Difference between revisions of "2007 AIME II Problems/Problem 10"
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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>. | 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{ | + | 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 | -MP8148 |
Revision as of 20:15, 10 March 2019
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
. 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
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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