AMC 10 · 2005 · #9
Grade 7 probabilityPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 Os 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.
Count every possible row
A row is fixed by where the two O tiles land, so counting rows is choosing 2 of 5 slots: C(5, 2)=10 rows.
Fixing the two O spots fixes the whole row, so counting rows is just choosing 2 slots out of 5.
Fixing where the two like letters sit fixes the whole row, so counting rows is choosing those two slots.
▸ Why?
Each choice of slots names exactly one row and each row names one choice, so counting either counts both.
▸ Why?
The two slots are chosen together out of one row of places, so the count is a plain combination.
Count the favorable rows
XOXOX pins the Os to positions 2 and 4, so the number of favorable rows among the ten is 1.
XOXOX pins down every letter, so only one of the listed rows can match it.
7.SP.C.8Identify SubproblemsDivide favorable by total
Every row is equally likely, so the one favorable row has probability 1/10 — 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.7Eliminate PossibilitiesOnly the two O spots decide the row, so there are C(5, 2)=10 equally likely rows and exactly one is XOXOX, making the probability 1/10.
- Count every possible row
- Count the favorable rows
- Divide favorable by total