AMC 10 · 2025 · #22

Grade 10 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Volume

geometric-probabilityvolume-pyramidsymmetry-argumentprobability-basic physical-representationidentify-subproblems ↑ Prerequisites: geometric-probability

📏 Long solution 💡 3 insights

Problem

We pick three numbers independently and uniformly at random from 0 to 1. We want the probability that the largest of the three is more than twice each of the other two. (This is the reading of the contest's condition that matches the listed answer choices.)

Pick an answer.

(A)

$\frac{1}{12}$

(B)

$\frac19$

(C)

$\frac18$

(D)

$\frac16$

(E)

$\frac14$

How to solve

Strategy Visualize Spatial Relationships

Three uniform picks are the same as one random point in the unit cube, so a probability becomes a volume. Splitting into the three symmetric cases (which number is the largest) turns the count into one solid — a pyramid — whose volume we read straight off the pyramid formula.

1STEP 1

Turn chance into volume

Read (X, Y, Z) as a random point spread evenly over the unit cube, so any event's probability equals the volume of its region.

P(event) = vol{(X,Y,Z)∈[0,1]³ : event holds}

2STEP 2

Split by which number is largest

Only one number can be largest, so by symmetry the event splits into three equal cases: the total is 3 times the X-dominant case.

P = P_X + P_Y + P_Z = 3 P_X

3STEP 3

Write the condition as inequalities

With X dominant, X beating twice each rival means Y and Z are both less than $\frac{X}{2}$ — which already forces X to be the largest.

X ＞ 2Y and X ＞ 2Z ⇔ Y ＜ X/2, Z ＜ X/2

4STEP 4

See the region as a pyramid

At height X = x the allowed (Y, Z) fill a square of side $\frac{x}{2}$ that grows from 0 to $\frac{1}{2}$ ; stacked, these form a pyramid tipped at the origin.

cross-section side = x/2, 0 ≤ x ≤ 1

5STEP 5

Volume of that pyramid

One third of base area times height gives one third times $\frac{1}{4}$ times 1, so the X-dominant case has probability $\frac{1}{12}$ .

V = 1/3 · (1/2)² · 1 = 1/3·1/4 = 1/12

6STEP 6

Triple it for the final answer

Adding the three equal, non-overlapping cases gives 3 times $\frac{1}{12}$ = $\frac{1}{4}$ , choice (E).

P = 3 · 1/12 = 1/4

Answer

1/4

\frac{1}{4}

is a sensible size: it sits well below

\frac{1}{2}

, the chance that the largest merely beats the plain sum of the other two. That ordering fits, because 'beat twice each rival' is a harder condition than 'beat the sum', so it must be less likely. The volume

\frac{1}{12}

for one case is clearly a small corner of the unit cube, and tripling a twelfth to a quarter is consistent.

💡Key takeaway

When you pick numbers at random, a probability is just how much of the shape they live in gets filled — here the winning region is a pyramid holding a twelfth of the cube, and three of them make a quarter.

Turn chance into volume
Split by which number is largest
Write the condition as inequalities
See the region as a pyramid
Volume of that pyramid
Triple it for the final answer