Question

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 $\frac{1}{3}$ 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?

Answer

Answer: 532

Concepts:independent events geometric sequence basic probability

Difficulty rating: 2200

Small Hint:

The ordering is decided at the first position where the two received strings differ

Big Hint:

At one position, compute the probabilities of equality and of receiving $a$ in $S_a$ and $b$ in $S_b$

Solution:

At any position, the received letters agree with probability $2\left(\frac23\right)\left(\frac13\right)=\frac49.$ The favorable first difference, $S_a$ receiving $a$ and $S_b$ receiving $b,$ has probability $(\frac{2}{3})^2=\frac{4}{9}.$ It can occur in the first, second, or third position after zero, one, or two agreements. Hence $\begin{aligned}p&=\frac49\left(1+\frac49+\left(\frac49\right)^2\right)\\&=\frac{532}{729}.\end{aligned}$ This fraction is in lowest terms, so its numerator is $532.$

1991 AIME Problem 10

Answer: 532

Difficulty rating: 2200

Small Hint:

Big Hint:

Solution:

Problem 10 in Other Years