2005 AMC 10A Problem 9

A row of five tiles is determined by which two of the five positions hold the $O$'s; the remaining three positions automatically get $X$'s. Every choice of positions is equally likely, and there are
$$\binom{5}{2} = 10$$
of them.

The pattern $XOXOX$ corresponds to exactly one choice: $O$'s in positions $2$ and $4$. So the probability is $\frac{1}{10}$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{10}}$.

Identical tiles collapse the sample space from $5!$ orderings to $\binom{5}{2}$ distinguishable patterns, each equally likely by symmetry. Alternatively, if you treat the tiles as distinct, both numerator and denominator get multiplied by $3!\,2!$, and the ratio $\frac{12}{120}$ is the same. Either bookkeeping works as long as it is consistent.

Sequential probability: the first tile is $X$ with probability $\frac{3}{5}$, then $O$ with $\frac{2}{4}$, then $X$ with $\frac{2}{3}$, then $O$ with $\frac{1}{2}$, then $X$ with certainty. Product: $\frac{3}{5}\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2} = \frac{1}{10}$.

• Using 5! = 120 or 2^5 = 32 as the total, or thinking XOXOX can be produced in more than one way.
• Counting the distinct-tile favorable cases ($12$) but dividing by the multiset total ($10$), mixing the two conventions.