Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 3)
Line 48: Line 48:
  
 
== Solution 3 ==
 
== 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 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 B. Thus, the probability that B is in A is the sum of all the values 6Cx({2^x}) for values of <math>x</math> ranging from 0 to 6. For the second probability, the ways to choose A stays the same but the ways to choose <math>B</math> 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 <math>B</math> is the null set. <math>A</math> can be any subset of <math>S</math>, so there are {2^6} 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 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 {(2+1)^6}. 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 {2^6} 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>.
 
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>.
 
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>.

Revision as of 13:09, 18 May 2017

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 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 6C$x$ 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 $6Cx({2^x})$ 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. $\dfrac{2\cdot 3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^{11}}=\dfrac{697}{2^{11}}$. 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 $697 + 2 + 11 = 710$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_10&oldid=85709"