AMC 10 · 2022 · #23

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basicgeometric-probabilityexpected-valuecomplementary-countingfraction-arithmetic complementary-countingeasier-related-problemidentify-subproblems ↑ Prerequisites: probability-basicgeometric-probability

📏 Long solution 💡 4 insights

Problem

Ant Amelia starts at 0. For each step n = 1, 2, 3 she picks a time t_n and a distance x_n, each independently and uniformly from (0,1). On step n she moves x_n to the right and uses t_n minutes. She stops at the end of a step whenever the total elapsed time first exceeds 1 minute, but she takes at most 3 steps. Find the probability that her final position exceeds 1.

Pick an answer.

(A)

$\frac{1}{3}$

(B)

$\frac{1}{2}$

(C)

$\frac{2}{3}$

(D)

$\frac{3}{4}$

(E)

$\frac{5}{6}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems) — the stopping rule splits into exactly two mutually exclusive scenarios (stop at step 2 vs. step 3), so the answer is a sum of two probabilities. Tool #9 (Easier) — first answer the cleaner sub-question "what is P(u₁ + u₂ ＞ 1) for two uniform (0,1) random variables?" with a small picture, before tackling three variables. Tool #1 (Diagram) — sketch the unit square (sum-of-two case) and unit cube with tetrahedron (sum-of-three case) to read the probabilities geometrically. Tool #16 (Complement) — for the sum-of-three, count the complement (sum ≤ 1, a tiny tetrahedron) rather than the bulky $\frac{5}{6}$ region directly.

1STEP 1

Two cases: S1 stops at step 2 (T₂,X₂ ＞ 1); S2 stops at step 3 (T₂ ≤ 1, X₃ ＞ 1). Independence factors each as P(time) · P(position).

P(pos ＞ 1) = P(T₂ ＞ 1) P(X₂ ＞ 1) + P(T₂ ≤ 1) P(X₃ ＞ 1)

2STEP 2

Warm-up: for uniforms u₁, u₂, the unit square has area 1 and u₁+u₂ ≤ 1 is a triangle of area $\frac{1}{2}$ , so P(u₁+u₂ ＞ 1) = $\frac{1}{2}$ .

P(u₁+u₂ ＞ 1) =

\frac{1}{2}

3STEP 3

Same square gives P(T₂ ＞ 1) = $\frac{1}{2}$ and P(X₂ ＞ 1) = $\frac{1}{2}$ , and P(T₂ ≤ 1) = $\frac{1}{2}$ . So P(S1) = $\frac{1}{4}$ .

P(S1) =

\frac{1}{2}

\frac{1}{2}

\frac{1}{4}

4STEP 4

For S2 need P(X₃ ＞ 1). Complement: x₁+x₂+x₃ ≤ 1 is a tetrahedron of volume $\frac{1}{6}$ in the unit cube, so P(X₃ ＞ 1) = $\frac{5}{6}$ .

P(X₃ ＞ 1) = 1 -

\frac{1}{6}

\frac{5}{6}

5STEP 5

S2 combines the time and position events: P(S2) = $\frac{1}{2}$ · $\frac{5}{6}$ = $\frac{5}{12}$ .

P(S2) =

\frac{1}{2}

\frac{5}{6}

\frac{5}{12}

6STEP 6

Add the exclusive scenarios: $\frac{1}{4}$ + $\frac{5}{12}$ = $\frac{8}{12}$ = $\frac{2}{3}$ , choice (C).

\frac{1}{4}

\frac{5}{12}

\frac{8}{12}

\frac{2}{3}

→ (C)

Answer

\frac{2}{3}

Sanity. The two scenarios partition the event "Amelia stops eventually" — and they must, because Amelia always stops by step 3 (the rules force it). Each scenario probability is non-negative and at most the marginal P(stop after that step). We computed

\frac{1}{4}

\frac{5}{12}

\frac{2}{3}

≈ 0.667 — squarely in (0,1), larger than

\frac{1}{2}

(consistent with the fact that taking three steps usually pushes Amelia past 1). Choice (C) matches. The other choices map to common miscounts: (B)

\frac{1}{2}

ignores that step 3 is more likely to land past 1; (E)

\frac{5}{6}

takes P(X₃ ＞ 1) without the time condition; (A)

\frac{1}{3}

is the complement guess.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability you already know — the stop rule cleanly splits the sample space into "stop at step 2" and "stop at step 3". For the first, both the time-sum and the position-sum live on a unit square, so each event has probability $\frac{1}{2}$ . For the second, the position-sum of three uniforms exceeds 1 with probability 1 - $\frac{1}{6}$ = $\frac{5}{6}$ (complement of a tetrahedron). Add: $\frac{1}{4}$ + $\frac{5}{12}$ = $\frac{2}{3}$ .

More like this