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Difference between revisions of "1991 AIME Problems/Problem 10"

(solution, but too lazy to finish typing)
(→‎Solution: format)
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== Solution ==
 
== Solution ==
−
Let us make a chart of values, 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>\displaystyle X_b</math> denoting the sum of all of the previous terms of <math>\displaystyle 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>
 +
\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}
 +
</cmath>
  
−
{| class= "wikitable" align="center"
+
The probability is <math>\sum P_a \cdot (27 - S_b)</math>, so the answer turns out to be <math>\frac{8\cdot 26 + 4 \cdot 24 \ldots 2 \cdot 8 + 1 \cdot 0}{27^2} = \frac{532}{729}</math>, and the solution is <math>\boxed{532}</math>.
−
| String || <math>\displaystyle P_a</math> || <math>\displaystyle P_b</math> || <math>\displaystyle X_b</math>
 
−
|-
 
−
| 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 <math>P_a \cdot (27 - X_b)</math> for each of the strings over <math>27^2</math>, so the answer turns out to be <math>\frac{8\cdot 26 + 4 \cdot 24 \ldots 2 \cdot 8 + 1 \cdot 0}{27^2} = \frac{532}{729}</math>, and the solution is <math>532</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 20:29, 11 April 2008

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 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 $\sum P_a \cdot (27 - S_b)$, 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 $\boxed{532}$.

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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