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Difference between revisions of "2007 AIME II Problems/Problem 10"

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== Problem ==
 
== Problem ==
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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 at least 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>)
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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 ==
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== Solution 3 ==
 
== Solution 3 ==
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<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>.
+
<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>.
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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>.
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 +
== 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>
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-MP8148
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== Solution 5 ==
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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.
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Case 1: B is contained in A.
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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>.
 +
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Case 2: B is contained in S-A.
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By symmetry, this case is the same as Case 1, once again giving us <math>3^6</math> possibilities.
 +
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Case 3: B is contained in both.
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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.
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 +
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>.
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- Spacesam
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 +
~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

Problem

Let $S$ be a set with six elements. Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$, not necessarily distinct, are chosen independently and at random from $\mathcal{P}$. The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$, $n$, and $r$ are positive integers, $n$ is prime, and $m$ and $n$ are relatively prime. Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$)

Solution 1

Use casework:

  • $B$ has 6 elements:
    • Probability: $\frac{1}{2^6} = \frac{1}{64}$
    • $A$ must have either 0 or 6 elements, probability: $\frac{2}{2^6} = \frac{2}{64}$.
  • $B$ has 5 elements:
    • Probability: ${6\choose5}/64 = \frac{6}{64}$
    • $A$ must have either 0, 6, or 1, 5 elements. The total probability is $\frac{2}{64} + \frac{2}{64} = \frac{4}{64}$.
  • $B$ has 4 elements:
    • Probability: ${6\choose4}/64 = \frac{15}{64}$
    • $A$ 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 $B$ and a fifth element out of the remaining $2$ numbers. The total probability is $\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{8}{64}$.

We could just continue our casework. In general, the probability of picking B with $n$ elements is $\frac{{6\choose n}}{64}$. Since the sum of the elements in the $k$th row of Pascal's Triangle is $2^k$, the probability of obtaining $A$ or $S-A$ which encompasses $B$ is $\frac{2^{7-n}}{64}$. In addition, we must count for when $B$ is the empty set (probability: $\frac{1}{64}$), of which all sets of $A$ will work (probability: $1$).

Thus, the solution we are looking for is $\left(\sum_{i=1}^6 \frac{{6\choose i}}{64} \cdot \frac{2^{7-i}}{64}\right) + \frac{1}{64} \cdot \frac{64}{64}$ $=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}$ $=\frac{1394}{2^{12}}$ $=\frac{697}{2^{11}}$.

The answer is $697 + 2 + 11 = 710$.

Solution 2

we need $B$ to be a subset of $A$ or $S-A$ we can divide each element of $S$ into 4 categories:

  • it is in $A$ and $B$
  • it is in $A$ but not in $B$
  • it is not in $A$ but is in $B$
  • or it is not in $A$ and not in $B$

these can be denoted as $+A+B$, $+A-B$,$-A+B$, and $-A-B$

we note that if all of the elements are in $+A+B$, $+A-B$ or $-A-B$ we have that $B$ is a subset of $A$ which can happen in $\dfrac{3^6}{4^6}$ ways

similarly if the elements are in $+A-B$,$-A+B$, or $-A-B$ we have that $B$ is a subset of $S-A$ which can happen in $\dfrac{3^6}{4^6}$ ways as well

but we need to make sure we don't over-count ways that are in both sets these are when $+A-B$ or $-A-B$ which can happen in $\dfrac{2^6}{4^6}$ ways so our probability is $\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}$.

so the final answer is $697 + 2 + 11 = 710$.

Solution 3

$B$ must be in $A$ or $B$ must be in $S-A$. This is equivalent to saying that $B$ must be in $A$ or $B$ is disjoint from $A$. The probability of this is the sum of the probabilities of each event individually minus the probability of each event occurring simultaneously. There are $\binom{6}{x}$ ways to choose $A$, where $x$ is the number of elements in $A$. From those $x$ elements, there are ${2^x}$ ways to choose $B$. Thus, the probability that $B$ is in $A$ is the sum of all the values $\binom{6}{x}({2^x})$ for values of $x$ ranging from $0$ to $6$. For the second probability, the ways to choose $A$ stays the same but the ways to choose $B$ is now ${2^{6-x}}$. 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 $B$ is the null set. $A$ can be any subset of $S$, so there are ${2^6}$ possibilities. Our final sum of possibilities is $2\cdot 3^6-2^6$. We have ${2^6}$ total possibilities for both $A$ and $B$, so there are ${2^{12}}$ total possibilities. This reduces down to $\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}$. The answer is thus $697 + 2 + 11 = 710$.

Solution 4

Let $|S|$ denote the number of elements in a general set $S$. We use complementary counting.

There is a total of $2^6$ elements in $P$, so the total number of ways to choose $A$ and $B$ is $(2^6)^2 = 2^{12}$.

Note that the number of $x$-element subset of $S$ is $\binom{6}{x}$. In general, for $0 \le |A| \le 6$, in order for $B$ to be in neither $A$ nor $S-A$, $B$ must have at least one element from both $A$ and $S-A$. In other words, $B$ must contain any subset of $A$ and $S-A$ except for the empty set $\{\}$. This can be done in $\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1)$ ways. As $|A|$ ranges from $0$ to $6$, we can calculate the total number of unsuccessful outcomes to be \[\sum_{|A| = 0}^{6} \binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1) = 2702.\] So our desired answer is \[1 - \dfrac{2702}{2^{12}} = \dfrac{697}{2^{11}} \Rightarrow \boxed{710}.\]

-MP8148

Solution 5

To begin with, we note that there are $2^6$ subsets of $S$(which we can assume is $\{1,2,3,4,5,6\}$), including the null set. This gives a total of $(2^6)^2 = 2^{12}$ total possibilities for A and B.

Case 1: B is contained in A. If B has $0$ elements, which occurs in $\binom{6}{0}$ ways, A can be anything, giving us $\binom{6}{0} \cdot 2^6$. If B has $1$ element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us $\binom{6}{1} \cdot 2^5$. Summing, we end up with the binomial expansion of $(2 + 1)^6 = 3^6$.

Case 2: B is contained in S-A. By symmetry, this case is the same as Case 1, once again giving us $3^6$ 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 $x$ in B which satisfies this restriction. Then, A must contain $x$ as well, but we also know that $S-A$ contains $x$, contradiction. Hence, B is the null set, whereas A can be anything. This gives us $2^6$ possibilities.

Since we have overcounted Case 3 in both of the other two cases, our final count is $2 \cdot 3^6 - 2^6$. This gives us the probability $\frac{2 \cdot 3^6 - 2^6}{2^{12}}$. Upon simplifying, we end up with $\frac{697}{2^{11}}$, giving the desired answer of $\boxed{710}$. - Spacesam

~clarifications by LeonidasTheConquerer

See also

2007 AIME II (Problems • Answer Key • Resources)
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