1991 AIME Problems/Problem 10

Problem

Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , 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 $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $\displaystyle p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $\displaystyle p$ is written as a fraction in lowest terms, what is its numerator?

Solution

Let us make a chart of values, where $P_a,\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$ , and $\displaystyle X_b$ denoting the sum of all of the previous terms of $\displaystyle P_b$ :

String	$\displaystyle P_a$	$\displaystyle P_b$	$\displaystyle X_b$
aaa	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

The probability is $P_a \cdot (27 - X_b)$ for each of the strings over $27^2$ , 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 $532$ .

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
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1991 AIME Problems/Problem 10

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