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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 elements. Let <math>P</math> be the set of all subsets 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 integers, <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>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>)
  
 
== Solution ==
 
== Solution ==

Revision as of 16:42, 30 March 2007

Problem

Let $S$ be a set with six elements. Let $P$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$, not necessarily distinct, are chosen independently and at random from $P$. The probability that $B$ is contained in at least 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

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 $\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{4}{64}$.

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(\displaystyle\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 $\displaystyle 697 + 2 + 11 = 710$.

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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