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== Problem == | == Problem == | ||
− | 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 | + | 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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so the final answer is <math>697 + 2 + 11 = 710</math>. | so the final answer is <math>697 + 2 + 11 = 710</math>. | ||
+ | |||
+ | == 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 <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>. | ||
+ | The answer is thus <math>697 + 2 + 11 = 710</math>. | ||
+ | |||
+ | == 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> | ||
+ | |||
+ | -MP8148 | ||
+ | |||
+ | == Solution 5 == | ||
+ | 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. | ||
+ | |||
+ | Case 1: B is contained in A only. | ||
+ | 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>. | ||
+ | |||
+ | Case 2: B is contained in S-A only. | ||
+ | By symmetry, this case is the same as Case 1, once again giving us <math>3^6</math> 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 <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. | ||
+ | |||
+ | 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>. | ||
+ | - Spacesam | ||
== 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}} |
Revision as of 11:48, 23 October 2022
Problem
Let be a set with six elements. Let
be the set of all subsets of
Subsets
and
of
, not necessarily distinct, are chosen independently and at random from
. The probability that
is contained in one of
or
is
where
,
, and
are positive integers,
is prime, and
and
are relatively prime. Find
(The set
is the set of all elements of
which are not in
)
Solution 1
Use casework:
has 6 elements:
- Probability:
must have either 0 or 6 elements, probability:
.
- Probability:
has 5 elements:
- Probability:
must have either 0, 6, or 1, 5 elements. The total probability is
.
- Probability:
has 4 elements:
- Probability:
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
and a fifth element out of the remaining
numbers. The total probability is
.
- Probability:
We could just continue our casework. In general, the probability of picking B with elements is
. Since the sum of the elements in the
th row of Pascal's Triangle is
, the probability of obtaining
or
which encompasses
is
. In addition, we must count for when
is the empty set (probability:
), of which all sets of
will work (probability:
).
Thus, the solution we are looking for is
.
The answer is .
Solution 2
we need to be a subset of
or
we can divide each element of
into 4 categories:
- it is in
and
- it is in
but not in
- it is not in
but is in
- or it is not in
and not in
these can be denoted as ,
,
, and
we note that if all of the elements are in ,
or
we have that
is a subset of
which can happen in
ways
similarly if the elements are in ,
, or
we have that
is a subset of
which can happen in
ways as well
but we need to make sure we don't over-count ways that are in both sets these are when or
which can happen in
ways
so our probability is
.
so the final answer is .
Solution 3
must be in
or
must be in
. This is equivalent to saying that
must be in
or
is disjoint from
. The probability of this is the sum of the probabilities of each event individually minus the probability of each event occurring simultaneously. There are
ways to choose
, where
is the number of elements in
. From those
elements, there are
ways to choose
. Thus, the probability that
is in
is the sum of all the values
for values of
ranging from
to
. For the second probability, the ways to choose
stays the same but the ways to choose
is now
. We see that these two summations are simply from the Binomial Theorem and that each of them is
. We subtract the case where both of them are true. This only happens when
is the null set.
can be any subset of
, so there are
possibilities. Our final sum of possibilities is
. We have
total possibilities for both
and
, so there are
total possibilities. This reduces down to
.
The answer is thus
.
Solution 4
Let denote the number of elements in a general set
. We use complementary counting.
There is a total of elements in
, so the total number of ways to choose
and
is
.
Note that the number of -element subset of
is
. In general, for
, in order for
to be in neither
nor
,
must have at least one element from both
and
. In other words,
must contain any subset of
and
except for the empty set
. This can be done in
ways. As
ranges from
to
, we can calculate the total number of unsuccessful outcomes to be
So our desired answer is
-MP8148
Solution 5
To begin with, we note that there are subsets of
(which we can assume is
), including the null set. This gives a total of
total possibilities for A and B.
Case 1: B is contained in A only.
If B has elements, which occurs in
ways, A can be anything, giving us
. If B has
element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us
. Summing, we end up with the binomial expansion of
.
Case 2: B is contained in S-A only.
By symmetry, this case is the same as Case 1, once again giving us 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 in B which satisfies this restriction. Then, A must contain
as well, but we also know that
contains
, contradiction. Hence, B is the null set, whereas A can be anything. This gives us
possibilities.
Since we have overcounted Case 3 in both of the other two cases, our final count is . This gives us the probability
. Upon simplifying, we end up with
, giving the desired answer of
.
- Spacesam
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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