AMC 10 · 2021 · #11

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split

complementary-countingindependent-eventsprime-factorizationprobability-basic complementary-countingcasework ↑ Prerequisites: probability-basiccomplementary-counting

📏 Medium solution 💡 3 insights

Problem

Six ordinary dice are thrown at once and the six numbers are multiplied together. Find the probability that the product is a multiple of 4. The size of the product does not matter — only whether the six faces together supply enough factors of two.

Pick an answer.

(A)

$\frac34$

(B)

$\frac{57}{64}$

(C)

$\frac{59}{64}$

(D)

$\frac{187}{192}$

(E)

$\frac{63}{64}$

How to solve

Strategy Change Focus / Count the Complement

Counting the rolls whose product IS divisible by 4 head-on is hopeless: almost every roll qualifies, and they qualify for many different reasons. That lopsidedness is exactly the signal for Tool #16 (Count the Complement) — find the probability of the rare failure and subtract from 1. Before the complement can be described, though, each die has to be read differently: Tool #15 (Organize Information in More Ways) says stop reading a die as a number from 1 to 6 and start reading it as "how many factors of 2 does this face contribute" — 0, 1, or 2. Tool #2 (Make a Systematic List) turns that into a three-row table of the six faces with their probabilities. Then failure means the six contributions total 0 or 1, which is exactly two cases, and Tool #7 (Identify Subproblems) handles them one at a time.

1STEP 1

Rewrite the goal as counting twos

Rewrite the goal as counting twos.

4 = 2² ⟹ product divisible by 4 ⇔ total factors of 2 ≥ 2

2STEP 2

Label each face by its twos

Label each face by its count of twos.

{1,3,5}: 0 twos, P = 3/6 = 1/2 {2,6}: 1 two, P = 2/6 = 1/3 {4}: 2 twos, P = 1/6

3STEP 3

Flip to the failing rolls

The failing rolls are far fewer.

P(divisible by 4) = 1 - P(total twos = 0) - P(total twos = 1)

4STEP 4

Case zero twos: all odd

Count the all-odd rolls.

P(total twos = 0) = (1/2)⁶ = 1/64

5STEP 5

Case one two: a single 2 or 6

Count the rolls with exactly one two.

P(total twos = 1) = 6 · 1/3 · (1/2)⁵ = 6 · 1/3 · 1/32 = 1/16 = 4/64

6STEP 6

Subtract the failures from one

Subtracting from one gives fifty-nine sixty-fourths.

P(not divisible by 4) = 1/64 + 4/64 = 5/64, P(divisible by 4) = 1 - 5/64 = 59/64 → (C)

Answer

59/64

Whole-number counting confirms the fractions. There are 6⁶ = 46,656 equally likely rolls. All-odd rolls number 3⁶ = 729. Rolls with exactly one die on 2 or 6 and the rest odd number 6 · 2 · 3⁵ = 6 · 2 · 243 = 2,916. Together that is 3,645 failures, and 3,645/46,656 = 5/64 exactly, so 59/64 stands. The size is also sensible: with six dice a product almost always picks up two factors of 2, so an answer just under 1 is expected, which rules out (A) 3/4 as far too small. The two near misses are the informative ones. Choice (E) 63/64 is what comes out if only the all-odd case is subtracted — that is the probability of being divisible by 2, not by 4. Choice (B) 57/64 comes from treating every even face as worth exactly one factor of 2: the failure probability then becomes 1/64 + 6 · 1/2 · (1/2)⁵ = 7/64, which forgets that a single 4 already finishes the job. Getting 59/64 means both traps were avoided.

💡Key takeaway

When a question asks for "at least" something and almost every case passes, count the few that fail and subtract from 1 — here only two failures exist: every die odd, or one lone 2 or 6.

Rewrite the goal as counting twos
Label each face by its twos
Flip to the failing rolls
Case zero twos: all odd
Case one two: a single 2 or 6
Subtract the failures from one