Difference between revisions of "2007 AIME II Problems/Problem 10"
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these can be denoted as <math>+A+B</math>, <math>+A-B</math>,<math>-A+B</math>, and <math>-A-B</math> | these can be denoted as <math>+A+B</math>, <math>+A-B</math>,<math>-A+B</math>, and <math>-A-B</math> | ||
− | we note that if all of the elements are in <math>+A+B</math>, <math>+A-B</math> | + | we note that if all of the elements are in <math>+A+B</math>, <math>+A-B</math> or <math>-A-B</math> |
we have that <math>B</math> is a subset of <math>A</math> | we have that <math>B</math> is a subset of <math>A</math> | ||
which can happen in <math>\dfrac{3^6}{4^6}</math> ways | which can happen in <math>\dfrac{3^6}{4^6}</math> ways | ||
− | similarly if the elements are in <math>+A-B</math>,<math>-A+B</math>, | + | similarly if the elements are in <math>+A-B</math>,<math>-A+B</math>, or <math>-A-B</math> |
we have that <math>B</math> is a subset of <math>S-A</math> | we have that <math>B</math> is a subset of <math>S-A</math> | ||
which can happen in <math>\dfrac{3^6}{4^6}</math> ways as well | which can happen in <math>\dfrac{3^6}{4^6}</math> ways as well | ||
− | but we need to make sure we don't over-count ways that are in both sets these are when <math>+A-B</math> | + | but we need to make sure we don't over-count ways that are in both sets these are when <math>+A-B</math> or <math>-A-B</math> |
which can happen in <math>\dfrac{2^6}{4^6}</math> ways | which can happen in <math>\dfrac{2^6}{4^6}</math> ways | ||
so our probability is <math>\dfrac{2*3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^11}=\dfrac{697}{2^{11}}</math>. | so our probability is <math>\dfrac{2*3^6-2^6}{4^6}= \dfrac{3^6-2^5}{2^11}=\dfrac{697}{2^{11}}</math>. |
Revision as of 18:05, 8 March 2014
Contents
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: .
- has 5 elements:
- Probability:
- must have either 0, 6, or 1, 5 elements. The total probability is .
- 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 .
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 .
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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