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1991 AIME Problems/Problem 10

Revision as of 21:39, 11 April 2008 by Azjps (talk | contribs) ([texed] solution by ehehheehee)

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 $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ 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 $P_a,\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$, and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \sum_{n=1}^{b} P_b$): 

\[\begin{tabular}{|r||r|r|r|}
\hline
\text{String}&P_a&P_b&S_b\\
\hline
aaa & 8 & 1 & 1 \\
aab & 4 & 2 & 3 \\
aba & 4 & 2 & 5 \\
abb & 2 & 4 & 9 \\
baa & 4 & 2 & 11 \\
bab & 2 & 4 & 15 \\
bba & 2 & 4 & 19 \\
bbb & 1 & 8 & 27 \\
\hline
\end{tabular}\] (Error compiling LaTeX. Unknown error_msg)

The probability is $p=\sum P_a \cdot (27 - S_b)$, 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 $\boxed{532}$.

Solution 2

Let $S(a,n)$ be the $n$th letter of string $S(a)$. Compare the first letter of the string $S(a)$ to the first letter of the string $S(b)$. There is a $(2/3)^2=4/9$ chance that $S(a,1)$ comes before $S(b,1)$. There is a $2(1/3)(2/3)=4/9$ that $S(a,1)$ is the same as $S(b,1)$.

If $S(a,1)=S(b,1)$, then you do the same for the second letters of the strings. But you have to multiply the $4/9$ chance that $S(a,2)$ comes before $S(b,2)$ as there is a $4/9$ chance we will get to this step.

Similarly, if $S(a,2)=S(b,2)$, then there is a $(4/9)^3$ chance that we will get to comparing the third letters and that $S(a)$ comes before $S(b)$.

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 (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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