Difference between revisions of "2007 AIME II Problems/Problem 10"
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**<math>A</math> 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 <math>B</math> and a fifth element out of the remaining <math>2</math> numbers. The total probability is <math>\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{4}{64}</math>. | **<math>A</math> 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 <math>B</math> and a fifth element out of the remaining <math>2</math> numbers. The total probability is <math>\frac{2}{64}\left({2\choose0} + {2\choose1} + {2\choose2}\right) = \frac{2}{64} + \frac{4}{64} + \frac{2}{64} = \frac{4}{64}</math>. | ||
− | We could just continue our casework. In general, the probability of picking B with <math>n</math> elements is <math>\frac{{6\choose n}}{64}</math>. Since the sum of the elements in the <math>k</math>th row of [[ | + | We could just continue our casework. In general, the probability of picking B with <math>n</math> elements is <math>\frac{{6\choose n}}{64}</math>. Since the sum of the elements in the <math>k</math>th row of [[Pascal's Triangle]] is <math>2^k</math>, the probability of obtaining <math>A</math> or <math>S-A</math> which encompasses <math>B</math> is <math>\frac{2^{7-n}}{64}</math>. In addition, we must count for when <math>B</math> is the empty set (probability: <math>\frac{1}{64}</math>), of which all sets of <math>A</math> will work (probability: <math>1</math>). |
Thus, the solution we are looking for is <math>\left(\displaystyle\sum_{i=1}^6 \frac{{6\choose i}}{64} \cdot \frac{2^{7-i}}{64}\right) + \frac{1}{64} \cdot \frac{64}{64}</math><math>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</math><math>=\frac{1394}{2^{12}}</math><math>=\frac{697}{2^{11}}</math>. | Thus, the solution we are looking for is <math>\left(\displaystyle\sum_{i=1}^6 \frac{{6\choose i}}{64} \cdot \frac{2^{7-i}}{64}\right) + \frac{1}{64} \cdot \frac{64}{64}</math><math>=\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}</math><math>=\frac{1394}{2^{12}}</math><math>=\frac{697}{2^{11}}</math>. |
Revision as of 15:41, 30 March 2007
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
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 .
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 |