AMC 10 · 2022 · #12

Grade 7 probability

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

probability-basiccomplementary-countingexponentsfraction-arithmeticbound-inequality-then-enumerate complementary-countingbound-inequality-then-enumerateguess-and-check ↑ Prerequisites: probability-basicfraction-arithmetic

📏 Medium solution 💡 2 insights

Problem

You roll two fair 6-sided dice n times. Each roll, the two faces are added. We want the smallest n for which the chance of seeing a sum of 7 at least once across the n rolls is greater than

\frac{1}{2}

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Tool #16 (Complement) is the standard trick for "at least once" problems. Counting all the ways a 7 could appear in any of the n rolls forces messy overlap cases; the opposite event — "no 7 in any roll" — is one clean number per n. Tool #2 (Systematic List) gives the 6 ordered pairs that sum to 7 out of 36, so a single-roll 7-probability is $\frac{6}{36}$ = $\frac{1}{6}$ . Tool #6 (Guess and Check) plugs the answer choices n=2,3,4,… into ( $\frac{5}{6}$ )ⁿ until the value drops below $\frac{1}{2}$ . Tool #3 (Eliminate) keeps us efficient — once a choice works and the smaller ones fail, we stop.

1STEP 1

Two dice give 36 equally likely outcomes; exactly 6 ordered pairs sum to 7, so one roll gives P(sum=7) = $\frac{1}{6}$ .

P(7) =

\frac{6}{36}

\frac{1}{6}

2STEP 2

Flip to the complement: one roll has P(not 7) = $\frac{5}{6}$ , since 1 − $\frac{1}{6}$ = $\frac{5}{6}$ .

P(not 7) = 1 -

\frac{1}{6}

\frac{5}{6}

3STEP 3

Rolls are independent, so no 7 across all n rolls is ( $\frac{5}{6}$ )ⁿ, giving P(at least one 7) = 1 - ( $\frac{5}{6}$ )ⁿ.

P(at least one 7) = 1 - (

\frac{5}{6}

)ⁿ

4STEP 4

"Greater than $\frac{1}{2}$ " rearranges to the cleaner test ( $\frac{5}{6}$ )ⁿ ＜ $\frac{1}{2}$ ; find the smallest n satisfying it.

(

\frac{5}{6}

)ⁿ ＜

\frac{1}{2}

5STEP 5

Test choices: ( $\frac{5}{6}$ )² ≈ 0.69 and ( $\frac{5}{6}$ )³ ≈ 0.58 stay above, but ( $\frac{5}{6}$ )⁴ ≈ 0.482 drops below $\frac{1}{2}$ .

(

\frac{5}{6}

)⁴ =

\frac{625}{1296}

≈ 0.482 ＜

\frac{1}{2}

6STEP 6

The least n that works is n = 4 — choice (C); n=5 and n=6 also satisfy it but are bigger than needed.

n = 4 → (C)

Answer

A single roll gives a 7 only about 16.7% of the time, so one or two rolls clearly cannot reach a 50% chance. Three rolls land at about 42% (still short), four rolls climb to about 52% (just over). This matches the gut feeling — you need a handful of rolls before "at least one 7" becomes more likely than not. Magnitudes also check out: 0.48 + 0.52 = 1, confirming the complement arithmetic is balanced.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability you already know — flipping "at least one 7" into its opposite "no 7 at all" turns the question into ( $\frac{5}{6}$ )ⁿ ＜ $\frac{1}{2}$ , and testing n=2,3,4 shows n=4 is the smallest that works.

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