Difference between revisions of "1991 AIME Problems/Problem 10"
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== Problem == | == Problem == | ||
+ | Two three-letter strings, <math>aaa^{}_{}</math> and <math>bbb^{}_{}</math>, 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 <math>a^{}_{}</math> when it should have been a <math>b^{}_{}</math>, or as a <math>b^{}_{}</math> when it should be an <math>a^{}_{}</math>. However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let <math>S_a^{}</math> be the three-letter string received when <math>aaa^{}_{}</math> is transmitted and let <math>S_b^{}</math> be the three-letter string received when <math>bbb^{}_{}</math> is transmitted. Let <math>p</math> be the [[probability]] that <math>S_a^{}</math> comes before <math>S_b^{}</math> in alphabetical order. When <math>p</math> is written as a [[fraction]] in [[irreducible fraction|lowest terms]], what is its [[numerator]]? | ||
+ | |||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
+ | 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 sums of <math>P_b</math> (in other words, <math>S_b = \sum_{n=1}^{b} P_b</math>): | ||
+ | <cmath> | ||
+ | \begin{array}{|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{array} | ||
+ | </cmath> | ||
+ | |||
+ | The probability is <math>p=\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>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>S(a,n)</math> be the <math>n</math>th letter of string <math>S(a)</math>. | ||
+ | Compare the first letter of the string <math>S(a)</math> to the first letter of the string <math>S(b)</math>. | ||
+ | There is a <math>(2/3)^2=4/9</math> chance that <math>S(a,1)</math> comes before <math>S(b,1)</math>. | ||
+ | There is a <math>2(1/3)(2/3)=4/9</math> that <math>S(a,1)</math> is the same as <math>S(b,1)</math>. | ||
+ | |||
+ | If <math>S(a,1)=S(b,1)</math>, then you do the same for the second letters of the strings. But you have to multiply the <math>4/9</math> chance that <math>S(a,2)</math> comes before <math>S(b,2)</math> as there is a <math>4/9</math> chance we will get to this step. | ||
+ | |||
+ | Similarly, if <math>S(a,2)=S(b,2)</math>, then there is a <math>(4/9)^3</math> chance that we will get to comparing the third letters and that <math>S(a)</math> comes before <math>S(b)</math>. | ||
+ | |||
+ | So we have <math>p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729</math>. Therefore, the answer is <math>\boxed{532}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Consider <math>n</math> letter strings instead. If the first letters all get transmitted correctly, then the <math>a</math> 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 <math>n-1</math> letter string following the first letter. This easily leads to a recursion: <math>p_n=\frac23\cdot\frac23+2\cdot\frac23\cdot\frac13p_{n-1}=\frac49+\frac49p_{n-1}</math>. Clearly, <math>p_0=0\implies p_1=\frac49\implies p_2=\frac{52}{81}\implies p_3=\frac{532}{729}</math>. Therefore, the answer is <math>\boxed{532}</math>. | ||
+ | |||
+ | ===Solution 4 (a more explicit Solution 3)=== | ||
+ | The probability that <math>S_a</math> will take the form <math>a</math> _ _ and that <math>S_b</math> will take the form <math>b</math> _ _ is <math>\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}</math>. Then, the probability that both <math>S_a</math> and <math>S_b</math> will share the same first digit is <math>2\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{9}</math>. Now if the first digits of either sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of <math>S_a</math> is <math>a</math> and that the second letter of <math>S_b</math> is <math>b</math> is <math>\frac{2}{3}\cdot\frac{2}{3}=\frac{4}{9}</math>. Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is <math>2\cdot\frac{2}{3}\cdot\frac{1}{3} = \frac{4}{9}</math>. Now, if the last case is true then the probability that <math>S_a</math> precedes <math>S_b</math> is <math>\frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}</math>. Therefore the total probability would be: <math>\frac{4}{9} + \frac{4}{9}\left(\frac{4}{9} + \frac{4}{9}\left(\frac{4}{9}\right)\right) = \frac{4}{9}+\frac{4}{9}\left(\frac{52}{81}\right) = \frac{4}{9} + \frac{208}{729} = \frac{532}{729}</math>. Therefore the answer is <math>\boxed{532}</math>. | ||
+ | |||
+ | |||
+ | ~qwertysri987 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1991|num-b=9|num-a=11}} | |
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 06:10, 25 February 2024
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,
):
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 . Therefore, the answer is
.
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,
. Therefore, the answer is
.
Solution 4 (a more explicit Solution 3)
The probability that will take the form
_ _ and that
will take the form
_ _ is
. Then, the probability that both
and
will share the same first digit is
. Now if the first digits of either sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of
is
and that the second letter of
is
is
. Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is
. Now, if the last case is true then the probability that
precedes
is
. Therefore the total probability would be:
. Therefore the answer is
.
~qwertysri987
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.