AMC 8 · 2024 · #25
Grade 7 probabilityProblem
A small airplane has rows of seats with seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A small airplane has $4$ rows of $3$ seats each, so $12$ seats total. Eight passengers are already seated at random, leaving $4$ empty seats. We want the probability that the next-to-board married couple can find $2$ **adjacent** seats in the **same row**.
Givens: Plane layout: $4$ rows $\times$ $3$ seats $= 12$ seats; Passengers already seated: $8$; Empty seats remaining: $12 - 8 = 4$; The $8$ passengers occupy seats **uniformly at random**; Answer choices: (A) $\tfrac{8}{15}$, (B) $\tfrac{32}{55}$, (C) $\tfrac{20}{33}$, (D) $\tfrac{34}{55}$, (E) $\tfrac{8}{11}$
Unknowns: The probability that there exist $2$ adjacent empty seats in some row for the couple
Understand
Restated: A small airplane has $4$ rows of $3$ seats each, so $12$ seats total. Eight passengers are already seated at random, leaving $4$ empty seats. We want the probability that the next-to-board married couple can find $2$ **adjacent** seats in the **same row**.
Givens: Plane layout: $4$ rows $\times$ $3$ seats $= 12$ seats; Passengers already seated: $8$; Empty seats remaining: $12 - 8 = 4$; The $8$ passengers occupy seats **uniformly at random**; Answer choices: (A) $\tfrac{8}{15}$, (B) $\tfrac{32}{55}$, (C) $\tfrac{20}{33}$, (D) $\tfrac{34}{55}$, (E) $\tfrac{8}{11}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #2 Make a Systematic List, #1 Draw a Diagram, #9 Solve an Easier Related Problem
"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 $4\times 3$ 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).
Execute — Answer: C
K.G.A.1 Step 1 - Draw the $4\times 3$ grid and zoom in on just one row of $3$ seats.
- Count the ways to place empty seats in that single row with no two adjacent: $0$ empties $\to 1$ way; $1$ empty $\to 3$ positions; $2$ empties $\to$ "left-middle" and "middle-right" are adjacent (forbidden), only **"left and right" end seats** work, so $1$ way; $3$ empties $\to$ always adjacent, $0$ ways.
- Per row, the counts for $(0,1,2,3)$ empties are $(1,3,1,0)$.
💡 Naming the seats by their position — "left, middle, right" — is the basic position vocabulary from Kindergarten geometry.
7.SP.C.8 Step 2 - Count the **total** number of ways to pick which $4$ of the $12$ seats are empty (the denominator).
- Order does not matter, so it is a combination: $\binom{12}{4}$.
- This becomes the common denominator that lets us compare "couple can sit" vs.
- "cannot sit" cleanly.
💡 Counting all outcomes with a single combination $\binom{12}{4}$ is the standard Grade 7 "organized list / compound event" move.
7.SP.C.8 Step 3 - Now count the **opposite** event: no two empty seats are adjacent.
- Use Tool #2 to systematically list how the $4$ empties split across the $4$ rows.
- Since each row can hold at most $2$ non-adjacent empties (a row of $3$ empties always has an adjacent pair), the only partitions of $4$ into parts $\le 2$ are exactly three: $(1,1,1,1)$, $(2,1,1,0)$, $(2,2,0,0)$.
💡 Listing every valid distribution of empties across rows is exactly the organized-list technique Grade 7 uses for compound events.
7.SP.C.8 Step 4 - Multiply the per-row counts $(1,3,1)$ from Step 1 for each partition.
- Partition $(1,1,1,1)$: every row has $1$ empty $\to 3 \times 3 \times 3 \times 3 = 81$.
- Partition $(2,1,1,0)$: choose which row gets $2$ ($\binom{4}{1}=4$), arrange that row ($1$ way), choose which $2$ of the remaining $3$ rows get $1$ ($\binom{3}{2}=3$), arrange each of those rows ($3 \times 3$).
- Total $= 4 \times 1 \times 3 \times 3 \times 3 = 108$.
