2007 AIME II Problems/Problem 10

From AoPSWiki

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

2007 AIME II (Problems • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

2007 AIME II Problems/Problem 10

From AoPSWiki

Problem

Solution

See also