in <math>P</math>, so the total number of ways to choose <math>A</math> and <math>B</math> is <math>(2^6)^2 = 2^{12}</math>. | ||
+ | |||
+ | Note that the number of <math>x</math>-element subset of <math>S</math> is <math>\binom{6}{x}</math>. In general, for <math>0 \le |A| \le 6</math>, in order for <math>B</math> to be in neither <math>A</math> nor <math>S-A</math>, <math>B</math> must have at least one element from both <math>A</math> and <math>S-A</math>. In other words, <math>B</math> must contain any subset of <math>A</math> and <math>S-A</math> except for the empty set <math>\{\}</math>. This can be done in <math>\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1)</math> ways. As <math>|A|</math> ranges from <math>0</math> to <math>6</math>, we can calculate the total number of unsuccessful outcomes to be <cmath>\sum_{|A| = 0}^{6} \binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1) = 2702.</cmath> So our desired answer is <cmath>1 - \dfrac{2702}{2^{12}} = \dfrac{697}{2^{11}} \Rightarrow \boxed{710}.</cmath> | ||
+ | |||
+ | -MP8148 | ||
+ | |||
+ | == Solution 5 == | ||
+ | To begin with, we note that there are <math>2^6</math> subsets of <math>S</math>(which we can assume is <math>\{1,2,3,4,5,6\}</math>), including the null set. This gives a total of <math>(2^6)^2 = 2^{12}</math> total possibilities for A and B. | ||
+ | |||
+ | Case 1: B is contained in A. | ||
+ | If B has <math>0</math> elements, which occurs in <math>\binom{6}{0}</math> ways, A can be anything, giving us <math>\binom{6}{0} \cdot 2^6</math>. If B has <math>1</math> element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us <math>\binom{6}{1} \cdot 2^5</math>. Summing, we end up with the binomial expansion of <math>(2 + 1)^6 = 3^6</math>. | ||
+ | |||
+ | Case 2: B is contained in S-A. | ||
+ | By symmetry, this case is the same as Case 1, once again giving us <math>3^6</math> 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 <math>x</math> in B which satisfies this restriction. Then, A must contain <math>x</math> as well, but we also know that <math>S-A</math> contains <math>x</math>, contradiction. Hence, B is the null set, whereas A can be anything. This gives us <math>2^6</math> possibilities. | ||
+ | |||
+ | Since we have overcounted Case 3 in both of the other two cases, our final count is <math>2 \cdot 3^6 - 2^6</math>. This gives us the probability <math>\frac{2 \cdot 3^6 - 2^6}{2^{12}}</math>. Upon simplifying, we end up with <math>\frac{697}{2^{11}}</math>, giving the desired answer of <math>\boxed{710}</math>. | ||
+ | - Spacesam | ||
+ | |||
+ | ~clarifications by LeonidasTheConquerer | ||
== See also == | == See also == | ||
{{AIME box|year=2007|n=II|num-b=9|num-a=11}} | {{AIME box|year=2007|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:02, 20 July 2024
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 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. 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. 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
~clarifications by LeonidasTheConquerer
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.