Difference between revisions of "1991 AIME Problems/Problem 10"
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Let us make a chart of values in alphabetical order, where <math>P_a,\ P_b</math> are the probabilities that each string comes from <math>aaa</math> and <math>bbb</math> multiplied by <math>27</math>, and <math>S_b</math> denotes the [[partial sum]]s of <math>P_b</math> (in other words, <math>S_b = \sum_{n=1}^{b} P_b</math>): | Let us make a chart of values in alphabetical order, where <math>P_a,\ P_b</math> are the probabilities that each string comes from <math>aaa</math> and <math>bbb</math> multiplied by <math>27</math>, and <math>S_b</math> denotes the [[partial sum]]s of <math>P_b</math> (in other words, <math>S_b = \sum_{n=1}^{b} P_b</math>): | ||
<cmath> | <cmath> | ||
− | \begin{ | + | \begin{array}{|r||r|r|r|} |
\hline | \hline | ||
\text{String}&P_a&P_b&S_b\\ | \text{String}&P_a&P_b&S_b\\ | ||
Line 20: | Line 20: | ||
bbb & 1 & 8 & 27 \\ | bbb & 1 & 8 & 27 \\ | ||
\hline | \hline | ||
− | \end{ | + | \end{array} |
</cmath> | </cmath> | ||
Revision as of 19:19, 10 March 2015
Problem
Two three-letter strings, and , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an when it should have been a , or as a when it should be an . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let be the three-letter string received when is transmitted and let be the three-letter string received when is transmitted. Let be the probability that comes before in alphabetical order. When is written as a fraction in lowest terms, what is its numerator?
Solution
Solution 1
Let us make a chart of values in alphabetical order, where are the probabilities that each string comes from and multiplied by , and denotes the partial sums of (in other words, ):
The probability is , so the answer turns out to be , and the solution is .
Solution 2
Let be the th letter of string . Compare the first letter of the string to the first letter of the string . There is a chance that comes before . There is a that is the same as .
If , then you do the same for the second letters of the strings. But you have to multiply the chance that comes before as there is a chance we will get to this step.
Similarly, if , then there is a chance that we will get to comparing the third letters and that comes before .
So we have .
Solution 3
Consider letter strings instead. If the first letters all get transmitted correctly, then the string will be first. Otherwise, the only way is for both of the first letters to be the same, and then we consider the next letter string following the first letter. This easily leads to a recursion: . Clearly, .
See also
1991 AIME (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.