AMC 10 · 2002 · #16

Grade 7 probability

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

probability-basicindependent-eventsmultiples complementary-counting ↑ Prerequisites: probability-basic

📏 Medium solution 💡 2 insights

Problem

Juan rolls a fair 8-sided die showing 1 through 8, then Amal rolls a fair ordinary 6-sided die. Find the probability that the two numbers multiply to a multiple of 3.

Pick an answer.

(A)

$\frac 1{12}$

(B)

$\frac 13$

(C)

$\frac 12$

(D)

$\frac 7{12}$

(E)

$\frac 23$

How to solve

Strategy Change Focus / Count the Complement

"The product is a multiple of 3" is really an "at least one" statement in disguise, and "at least one" events are almost always cheaper to count backwards. Tool #7 (Identify Subproblems) does the translation first, using the fact that 3 is prime to replace a statement about the product with a statement about the two rolls separately. Tool #2 (Make a Systematic List) then reads off how many faces on each die are multiples of 3. Tool #16 (Change Focus / Count the Complement) computes the probability that neither roll is a multiple of 3 — a single clean product — and subtracts from 1. Tool #12 (Draw a Venn Diagram) is kept as the cross-check, since the same answer must come out of adding two overlapping events.

1STEP 1

Turn the product into two rolls

Since 3 is prime, the product is a multiple of 3 exactly when at least one roll is.

3 ∣ jm ⇔ 3 ∣ j or 3 ∣ m

2STEP 2

Count multiples of three on each die

Each die has two multiples of three, giving chances 1/4 and 1/3.

P(Juan) = 2/8 = 1/4, P(Amal) = 2/6 = 1/3

3STEP 3

Flip to neither roll working

The rolls are independent, so neither qualifying has probability 1/2.

P(neither) = 3/4 · 2/3 = 6/12 = 1/2

4STEP 4

Subtract from one

Subtracting from one gives 1/2, choice (C).

1 - 1/2 = 1/2 → (C)

Answer

1/2

Count outcomes directly instead of using probabilities. There are 8 × 6 = 48 equally likely pairs of rolls. Neither roll is a multiple of 3 in 6 × 4 = 24 of them (Juan has 6 safe faces, Amal has 4). So 48 - 24 = 24 pairs give a product divisible by 3, and 24/48 = 1/2, matching. The answer is also sensible in size: Amal alone supplies a 1/3 chance, so the total must exceed 1/3, ruling out (A) and (B); and it must be below 1/4 + 1/3 = 7/12, since simply adding the two chances double-counts the rolls where both are multiples of 3 — which is exactly why (D) 7/12 is the trap for forgetting the overlap.

💡Key takeaway

For "at least one", find the chance it never happens and subtract from 1 — one multiplication beats a pile of cases.

Turn the product into two rolls
Count multiples of three on each die
Flip to neither roll working
Subtract from one

Turn the product into two rolls

Count multiples of three on each die

Flip to neither roll working

Subtract from one

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