AMC 10 · 2022 · #22

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

An ant starts at zero. For each of three steps it picks a time and a distance, each independently and uniformly from zero to one. On each step it moves that distance right and uses that time. It stops at the end of the step where the total time first exceeds one minute, but takes at most three steps. Find the probability that its final position exceeds one.

Pick an answer.

(A)

$\frac{1}{3}$

(B)

$\frac{1}{2}$

(C)

$\frac{2}{3}$

(D)

$\frac{3}{4}$

(E)

$\frac{5}{6}$

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 5/6 region directly.

1STEP 1

Split into two cases

There are only two stopping cases.

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

2STEP 2

Two numbers exceeding one

Two uniform numbers exceed one half the time.

P(u₁+u₂ ＞ 1) = 1/2

3STEP 3

Probability of the first case

Multiply the two probabilities.

P(S1) = 1/2 · 1/2 = 1/4

4STEP 4

Three numbers exceeding one

Three uniform numbers almost always exceed one.

P(X₃ ＞ 1) = 1 - 1/6 = 5/6

5STEP 5

Probability of the second case

Compute the second case too.

P(S2) = 1/2 · 5/6 = 5/12

6STEP 6

Add the two cases

Adding gives two thirds.

1/4 + 5/12 = 8/12 = 2/3 → (C)

Answer

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 1/4 + 5/12 = 2/3 ≈ 0.667 — squarely in (0,1), larger than 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) 1/2 ignores that step 3 is more likely to land past 1; (E) 5/6 takes P(X₃ ＞ 1) without the time condition; (A) 1/3 is the complement guess.

💡Key takeaway

This AMC 12 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 1/2. For the second, the position-sum of three uniforms exceeds 1 with probability 1 - 1/6 = 5/6 (complement of a tetrahedron). Add: 1/4 + 5/12 = 2/3.