AMC 10 · 2019 · #22

Grade 7 probability

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

probability-basicgeometric-probabilitycaseworkconditional-probabilityfraction-arithmetic caseworkidentify-subproblems ↑ Prerequisites: probability-basicgeometric-probabilityconditional-probability

📏 Medium solution 💡 4 insights

Problem

A number x in [0, 1] is built as follows: flip a fair coin. If heads, flip again — get 0 on heads, 1 on tails. If tails, draw x uniformly from [0, 1]. Pick y independently the same way. What is the probability that |x - y| ＞

\frac{1}{2}

Pick an answer.

(A)

$\frac{1}{3}$

(B)

$\frac{7}{16}$

(C)

$\frac{1}{2}$

(D)

$\frac{9}{16}$

(E)

$\frac{2}{3}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): split into four cases by (branch for x) × (branch for y) — each independent and equally likely with probability $\frac{1}{4}$ . Tool #2 (Systematic List): enumerate all four cases (Discrete-Discrete, Discrete-Uniform, Uniform-Discrete, Uniform-Uniform). Tool #1 (Diagram): for the Uniform-Uniform case, draw the unit square and shade |x-y| ＞ $\frac{1}{2}$ — two right-triangle corners. Tool #9 (Easier Problem): each sub-case is a small problem solvable by counting or by a clean area; combining them is just a weighted sum.

1STEP 1

Split by which branch made x and y: the four joint cases DD, DU, UD, UU are independent, each with probability $\frac{1}{4}$ .

P(DD) = P(DU) = P(UD) = P(UU) =

\frac{1}{4}

2STEP 2

Case DD: both in {0, 1}, so |x-y| ＞ $\frac{1}{2}$ needs x ≠ y — two of four outcomes give conditional probability $\frac{1}{2}$ .

P(· ∣ DD) =

\frac{1}{2}

3STEP 3

Case DU: x is 0 or 1, y uniform. Either endpoint leaves half the interval more than $\frac{1}{2}$ away, so the conditional probability is $\frac{1}{2}$ .

P(· ∣ DU) =

\frac{1}{2}

4STEP 4

Case UD is DU with x and y swapped, so by symmetry the conditional probability is $\frac{1}{2}$ .

P(· ∣ UD) =

\frac{1}{2}

5STEP 5

Case UU: both uniform. On the unit square, |x - y| ＞ $\frac{1}{2}$ forms two corner triangles of area $\frac{1}{8}$ each — conditional probability $\frac{1}{4}$ .

P(· ∣ UU) = 2 ·

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

\frac{1}{4}

6STEP 6

Combine by the law of total probability with weight $\frac{1}{4}$ each: $\frac{1}{4}$ ( $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{4}$ ) = $\frac{7}{16}$ , choice (B).

P =

\frac{1}{4}

(

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

\frac{1}{4}

) =

\frac{7}{16}

Answer

\frac{7}{16}

\frac{7}{16}

≈ 0.44 is less than

\frac{1}{2}

. Intuitively if both x and y were uniform we'd get only

\frac{1}{4}

, but the discrete branch boosts the chance — when x or y snaps to an endpoint 0 or 1, escaping the other by more than

\frac{1}{2}

is easier. So a number between

\frac{1}{4}

and

\frac{1}{2}

is exactly what we expect. Conversion check:

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

\frac{1}{4}

\frac{2+2+2+1}{4}

\frac{7}{4}

, and

\frac{1}{4}

\frac{7}{4}

\frac{7}{16}

. Choice (D)

\frac{9}{16}

is the complementary probability P(|x-y| ≤

\frac{1}{2}

) trap.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability and unit-square geometric probability you already know — split into 4 equally likely cases (Discrete-Discrete, Discrete-Uniform, Uniform-Discrete, Uniform-Uniform), get conditional chances $\frac{1}{2}$ , $\frac{1}{2}$ , $\frac{1}{2}$ , $\frac{1}{4}$ , then average: $\frac{1}{4}$ ( $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{4}$ ) = $\frac{7}{16}$ .

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