AMC 10 · 2005 · #12
Grade 7 probabilityTwelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Twelve fair six-sided dice are rolled at once. Multiply the twelve top numbers together. Find the probability that this product is a prime number.
Givens: Twelve dice are rolled, each independently showing $1,2,3,4,5,$ or $6$; Every face of every die is equally likely; The product is the twelve top numbers multiplied together; Answer choices: (A) $\left(\frac{1}{12}\right)^{12}$, (B) $\left(\frac{1}{6}\right)^{12}$, (C) $2\left(\frac{1}{6}\right)^{11}$, (D) $\frac{5}{2}\left(\frac{1}{6}\right)^{11}$, (E) $\left(\frac{1}{6}\right)^{10}$
Unknowns: The probability that the product of the twelve top faces is prime
Understand
Restated: Twelve fair six-sided dice are rolled at once. Multiply the twelve top numbers together. Find the probability that this product is a prime number.
Givens: Twelve dice are rolled, each independently showing $1,2,3,4,5,$ or $6$; Every face of every die is equally likely; The product is the twelve top numbers multiplied together; Answer choices: (A) $\left(\frac{1}{12}\right)^{12}$, (B) $\left(\frac{1}{6}\right)^{12}$, (C) $2\left(\frac{1}{6}\right)^{11}$, (D) $\frac{5}{2}\left(\frac{1}{6}\right)^{11}$, (E) $\left(\frac{1}{6}\right)^{10}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The word "prime" is the whole problem, so Tool #7 (Identify Subproblems) splits the work into three clean pieces: first decide exactly which rolls make the product prime, then count those winning rolls, then divide by the total number of rolls. The first piece is pure number theory — a prime has a single prime factor, which forces eleven of the dice to be $1$ and the last to be a prime face. Tool #2 (Make a Systematic List) handles the counting: pick which die is the prime and which prime it shows. Tool #3 (Eliminate Possibilities) is the final match — the count collapses to a single power of $\frac{1}{6}$, and only one answer choice has that clean form.
Execute — Answer: E
4.OA.B.4 Step 1 When is the product prime?
- A prime number is built from exactly one prime factor, used a single time — like $2$, $3$, or $5$.
- Multiplying introduces every prime factor from every die, so to keep the product prime, eleven of the twelve dice must contribute no prime factors at all, meaning they show $1$.
- The one remaining die must contribute exactly one prime factor, so it shows a prime face: $2$, $3$, or $5$.
- A face of $4=2\times2$ or $6=2\times3$ would add extra factors and spoil the prime, and any second non-$1$ die would too.
💡 A prime cannot be split, so only one die may carry a single prime while all the rest sit at $1$.
7.SP.C.8 Step 2 Count the winning rolls
- Build a winning roll by two independent choices.
- First choose which one of the twelve dice is the prime one: $12$ ways.
- Then choose what prime that die shows — $2$, $3$, or $5$: $3$ ways.
- Every other die is forced to be $1$, which is just $1$ way each.
- Multiply the choices together to count all winning rolls.
💡 Independent choices multiply, so $12$ positions times $3$ prime faces gives every good roll.
6.EE.A.1 Step 3 Count all possible rolls
- Each of the twelve dice independently shows one of six faces, so the total number of equally likely rolls multiplies six by itself twelve times.
- That is $6^{12}$.
💡 Twelve independent six-way choices stack up as the power $6^{12}$.
6.EE.A.1 Step 4 Divide and simplify
- The probability is the winning rolls over all rolls: $\frac{36}{6^{12}}$.
- Since $36 = 6\times6 = 6^{2}$, two of the twelve sixes in the bottom cancel, leaving $6^{10}$ underneath.
- So the probability is $\frac{1}{6^{10}} = \left(\frac{1}{6}\right)^{10}$, which is choice (E).
💡 The $36$ on top is just two sixes, so it erases two sixes from the bottom.
4.OA.B.4 A prime number is built from exactly one prime factor, used a single time — like 7.SP.C.8 Build a winning roll by two independent choices. First choose which one of the t 6.EE.A.1 Each of the twelve dice independently shows one of six faces, so the total numbe 6.EE.A.1 The probability is the winning rolls over all rolls: $\frac{36}{6^{12}}$. Since Review
Reasonableness: The result $\left(\frac{1}{6}\right)^{10}$ is a positive number far below $1$, which fits — a prime product is a rare event. It sits neatly between the other tiny choices: it is exactly $6\left(\frac{1}{6}\right)^{11}$, larger than (C) $2\left(\frac{1}{6}\right)^{11}$ and (D) $\frac{5}{2}\left(\frac{1}{6}\right)^{11}$, which are the traps for a solver who counts only one or two prime faces instead of all three, or who forgets to place the prime on any of the twelve dice. Choice (B) $\left(\frac{1}{6}\right)^{12}$ is the trap for counting a single fixed winning roll and forgetting the $36$ ways it can happen.
Alternative: Work with probabilities directly instead of counts (Tool #4-style setup). The chance a specific chosen die shows a prime is $\frac{3}{6} = \frac{1}{2}$, and each of the other eleven dice shows $1$ with chance $\frac{1}{6}$, so one labeled arrangement has probability $\frac{1}{2}\left(\frac{1}{6}\right)^{11}$. There are $12$ choices for which die is the prime one, giving $12\cdot\frac{1}{2}\left(\frac{1}{6}\right)^{11} = 6\left(\frac{1}{6}\right)^{11} = \left(\frac{1}{6}\right)^{10}$ — the same (E).
CCSS standards used (min grade 7)
4.OA.B.4Find factor pairs and recognize that a whole number is prime or composite (Characterizing which rolls give a prime product: eleven dice at $1$ and one die at a prime face $2,3,$ or $5$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $36$ winning rolls as a compound event: $12$ positions times $3$ prime faces.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Writing the total as $6^{12}$ and simplifying $\frac{36}{6^{12}} = \left(\frac{1}{6}\right)^{10}$ by cancelling.)
⭐ A product is prime only when one die shows a prime ($2,3,$ or $5$) and the rest all show $1$, so count $12\times3=36$ winning rolls out of $6^{12}$ to get $\left(\frac{1}{6}\right)^{10}$.
⭐ A product is prime only when one die shows a prime ($2,3,$ or $5$) and the rest all show $1$, so count $12\times3=36$ winning rolls out of $6^{12}$ to get $\left(\frac{1}{6}\right)^{10}$.
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