Competition · AMC preparation · step 4 of 4
AMC 8 · 2011 · #12
Grade 7 probabilityPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Probability with equally likely outcomes is favorable/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.
Label the four seats
Label the square's sides Top, Right, Bottom, Left: Top faces Bottom, Left faces Right, so every side has exactly one opposite side.
Recognizing the pairs of parallel sides of a square — and that "opposite" means "across, not adjacent" — is the Grade 4 shape-classification skill.
4.G.A.2Draw A DiagramFix Angie's seat
Rotating the table changes nothing, so fix Angie at the Top; only the other three shuffle, leaving 3! = 6 arrangements.
Shrinking the sample space using a symmetry — without changing any probability — is the Grade 7 "build a fair probability model" move.
We may fix Angie on one chosen side of the table without changing the probability that Carlos ends up across from her.
▸ Why?
Turning the whole table a quarter-turn moves each person to the next side, so it changes one seating into another and shifts Angie to a new side.
▸ Why?
A quarter-turn slides the square exactly onto itself, so each side lands on the next side and two people who were across from each other stay across from each other.
▸ Why?
Four quarter-turns in a row bring the table back to its start, so the turns carry Angie once through every side.
▸ Why?
Those turns match every seating that has Angie on one side with exactly one seating that has her on any other side, so each side gives the same number of arrangements and the same share with Carlos across from her.
List the six arrangements
List all 6 orderings of (Bridget, Carlos, Diego) into (Right, Bottom, Left), alphabetically by who takes the first seat.
Writing out the 3! = 6 orderings systematically — first letter first — is the Grade 7 "organized list" outcome-counting method.
7.SP.C.8Make A Systematic ListCount the opposite cases
Carlos faces Angie only in the Bottom (middle) slot: (B,C,D) and (D,C,B) — that's 2 of 6 cases.
Probability from a finite, equally likely sample space is favorable ÷ total — Grade 7 probability fundamentals.
7.SP.C.7Make A Systematic ListFix 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.
- Label the four seats
- Fix Angie's seat
- List the six arrangements
- Count the opposite cases
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