AMC 8 · 2001 · #18
Grade 7 probabilitycountingProblem
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
Pick an answer.
AMC 8 2001 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: Two fair $6$-sided dice are thrown. What is the probability that the product of the two numbers showing is a multiple of $5$?
Givens: Two fair dice are thrown, each showing a number in $\{1,2,3,4,5,6\}$; The two dice are independent, so each of the $6 \times 6 = 36$ ordered outcomes is equally likely; We want $P(\text{product is a multiple of } 5)$; Answer choices: (A) $\tfrac{1}{36}$, (B) $\tfrac{1}{18}$, (C) $\tfrac{1}{6}$, (D) $\tfrac{11}{36}$, (E) $\tfrac{1}{3}$
Unknowns: The probability that the product of the two rolls is a multiple of $5$
Understand
Restated: Two fair $6$-sided dice are thrown. What is the probability that the product of the two numbers showing is a multiple of $5$?
Givens: Two fair dice are thrown, each showing a number in $\{1,2,3,4,5,6\}$; The two dice are independent, so each of the $6 \times 6 = 36$ ordered outcomes is equally likely; We want $P(\text{product is a multiple of } 5)$; Answer choices: (A) $\tfrac{1}{36}$, (B) $\tfrac{1}{18}$, (C) $\tfrac{1}{6}$, (D) $\tfrac{11}{36}$, (E) $\tfrac{1}{3}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #2 Make a Systematic List
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 $5$s' separately, count the easier opposite event 'neither die shows a $5$' and subtract from $1$. Tool #2 (Make a Systematic List) supplies the $6 \times 6 = 36$ equally likely ordered outcomes that anchor the probability fractions.
Execute — Answer: D
6.NS.B.4 Step 1 - Rewrite the event in simpler form.
- Because $5$ is prime, a product $a \cdot b$ is a multiple of $5$ exactly when $a$ or $b$ is a multiple of $5$.
- On a standard die, the only face that is a multiple of $5$ is $5$ itself, so the event becomes 'at least one die shows a $5$'.
💡 Replacing a hard event with a simpler equivalent event is the Grade 6 'factor / prime' move applied to probability.
7.SP.C.8 Step 2 - Count the sample space.
- The two dice are independent, so the equally likely ordered outcomes form a $6 \times 6$ grid.
💡 Listing the ordered pairs $(a,b)$ in a $6 \times 6$ grid is the standard Grade 7 set-up for two independent trials.
7.SP.C.8 Step 3 - Switch to the complement.
- Instead of counting outcomes with at least one $5$, count the easier opposite: outcomes where neither die shows a $5$.
- Each die then has $5$ allowed faces $\{1,2,3,4,6\}$, and the two dice are independent.
💡 'At least one' is almost always easier as $1 - P(\text{none})$ — that is exactly Tool #16's complement trick.
7.SP.C.7 Step 4 Subtract from $1$ to get the asked probability.
💡 Every outcome either has a $5$ or has no $5$, so the two probabilities add to $1$.
6.NS.B.4 Rewrite the event in simpler form. Because $5$ is prime, a product $a \cdot b$ i 7.SP.C.8 Count the sample space. The two dice are independent, so the equally likely orde 7.SP.C.8 Switch to the complement. Instead of counting outcomes with at least one $5$, co 7.SP.C.7 Subtract from $1$ to get the asked probability. Review
Reasonableness: Cross-check by direct count in the $6 \times 6$ 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 $\dfrac{11}{36}$ — matches (D). The answer also fits a quick sanity check: the chance of getting a $5$ on either die alone is $\tfrac{1}{6} \approx 0.167$, so 'at least one $5$' should be a little less than $\tfrac{2}{6} \approx 0.333$ (because of the shared $(5,5)$), and $\tfrac{11}{36} \approx 0.306$ lands right there.
Alternative: Tool #2 (Make a Systematic List) without complement. Split the favorable event into two disjoint cases by the number of $5$s rolled. Exactly two $5$s: only $(5,5)$, so $1$ outcome. Exactly one $5$: the $5$ is on die A or die B; the other die has $5$ allowed faces $\{1,2,3,4,6\}$; that is $2 \times 5 = 10$ outcomes. Total favorable $= 1 + 10 = 11$, probability $\dfrac{11}{36}$. Same answer (D).
CCSS standards used (min grade 7)
7.SP.C.7Develop a probability model and use it to find probabilities of events; compare probabilities from a model to observed frequencies (Treating each of the $36$ ordered $(a,b)$ outcomes as equally likely, so probability $=$ favorable count $\div 36$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Using independence of the two dice to compute $P(\text{no } 5) = \tfrac{5}{6} \cdot \tfrac{5}{6} = \tfrac{25}{36}$ on the $6 \times 6$ grid of ordered outcomes.)6.NS.B.4Find the greatest common factor and least common multiple; apply properties of primes and factors (Using the fact that $5$ is prime to conclude that a product is a multiple of $5$ exactly when one of the factors is a multiple of $5$ — i.e. when one of the dice shows $5$.)
⭐ A product is a multiple of $5$ only when at least one die rolls a $5$. Count the easy opposite — neither die is a $5$, $\tfrac{5}{6} \cdot \tfrac{5}{6} = \tfrac{25}{36}$ — and the answer is $1 - \tfrac{25}{36} = \tfrac{11}{36}$.
⭐ A product is a multiple of $5$ only when at least one die rolls a $5$. Count the easy opposite — neither die is a $5$, $\tfrac{5}{6} \cdot \tfrac{5}{6} = \tfrac{25}{36}$ — and the answer is $1 - \tfrac{25}{36} = \tfrac{11}{36}$.
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