Competition · AMC preparation · step 4 of 4
AMC 8 · 2024 · #25
Grade 7 probability
Pick an answer.
AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
"Probability the couple can sit together" = "probability that among 4 empty seats, at least one adjacent pair lies in the same row." The phrase "at least one" is the textbook trigger for Tool #16 (Complement) — count the opposite event (no two empty seats are adjacent) instead, because it has far fewer cases. To count the opposite, use Tool #2 (Systematic List) on how the 4 empties split across the 4 rows. Tool #1 (Diagram) keeps the 4 × 3 grid visible, and Tool #9 (Easier Problem) lets us first work out the tiny one-row subproblem (how many non-adjacent placements fit in a single 3-seat row).
Count non-adjacent placements in one row
In one row of 3 seats with no two empties adjacent, the counts for (0,1,2,3) empties are (1,3,1,0) — two empties must be the end seats.
Naming the seats by their position — "left, middle, right" — is the basic position vocabulary from Kindergarten geometry.
K.G.A.1Draw A DiagramCount all ways to empty four seats
The denominator is the ways to pick which 4 of 12 seats are empty; order-free, so the combination C(12,4) = 495.
Counting all outcomes with a single combination C(12, 4) is the standard Grade 7 "organized list / compound event" move.
7.SP.C.8Change Focus Count The ComplementCount the opposite event
For the opposite event, list how 4 empties split across rows; with ≤2 non-adjacent per row, only (1,1,1,1), (2,1,1,0), (2,2,0,0) qualify.
Listing every valid distribution of empties across rows is exactly the organized-list technique Grade 7 uses for compound events.
7.SP.C.8Make A Systematic ListMultiply the per-row counts
Multiply the per-row (1,3,1) counts within each partition: 81, 108, and 6, which sum to 195 non-adjacent arrangements.
Multiplying per-row counts and summing across partitions is the Grade 7 organized-list / multiplication-principle workflow for compound events.
7.SP.C.8Make A Systematic ListFind the opposite probability
The opposite event's probability is ; dividing by the gcd 15 reduces it to — the chance the couple cannot sit together.
Forming a probability as (favorable outcomes)/(all outcomes) is the Grade 7 probability-model recipe.
7.SP.C.7Change Focus Count The ComplementSubtract from one
Apply the complement rule: 1 - = , which matches choice (C); none of the other four fractions equal it.
Subtracting a fraction from 1 = 33/33 over a common denominator is exactly the Grade 5 fraction subtraction skill.
The probability the couple can find two side-by-side empty seats in some row equals 1 minus the probability that no two of the four empty seats are side by side.
▸ Why?
Every way of leaving four seats empty either gives the couple a side-by-side pair in some row or it does not, and no single arrangement does both, so the chance of one case is just a full whole (1) with the other case's chance taken away.
▸ Why?
The arrangements 'with a side-by-side empty pair' and 'with none' together make up every arrangement, none shared and none left out, so their counts add back to the total number of arrangements.
▸ Why?
Sorting each arrangement by whether it holds a side-by-side empty pair drops it into exactly one of the two piles, with nothing counted twice and nothing missing, so the two piles rebuild the complete set of arrangements.
▸ Why?
Because all of the arrangements are equally likely, each case's probability is simply its share of that one complete set, so the two shares must fill up a single whole (1).
▸ Why?
Every arrangement of the four empty seats is just as likely as any other, so a case's probability is exactly its own count of arrangements divided by the total count; the 'pair' share and the 'no pair' share are those two counts over the same total, and since the counts add to that total the two shares add to a full whole (1).
This last AMC 8 problem really only needs the Grade 7 "organized list for compound events" and "complement rule" you already know!
- Count non-adjacent placements in one row
- Count all ways to empty four seats
- Count the opposite event
- Multiply the per-row counts
- Find the opposite probability
- Subtract from one
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