AMC 10 · 2022 · #12

Grade 7 probability

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

📏 Medium solution 💡 2 insights

Problem

A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$ ?

$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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 $\tfrac{1}{2}$.

Givens: Two fair 6-sided dice, rolled $n$ times; Each roll has $6 \times 6 = 36$ equally likely outcomes; Rolls are independent of each other; Answer choices: (A) 2, (B) 3, (C) 4, (D) 5, (E) 6

Unknowns: The smallest integer $n$ with $P(\text{at least one sum-of-7 in } n \text{ rolls}) > \tfrac{1}{2}$

Understand

Givens: Two fair 6-sided dice, rolled $n$ times; Each roll has $6 \times 6 = 36$ equally likely outcomes; Rolls are independent of each other; Answer choices: (A) 2, (B) 3, (C) 4, (D) 5, (E) 6

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #2 Make a Systematic List, #6 Guess and Check, #3 Eliminate Possibilities

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 $\tfrac{6}{36} = \tfrac{1}{6}$. Tool #6 (Guess and Check) plugs the answer choices $n=2,3,4,\ldots$ into $(\tfrac{5}{6})^n$ until the value drops below $\tfrac{1}{2}$. Tool #3 (Eliminate) keeps us efficient — once a choice works and the smaller ones fail, we stop.

Execute — Answer: C

#2 Make a Systematic List 7.SP.C.7 Step 1

Find the chance of rolling a sum of 7 on one roll.
With two dice there are $6 \times 6 = 36$ equally likely outcomes.
List the ordered pairs that sum to 7: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — exactly 6 pairs.
So one-roll $P(\text{sum}=7) = \tfrac{6}{36} = \tfrac{1}{6}$.

$$P(7) = \dfrac{6}{36} = \dfrac{1}{6}$$

💡 Grade 7: a probability is just (favorable outcomes) / (total outcomes), and listing is the safe way to count.

#16 Change Focus / Count the Complement 7.SP.C.5 Step 2

Flip to the complement.
The event "at least one 7 in $n$ rolls" is the opposite of "no 7 in any roll." One-roll $P(\text{not } 7) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$.

$$P(\text{not } 7) = 1 - \tfrac{1}{6} = \tfrac{5}{6}$$

💡 Grade 7: probabilities of an event and its complement add to 1, so subtracting is faster than re-listing.

#16 Change Focus / Count the Complement 7.SP.C.8 Step 3

Multiply across independent rolls.
The rolls are independent, so the chance of getting no 7 across all $n$ rolls is $\left(\tfrac{5}{6}\right)^n$.
Therefore the chance of at least one 7 is $1 - \left(\tfrac{5}{6}\right)^n$.

$$P(\text{at least one } 7) = 1 - \left(\dfrac{5}{6}\right)^n$$

💡 Grade 7: independent events multiply, so "no 7" $n$ times in a row is $(\tfrac{5}{6})^n$.

#16 Change Focus / Count the Complement 6.EE.B.8 Step 4

Translate "greater than $\tfrac{1}{2}$" into a clean inequality.
We need $1 - (\tfrac{5}{6})^n > \tfrac{1}{2}$, i.e.
$(\tfrac{5}{6})^n < \tfrac{1}{2}$.
So we look for the smallest $n$ that pushes $(\tfrac{5}{6})^n$ below one-half.

$$\left(\dfrac{5}{6}\right)^n < \dfrac{1}{2}$$

💡 Grade 6: rearranging the inequality keeps the same meaning, just easier to test.

#6 Guess and Check 6.EE.A.1 Step 5

Test the answer choices in order.
$n=2$: $(\tfrac{5}{6})^2 = \tfrac{25}{36} \approx 0.69$ — too big.
$n=3$: $\tfrac{125}{216} \approx 0.58$ — still too big.
$n=4$: $\tfrac{625}{1296} \approx 0.48$ — finally below $0.5$.
So $n=4$ is the smallest.

$$\left(\dfrac{5}{6}\right)^4 = \dfrac{625}{1296} \approx 0.482 < \dfrac{1}{2}$$

💡 Grade 6 exponents: each new roll multiplies by another $\tfrac{5}{6}$, so the value shrinks step by step.

#3 Eliminate Possibilities 6.EE.B.5 Step 6

Match $n=4$ to the answer choices: that is (C).
Choices (D) $n=5$ and (E) $n=6$ also satisfy the inequality but are bigger than needed; (A) $n=2$ and (B) $n=3$ fall short.

$$n = 4 \;\Rightarrow\; \textbf{(C)}$$

💡 Final compare: the question asks for the LEAST $n$, so we stop at the first success.

[1] #2 7.SP.C.7 Find the chance of rolling a sum of 7 on one roll. With two dice there are $6 \t

[2] #16 7.SP.C.5 Flip to the complement. The event "at least one 7 in $n$ rolls" is the opposite

[3] #16 7.SP.C.8 Multiply across independent rolls. The rolls are independent, so the chance of g

[4] #16 6.EE.B.8 Translate "greater than $\tfrac{1}{2}$" into a clean inequality. We need $1 - (\

[5] #6 6.EE.A.1 Test the answer choices in order. $n=2$: $(\tfrac{5}{6})^2 = \tfrac{25}{36} \app

[6] #3 6.EE.B.5 Match $n=4$ to the answer choices: that is (C). Choices (D) $n=5$ and (E) $n=6$

Review

Reasonableness: 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.

Alternative: Tool #2 (Systematic List) without the complement: count ordered $n$-roll sequences that contain at least one 7. For $n=4$ this means summing inclusion-exclusion terms ($\binom{4}{1} \cdot 6 \cdot 30^3 - \binom{4}{2} \cdot 6^2 \cdot 30^2 + \cdots$) over $36^4 = 1{,}679{,}616$ total. Same answer, far more arithmetic — which is exactly why the complement tool wins.

CCSS standards used (min grade 7)

6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Computing $(\tfrac{5}{6})^2, (\tfrac{5}{6})^3, (\tfrac{5}{6})^4$ to test the inequality.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Selecting the smallest integer $n$ from the answer choices that satisfies $(\tfrac{5}{6})^n < \tfrac{1}{2}$.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Rewriting "chance greater than $\tfrac{1}{2}$" as $(\tfrac{5}{6})^n < \tfrac{1}{2}$ to test.)
7.SP.C.5 Understand that the probability of a chance event is between 0 and 1 (Using $P(\text{not } 7) = 1 - P(7) = \tfrac{5}{6}$ from the complement rule.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Computing $P(\text{sum}=7) = \tfrac{6}{36}$ from the uniform-outcome model on two dice.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Multiplying independent-roll probabilities to get $P(\text{no 7 in } n \text{ rolls}) = (\tfrac{5}{6})^n$.)

⭐ 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 $(\tfrac{5}{6})^n < \tfrac{1}{2}$, and testing $n=2,3,4$ shows $n=4$ is the smallest that works.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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