AMC 10 · 2008 · #22
Grade 11 probabilityPick an answer.
Tool #15 (Organize Information in More Ways) does the work everyone skips. The problem describes a process — twelve drivers choosing one at a time from a shrinking menu — but every clean solution silently treats the four leftover spaces as a uniformly random 4-element subset. Those are different statements, and the second one has to be earned: re-record the process as an ordered list of choices, then collapse the list to a set and count how many lists give the same set. Tool #16 (Change Focus / Count the Complement) then flips the event, because "at least two empty spaces touch" is a mess of overlapping cases while "no two empty spaces touch" is one clean shape. Tool #4 (Introduce a Variable) names the four empty positions a₁ ＜ a₂ ＜ a₃ ＜ a₄ and converts "no two touch" into the inequality a_i+1 ≥ a_i + 2, which a shift map turns into an ordinary choice of 4 from 13 — and the shift map is exhibited with its inverse, so it is a genuine bijection rather than a plausible-looking pairing. Tool #3 (Eliminate Possibilities) closes: the five choices lie within 0.06 of each other, so only the exact fraction decides between them.
Earn the uniform model
Every set of empty spaces is equally likely, which must be proved.
Every ordering of the same twelve occupied spaces is one equally likely story, and each set of twelve owns exactly the same number of stories, so no set is favoured.
7.SP.C.7Organize Information In More WaysTurn parking into adjacency
Parking then means the empty set contains two neighbours.
One free pair side by side is both necessary and sufficient, so the question is purely about whether the four gaps touch.
10.S-CP.A.1Change Focus Count The ComplementSlide the four empties together
A shift makes counting the non-adjacent sets easy.
Deleting one space from each gap squeezes the row from 16 down to 13 and makes the forced separations vanish, and re-inserting them puts everything back exactly.
8.F.A.1Organize Information In More WaysCompute the failure probability
That gives the failure probability as 11/28.
Both events are counts of equally likely subsets, so the probability is just one count divided by the other.
11.S-CP.B.9Introduce A VariableSubtract to get the parking probability
Subtracting gives 17/28.
Every arrangement either has two empty spaces touching or does not, so one subtraction from 1 finishes it.
Every arrangement either has two empty spaces touching or it does not, so one subtraction from one finishes it.
▸ Why?
The wanted event and its opposite together account for everything, so their chances add to one.
▸ Why?
No arrangement can be in both, so nothing is counted twice when the two are separated.
Separate the five choices
Decimals separate the five choices cleanly, confirming 17/28, choice (E).
When five choices sit within a few hundredths of each other, only exact arithmetic decides, and the exact value lands on one of them.
7.NS.A.2Eliminate PossibilitiesCars parking one at a time leave every set of four empty spaces equally likely, so the whole question becomes: how often do four spots chosen from sixteen all avoid touching? Sliding them left by 0, 1, 2, 3 removes the forced gaps and turns that into a plain choice of four from thirteen.
- Earn the uniform model
- Turn parking into adjacency
- Slide the four empties together
- Compute the failure probability
- Subtract to get the parking probability
- Separate the five choices