- Partition $(2,2,0,0)$: choose which $2$ rows get $2$ ($\binom{4}{2}=6$), each placed in $1$ way.
- Total $= 6$.
💡 Multiplying per-row counts and summing across partitions is the Grade 7 organized-list / multiplication-principle workflow for compound events.
7.SP.C.7 Step 5 - Compute the probability of the opposite event.
- Numerator $= 195$, denominator $= \binom{12}{4} = 495$.
- Reduce by the gcd $15$: $\dfrac{195}{495} = \dfrac{13}{33}$.
- This is the probability that the couple **cannot** sit together.
💡 Forming a probability as $\dfrac{\text{favorable outcomes}}{\text{all outcomes}}$ is the Grade 7 probability-model recipe.
5.NF.A.1 Step 6 - Apply the complement rule $P(A) = 1 - P(\text{not } A)$ to get the probability the couple **can** sit together: $1 - \dfrac{13}{33} = \dfrac{33 - 13}{33} = \dfrac{20}{33}$.
- Among the choices, $\dfrac{20}{33}$ matches **(C)** exactly; the other options $\dfrac{8}{15}, \dfrac{32}{55}, \dfrac{34}{55}, \dfrac{8}{11}$ all differ from $\dfrac{20}{33}$, so Tool #3 (eliminate) confirms (C).
💡 Subtracting a fraction from $1 = \dfrac{33}{33}$ over a common denominator is exactly the Grade 5 fraction subtraction skill.
K.G.A.1 Draw the $4\times 3$ grid and zoom in on just one row of $3$ seats. Count the wa 7.SP.C.8 Count the **total** number of ways to pick which $4$ of the $12$ seats are empty 7.SP.C.8 Now count the **opposite** event: no two empty seats are adjacent. Use Tool #2 t 7.SP.C.8 Multiply the per-row counts $(1,3,1)$ from Step 1 for each partition. Partition 7.SP.C.7 Compute the probability of the opposite event. Numerator $= 195$, denominator $= 5.NF.A.1 Apply the complement rule $P(A) = 1 - P(\text{not } A)$ to get the probability t Review
Reasonableness: The answer $\dfrac{20}{33} \approx 0.606$ is just over $\tfrac{1}{2}$ and well below $1$ — a sensible probability. With $4$ empty seats spread across $4$ rows of $3$, and each row offering $2$ possible adjacent pairs, it is reasonable that the couple usually finds a pair. A quick second check: the number of "good" (couple-can-sit) arrangements is $495 - 195 = 300$, and $\dfrac{300}{495} = \dfrac{20}{33}$ — same answer.
Alternative: An alternative is to apply Tool #2 (Systematic List) **directly** to the favorable event: split into cases by how the adjacent empty pairs appear (e.g., exactly one adjacent pair: $240$, a whole row empty: $36$, two independent adjacent pairs: $24$, total $300$). That also yields $\dfrac{300}{495} = \dfrac{20}{33}$, but you have to be careful not to double-count, which is why the complement-first approach (Tool #16) is cleaner.
CCSS standards used (min grade 7)
K.G.A.1Describe positions of objects using above, below, beside, in front of (Naming "left, middle, right" within a $3$-seat row to identify which pairs are adjacent vs. the two end seats.)5.NF.A.1Add and subtract fractions with unlike denominators (Subtracting $1 - \dfrac{13}{33} = \dfrac{20}{33}$ over a common denominator and reducing $\dfrac{195}{495}$.)7.SP.C.7Develop probability models and use them to find probabilities of events (Defining the probability as (favorable outcomes) / (total outcomes) and computing $P(\text{no adjacent}) = \dfrac{195}{495}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting all outcomes $\binom{12}{4}=495$ and counting the favorable partitions $(1,1,1,1), (2,1,1,0), (2,2,0,0)$ giving $81+108+6=195$ via the multiplication principle.)
⭐ This last AMC 8 problem really only needs the Grade 7 "organized list for compound events" and "complement rule" you already know!
⭐ This last AMC 8 problem really only needs the Grade 7 "organized list for compound events" and "complement rule" you already know!
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