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1991 AIME Problems/Problem 10

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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?

Contents

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, ): \begin{tabular}{|r||r|r|r|}\hline\text{String}&P_a&P_b&S_b\\\hlineaaa & 8 & 1 & 1 \\aab & 4 & 2 & 3 \\aba & 4 & 2 & 5 \\abb & 2 & 4 & 9 \\baa & 4 & 2 & 11 \\bab & 2 & 4 & 15 \\bba & 2 & 4 & 19 \\bbb & 1 & 8 & 27 \\\hline\end{tabular}

The probability is , so the answer turns out to be \frac{8\cdot 26 + 4 \cdot 24 \ldots 2 \cdot 8 + 1 \cdot 0}{27^2} = \frac{532}{729}, 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 p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=\boxed{532}/729.

See also

1991 AIME (ProblemsResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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