AMC 8 · 2011 · #12

Grade 7 probability
probability-basicsystematic-enumerationpermutations-basic systematic-enumerationcasework ↑ Prerequisites: probability-basic
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Problem

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?

(A) 14(B) 13(C) 12(D) 23(E) 34\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34

Pick an answer.

(A)
frac14
(B)
frac13
(C)
frac12
(D)
frac23
(E)
frac34
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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

#1 Draw a Diagram 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.
$$\text{opposite pairs} = \{\text{Top}, \text{Bottom}\},\ \{\text{Left}, \text{Right}\}$$

💡 Recognizing the pairs of parallel sides of a square — and that "opposite" means "across, not adjacent" — is the Grade 4 shape-classification skill.

#9 Solve an Easier Related Problem 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.
$$4! = 24 \;\xrightarrow{\text{fix Angie at Top}}\; 3! = 6 \text{ arrangements}$$

💡 Shrinking the sample space using a symmetry — without changing any probability — is the Grade 7 "build a fair probability model" move.

#2 Make a Systematic List 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.

$$\begin{array}{l} (B,C,D),\ (B,D,C),\ (C,B,D),\ (C,D,B),\ (D,B,C),\ (D,C,B) \end{array}$$

💡 Writing out the $3! = 6$ orderings systematically — first letter first — is the Grade 7 "organized list" outcome-counting method.

#2 Make a Systematic List 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 $6$.
$$P(\text{Carlos opposite Angie}) = \dfrac{2}{6} = \dfrac{1}{3} \;\Rightarrow\; \textbf{(B)}$$

💡 Probability from a finite, equally likely sample space is favorable $\div$ total — Grade 7 probability fundamentals.

[1] #1 4.G.A.2 Draw the square table and label its four sides Top, Right, Bottom, Left. The key
[2] #9 7.SP.C.7 Use the square's symmetry to simplify. Rotating the table doesn't change who sit
[3] #2 7.SP.C.8 List the $6$ ways to seat (Bridget, Carlos, Diego) into (Right, Bottom, Left), i
[4] #2 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.2 Classify 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.7 Develop 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.8 Find 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.