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

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*<math>B</math> has 4 elements:
 
*<math>B</math> has 4 elements:
 
**Probability: <math>{6\choose4}/64 = \frac{15}{64}</math>
 
**Probability: <math>{6\choose4}/64 = \frac{15}{64}</math>
−
**<math>A</math> 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 <math>B</math> and a fifth element out of the remaining <math>2</math> numbers. The total probability is <math>\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{4}{64}</math>.
+
**<math>A</math> 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 <math>B</math> and a fifth element out of the remaining <math>2</math> numbers. The total probability is <math>\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{8}{64}</math>.
  
 
We could just continue our casework. In general, the probability of picking B with <math>n</math> elements is <math>\frac{{6\choose n}}{64}</math>. Since the sum of the elements in the <math>k</math>th row of [[Pascal's Triangle]] is <math>2^k</math>, the probability of obtaining <math>A</math> or <math>S-A</math> which encompasses <math>B</math> is <math>\frac{2^{7-n}}{64}</math>. In addition, we must count for when <math>B</math> is the empty set (probability: <math>\frac{1}{64}</math>), of which all sets of <math>A</math> will work (probability: <math>1</math>).
 
We could just continue our casework. In general, the probability of picking B with <math>n</math> elements is <math>\frac{{6\choose n}}{64}</math>. Since the sum of the elements in the <math>k</math>th row of [[Pascal's Triangle]] is <math>2^k</math>, the probability of obtaining <math>A</math> or <math>S-A</math> which encompasses <math>B</math> is <math>\frac{2^{7-n}}{64}</math>. In addition, we must count for when <math>B</math> is the empty set (probability: <math>\frac{1}{64}</math>), of which all sets of <math>A</math> will work (probability: <math>1</math>).
  
−
Thus, the solution we are looking for is <math>\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}</math><math>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</math><math>=\frac{1394}{2^{12}}</math><math>=\frac{697}{2^{11}}</math>.
+
Thus, the solution we are looking for is <math>\left(\sum_{i=1}^6 \frac{{6\choose i}}{64} \cdot \frac{2^{7-i}}{64}\right) + \frac{1}{64} \cdot \frac{64}{64}<cmath>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</cmath>=\frac{1394}{2^{12}}</math><math>=\frac{697}{2^{11}}</math>.
  
−
The answer is <math>\displaystyle 697 + 2 + 11 = 710</math>.
+
The answer is <math>697 + 2 + 11 = 710</math>.
  
 
== 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}}

Revision as of 10:42, 10 March 2012

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 $\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{8}{64}$.

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}<cmath>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</cmath>=\frac{1394}{2^{12}}$$=\frac{697}{2^{11}}$.

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

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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