AMC 8 · 2001 · #18

Grade 7 probabilitycounting

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

probability-basiccomplementary-countingmultiplesfraction-arithmetic complementary-countingidentify-subproblems ↑ Prerequisites: probability-basicmultiples

📏 Short solution 💡 2 insights

Problem

Two fair 6-sided dice are thrown. What is the probability that the product of the two numbers showing is a multiple of 5?

Pick an answer.

(A)

$\dfrac{1}{36}$

(B)

$\dfrac{1}{18}$

(C)

$\dfrac{1}{6}$

(D)

$\dfrac{11}{36}$

(E)

$\dfrac{1}{3}$

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

First reduce 'product is a multiple of 5' to the cleaner event 'at least one die shows a 5' — true because 5 is prime and 5 is the only multiple of 5 on a die. The phrase 'at least one' is the classic flag for Tool #16 (Count the Complement): instead of summing 'exactly one 5' and 'exactly two 5s' separately, count the easier opposite event 'neither die shows a 5' and subtract from 1. Tool #2 (Make a Systematic List) supplies the 6 × 6 = 36 equally likely ordered outcomes that anchor the probability fractions.

1STEP 1

Because 5 is prime and 5 is the only multiple of 5 on a die, the event becomes 'at least one die shows a 5'.

{a · b ≡ 0 (mod 5)} = {a = 5 or b = 5}

2STEP 2

The two dice are independent, so the equally likely ordered pairs fill a 6 × 6 = 36 grid.

|sample space| = 6 × 6 = 36

3STEP 3

Count the easy opposite instead: neither die shows a 5, with probability $\frac{5}{6}$ · $\frac{5}{6}$ = $\frac{25}{36}$ .

P(no 5) =

\frac{5}{6}

\frac{5}{6}

\frac{25}{36}

4STEP 4

Subtract from 1: the probability is 1 - $\frac{25}{36}$ = $\frac{11}{36}$ .

P(at least one 5) = 1 -

\frac{25}{36}

\frac{11}{36}

→ (D)

Answer

\frac{11}{36}

Cross-check by direct count in the 6 × 6 grid. Outcomes with a 5 on the first die: (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) — that is 6. Outcomes with a 5 on the second die: (1,5),(2,5),(3,5),(4,5),(5,5),(6,5) — another 6. The pair (5,5) is in both lists, so by inclusion-exclusion the favorable count is 6 + 6 - 1 = 11, giving

\frac{11}{36}

— matches (D). The answer also fits a quick sanity check: the chance of getting a 5 on either die alone is

\frac{1}{6}

≈ 0.167, so 'at least one 5' should be a little less than

\frac{2}{6}

≈ 0.333 (because of the shared (5,5)), and

\frac{11}{36}

≈ 0.306 lands right there.

💡Key takeaway

A product is a multiple of 5 only when at least one die rolls a 5. Count the easy opposite — neither die is a 5, $\frac{5}{6}$ · $\frac{5}{6}$ = $\frac{25}{36}$ — and the answer is 1 - $\frac{25}{36}$ = $\frac{11}{36}$ .

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