AMC 8 · 2015 · #7

Grade 7 probability

probability-basicparitysystematic-enumeration complementary-countingsystematic-enumeration ↑ Prerequisites: fraction-arithmeticparity

📏 Short solution 💡 3 insights

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Problem

Each of two boxes contains three chips numbered $1$ , $2$ , $3$ . A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?

$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}$

Pick an answer.

(A)

$frac{1}{9}$

(B)

$frac{2}{9}$

(C)

$frac{4}{9}$

(D)

$frac{1}{2}$

(E)

$frac{5}{9}$

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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?

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

#2 Make a Systematic List 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 $(a, b)$.

$$3 \times 3 = 9 \text{ outcomes}$$

💡 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.

#16 Change Focus / Count the Complement 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.

$$a \times b \text{ is odd} \iff a \text{ is odd AND } b \text{ is odd}$$

💡 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.

#2 Make a Systematic List 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 $(1,1), (1,3), (3,1), (3,3)$.

$$2 \times 2 = 4 \text{ odd-product outcomes}$$

💡 The systematic list keeps the count honest — exactly $4$ ordered pairs of odd numbers, no more, no less.

#16 Change Focus / Count the Complement 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 $(1,2), (2,1), (2,2), (2,3), (3,2)$.

$$9 - 4 = 5 \text{ even-product outcomes}$$

💡 Complement counting: total $-$ unwanted $=$ wanted. This avoided having to enumerate five separate even-product cases from scratch.

#3 Eliminate Possibilities 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).

$$P(\text{even product}) = \dfrac{5}{9} \;\Rightarrow\; \textbf{(E)}$$

💡 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.

[1] #2 7.SP.C.8 Count the total number of outcomes using the multiplication principle. Box 1 has

[2] #16 3.OA.B.5 Apply the parity rule: a product $a \times b$ is even unless both $a$ and $b$ ar

[3] #2 7.SP.C.8 Count the odd-product outcomes by listing. The odd chips in each box are ${1, 3

[4] #16 7.SP.C.8 Subtract from the total to get the number of even-product outcomes. Total $9$ mi

[5] #3 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.5 Apply 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.7 Develop 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.8 Find 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 7)

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