AMC 8 · 2015 · #7
Easy mode Grade 7Problem
Imagine two boxes. Each box has three chips inside, numbered 1, 2, and 3.
You reach into the first box and pull out one chip at random. Then you reach into the second box and pull out one chip at random. You multiply the two numbers you got.
What is the probability that the product is an even number?
Pick an answer.
AMC 8 2015 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 boxes each hold three chips numbered $1, 2, 3$. Pull one chip from each box at random and multiply the two numbers. What is the probability that the product is even?
Givens: Box 1 chips: $\{1, 2, 3\}$; Box 2 chips: $\{1, 2, 3\}$; One chip is drawn from each box, independently and uniformly at random; The two numbers are multiplied; Answer choices: (A) $\tfrac{1}{9}$, (B) $\tfrac{2}{9}$, (C) $\tfrac{4}{9}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{5}{9}$
Unknowns: The probability that the product of the two drawn numbers is even
Understand
Restated: Two boxes each hold three chips numbered $1, 2, 3$. Pull one chip from each box at random and multiply the two numbers. What is the probability that the product is even?
Givens: Box 1 chips: $\{1, 2, 3\}$; Box 2 chips: $\{1, 2, 3\}$; One chip is drawn from each box, independently and uniformly at random; The two numbers are multiplied; Answer choices: (A) $\tfrac{1}{9}$, (B) $\tfrac{2}{9}$, (C) $\tfrac{4}{9}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{5}{9}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
There are only $3 \times 3 = 9$ outcomes, small enough to list every pair explicitly — Tool #2 (Systematic List) is the cleanest fit for this size. The wording "product is even" is a classic trigger for Tool #16 (Complement): "even product" splits into many cases, but "odd product" needs both factors odd — only $2 \times 2 = 4$ pairs. We compute $1 - \tfrac{4}{9}$ instead of summing five cases. Tool #3 (Eliminate) is a check: only one of the five fractions can match a count out of $9$.
Execute — Answer: E
7.SP.C.8 Step 1 - Count the total number of outcomes using the multiplication principle.
- Box 1 has $3$ choices and Box 2 has $3$ choices, so there are $3 \times 3 = 9$ equally likely ordered pairs $(a, b)$.
💡 Listing $(1,1), (1,2), (1,3), (2,1), \ldots, (3,3)$ in row order gives all $9$ pairs with no repeats and no gaps — the heart of Tool #2.
3.OA.B.5 Step 2 - Apply the parity rule: a product $a \times b$ is even unless both $a$ and $b$ are odd.
- So instead of counting even-product pairs directly, count the easier complement — pairs where both numbers are odd.
💡 Even times anything is even, so the only way the product stays odd is if neither factor is even. That single observation turns the problem from $5$ cases into $1$ case.
7.SP.C.8 Step 3 - Count the odd-product outcomes by listing.
- The odd chips in each box are $\{1, 3\}$, giving $2$ choices in Box 1 and $2$ choices in Box 2.
- The four odd-product pairs are $(1,1), (1,3), (3,1), (3,3)$.
💡 The systematic list keeps the count honest — exactly $4$ ordered pairs of odd numbers, no more, no less.
7.SP.C.8 Step 4 - Subtract from the total to get the number of even-product outcomes.
- Total $9$ minus odd $4$ leaves $9 - 4 = 5$ pairs whose product is even, namely $(1,2), (2,1), (2,2), (2,3), (3,2)$.
💡 Complement counting: total $-$ unwanted $=$ wanted. This avoided having to enumerate five separate even-product cases from scratch.
7.SP.C.7 Step 5 - Form the probability as favorable over total.
- With $5$ even-product outcomes out of $9$ equally likely ones, the probability is $\tfrac{5}{9}$, which matches choice (E).
💡 Among the five choices, $\tfrac{5}{9}$ is the only fraction with denominator $9$ that comes from a count of $5$ out of $9$ equally likely outcomes — the others are eliminated.
7.SP.C.8 Count the total number of outcomes using the multiplication principle. Box 1 has 3.OA.B.5 Apply the parity rule: a product $a \times b$ is even unless both $a$ and $b$ ar 7.SP.C.8 Count the odd-product outcomes by listing. The odd chips in each box are ${1, 3 7.SP.C.8 Subtract from the total to get the number of even-product outcomes. Total $9$ mi 7.SP.C.7 Form the probability as favorable over total. With $5$ even-product outcomes out Review
Reasonableness: Sanity check the answer's size. The product is forced to be even whenever a $2$ is drawn from either box — that already covers a lot of pairs. The chance of drawing a $2$ from Box 1 is $\tfrac{1}{3}$, from Box 2 is $\tfrac{1}{3}$, so $P(\text{at least one } 2) = 1 - \tfrac{2}{3} \cdot \tfrac{2}{3} = 1 - \tfrac{4}{9} = \tfrac{5}{9}$. This independently confirms $\tfrac{5}{9}$ and rules out the smaller options (A) $\tfrac{1}{9}$, (B) $\tfrac{2}{9}$, (C) $\tfrac{4}{9}$, and the symmetric-looking (D) $\tfrac{1}{2}$.
Alternative: Tool #2 (Systematic List) alone, without the complement shortcut, also works: write all $9$ pairs and mark each product as E or O. Even products: $(1,2), (2,1), (2,2), (2,3), (3,2)$ — count is $5$, so $P = \tfrac{5}{9}$. Same answer, slightly more enumeration. For a $3 \times 3$ grid this is fast; for larger boxes the complement (Tool #16) becomes much more efficient.
CCSS standards used (min grade 7)
3.OA.B.5Apply properties of operations as strategies to multiply and divide (Using the parity rule that an even factor forces an even product, so a product is odd only when both factors are odd.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating the $9$ ordered pairs as equally likely outcomes and computing $P(\text{even product}) = \tfrac{\text{favorable}}{\text{total}} = \tfrac{5}{9}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Enumerating the $3 \times 3 = 9$ outcomes from two independent draws and counting the $4$ odd-product pairs via a systematic list.)
⭐ When a problem says "the product is even," flip it around and count when it is odd instead — both numbers odd is just one tiny case to check.
⭐ When a problem says "the product is even," flip it around and count when it is odd instead — both numbers odd is just one tiny case to check.
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