AMC 10 · 2005 · #9
Grade 7 probabilityThree tiles are marked X and two other tiles are marked O. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Three tiles say $X$ and two tiles say $O$. All five are shuffled into a random row. Find the probability that the row comes out exactly as $XOXOX$.
Givens: There are five tiles: three marked $X$ and two marked $O$; The tiles are placed in a row in a random order, every order equally likely; Tiles with the same letter are identical to each other; Answer choices: (A) $\frac{1}{12}$, (B) $\frac{1}{10}$, (C) $\frac{1}{6}$, (D) $\frac{1}{4}$, (E) $\frac{1}{3}$
Unknowns: The probability that the random row reads $XOXOX$
Understand
Restated: Three tiles say $X$ and two tiles say $O$. All five are shuffled into a random row. Find the probability that the row comes out exactly as $XOXOX$.
Givens: There are five tiles: three marked $X$ and two marked $O$; The tiles are placed in a row in a random order, every order equally likely; Tiles with the same letter are identical to each other; Answer choices: (A) $\frac{1}{12}$, (B) $\frac{1}{10}$, (C) $\frac{1}{6}$, (D) $\frac{1}{4}$, (E) $\frac{1}{3}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The question is a "how many ways" count wrapped in a probability, so Tool #2 (Make a Systematic List) is the engine: every distinct row is fixed once you decide which two of the five positions hold the $O$ tiles, so counting the ways to place two $O$s among five slots counts all the rows. Tool #7 (Identify Subproblems) splits the job into the two pieces a probability needs — the size of the whole sample space, then the number of favorable rows. Tool #3 (Eliminate Possibilities) closes it out: the count gives one exact fraction, and only one answer choice matches.
Execute — Answer: B
7.SP.C.8 Step 1 Count every possible row
- A row is completely decided by where the two $O$ tiles land, because every other slot must then hold an $X$.
- So counting the rows is the same as counting the ways to choose $2$ of the $5$ positions for the $O$s.
- That is the combination $\binom{5}{2}$, which equals $\frac{5\times4}{2\times1}=10$.
- So there are $10$ equally likely rows in all.
💡 Fixing the two $O$ spots fixes the whole row, so counting rows is just choosing $2$ slots out of $5$.
7.SP.C.8 Step 2 Count the favorable rows
- The target pattern $XOXOX$ is a single specific arrangement: $O$s sit in positions $2$ and $4$, and $X$s fill positions $1,3,5$.
- There is no other way to read $XOXOX$, so exactly one of the ten rows is favorable.
💡 $XOXOX$ pins down every letter, so only one of the listed rows can match it.
7.SP.C.7 Step 3 Divide favorable by total
- Since all $10$ rows are equally likely, the probability of the one favorable row is favorable over total, $\frac{1}{10}$.
- That is choice (B).
💡 With every row equally likely, the chance of the single winning row is just $1$ out of the $10$ rows.
7.SP.C.8 A row is completely decided by where the two $O$ tiles land, because every other 7.SP.C.8 The target pattern $XOXOX$ is a single specific arrangement: $O$s sit in positio 7.SP.C.7 Since all $10$ rows are equally likely, the probability of the one favorable row Review
Reasonableness: The answer $\frac{1}{10}$ is one specific outcome out of ten equally likely rows, which is exactly what a single named pattern should be worth, so the size feels right. The trap choices come from miscounting the sample space: (E) $\frac{1}{3}$ and (C) $\frac{1}{6}$ appear if a solver treats the tiles as all-distinct and mishandles the identical copies, and (A) $\frac{1}{12}$ appears from an over-count of the arrangements. Treating the like tiles as identical gives the clean $10$ rows and the matching $\frac{1}{10}$.
Alternative: Count with all five tiles labeled as distinct. Then there are $5!=120$ orderings. The favorable pattern $XOXOX$ can be produced by $3!$ orderings of the three $X$ tiles and $2!$ orderings of the two $O$ tiles, giving $3!\times2!=12$ favorable orderings. The probability is $\frac{12}{120}=\frac{1}{10}$ — the same (B).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $10$ possible rows as $\binom{5}{2}$ and identifying the single favorable row $XOXOX$.)7.SP.C.7Develop probability models and use them to find probabilities of events (Using the equally-likely (uniform) model to set the probability as favorable rows over total rows, $\frac{1}{10}$.)
⭐ Only the two $O$ spots decide the row, so there are $\binom{5}{2}=10$ equally likely rows and exactly one is $XOXOX$, making the probability $\frac{1}{10}$.
⭐ Only the two $O$ spots decide the row, so there are $\binom{5}{2}=10$ equally likely rows and exactly one is $XOXOX$, making the probability $\frac{1}{10}$.
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