AMC 8 · 2011 · #12
Grade 7 probabilityProblem
Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
Pick an answer.
AMC 8 2011 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: Four people — Angie, Bridget, Carlos, and Diego — sit at random around a square table, one person per side. What is the probability that Angie and Carlos end up on opposite sides?
Givens: $4$ people, $4$ sides of a square — one person per side; All seatings are equally likely ("at random"); On a square, each side has exactly one opposite side; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Unknowns: The probability that Angie and Carlos sit on opposite sides of the square
Understand
Restated: Four people — Angie, Bridget, Carlos, and Diego — sit at random around a square table, one person per side. What is the probability that Angie and Carlos end up on opposite sides?
Givens: $4$ people, $4$ sides of a square — one person per side; All seatings are equally likely ("at random"); On a square, each side has exactly one opposite side; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem
Probability with equally likely outcomes is $\dfrac{\text{favorable}}{\text{total}}$, so the job is just careful counting. Tool #1 (Draw a Diagram) pins down what "opposite" means on a square — a seat has one opposite seat (across) and two adjacent seats. Tool #9 (Easier Related Problem) shrinks the sample space: by the rotational symmetry of the square we can fix Angie on one side, turning $4! = 24$ arrangements into just $3! = 6$ arrangements of the other three people. Tool #2 (Systematic List) then enumerates those $6$ cases and counts how many put Carlos directly across from Angie.
Execute — Answer: B
4.G.A.2 Step 1 - Draw the square table and label its four sides Top, Right, Bottom, Left.
- The key feature: Top and Bottom are opposite, Left and Right are opposite.
- Every side has exactly one opposite side and two adjacent sides.
💡 Recognizing the pairs of parallel sides of a square — and that "opposite" means "across, not adjacent" — is the Grade 4 shape-classification skill.
7.SP.C.7 Step 2 - Use the square's symmetry to simplify.
- Rotating the table doesn't change who sits opposite whom, so we lose nothing by placing Angie at the Top.
- Now only the other three people (Bridget, Carlos, Diego) are random, in the three remaining seats.
💡 Shrinking the sample space using a symmetry — without changing any probability — is the Grade 7 "build a fair probability model" move.
7.SP.C.8 Step 3 List the $6$ ways to seat (Bridget, Carlos, Diego) into (Right, Bottom, Left), in alphabetical order of the first seat to keep the list disciplined.
💡 Writing out the $3! = 6$ orderings systematically — first letter first — is the Grade 7 "organized list" outcome-counting method.
7.SP.C.7 Step 4 - Carlos is opposite Angie exactly when Carlos sits at the Bottom (the middle slot in our tuple).
- Highlight those arrangements: $(B,C,D)$ and $(D,C,B)$.
- That's $2$ favorable cases out of $6$.
💡 Probability from a finite, equally likely sample space is favorable $\div$ total — Grade 7 probability fundamentals.
4.G.A.2 Draw the square table and label its four sides Top, Right, Bottom, Left. The key 7.SP.C.7 Use the square's symmetry to simplify. Rotating the table doesn't change who sit 7.SP.C.8 List the $6$ ways to seat (Bridget, Carlos, Diego) into (Right, Bottom, Left), i 7.SP.C.7 Carlos is opposite Angie exactly when Carlos sits at the Bottom (the middle slot Review
Reasonableness: Quick sanity check by direct reasoning: after Angie picks any side, Carlos is equally likely to land on any of the $3$ remaining sides. Exactly $1$ of those $3$ sides is opposite Angie, so $P = \tfrac{1}{3}$. This matches the listed count $\tfrac{2}{6} = \tfrac{1}{3}$ and lands on choice (B). The answer also passes the gut-check: "opposite" is the strictest of the three positions Carlos can take (opposite, left of Angie, right of Angie), so a probability near $\tfrac{1}{3}$ — not $\tfrac{1}{2}$ or higher — is what we expect.
Alternative: Tool #16 (Change Focus): instead of listing all $24$ full arrangements, just think about Carlos's seat. Once Angie is seated, Carlos's seat is uniformly random among the remaining $3$ — Bottom, Left, or Right. Only Bottom is opposite Angie, so $P = \tfrac{1}{3}$ directly, no list needed. Same answer (B).
CCSS standards used (min grade 7)
4.G.A.2Classify two-dimensional figures by properties of their sides and angles (Identifying the pairs of opposite sides of the square table — the geometric meaning of "opposite each other".)7.SP.C.7Develop a probability model and use it to find probabilities of events (Setting up the uniform probability model on the $6$ equally likely arrangements after fixing Angie's seat, then computing $P = \tfrac{\text{favorable}}{\text{total}} = \tfrac{2}{6}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Listing the $3! = 6$ seatings of Bridget, Carlos, and Diego systematically and counting the ones with Carlos opposite Angie.)
⭐ Fix Angie at one side, list where the other three can sit, and count: this AMC 8 probability question is a clean Grade 7 "favorable over total" calculation.
⭐ Fix Angie at one side, list where the other three can sit, and count: this AMC 8 probability question is a clean Grade 7 "favorable over total" calculation.
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