Difference between revisions of "2007 AIME II Problems/Problem 10"
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− | <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 14:09, 18 May 2017
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 at least 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 6C
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 {(2+1)^6}. 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 {2^6} 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
.
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.