AMC 8 · 2004 · #21
Grade 7 probabilityProblem
Spinners and are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Spinner $A$ has four equal regions labeled $1, 2, 3, 4$ and spinner $B$ has three equal regions labeled $1, 2, 3$. Both spinners are spun once. What is the probability that the product of the two numbers shown is even?
Givens: Spinner $A$: $\{1, 2, 3, 4\}$, four equal regions; Spinner $B$: $\{1, 2, 3\}$, three equal regions; Each region on a given spinner is equally likely; The two spinners are independent; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Unknowns: The probability that the product $A \times B$ is an even number
Understand
Restated: Spinner $A$ has four equal regions labeled $1, 2, 3, 4$ and spinner $B$ has three equal regions labeled $1, 2, 3$. Both spinners are spun once. What is the probability that the product of the two numbers shown is even?
Givens: Spinner $A$: $\{1, 2, 3, 4\}$, four equal regions; Spinner $B$: $\{1, 2, 3\}$, three equal regions; Each region on a given spinner is equally likely; The two spinners are independent; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #7 Identify Subproblems
The event "product is even" breaks into three favorable cases (even$\times$even, even$\times$odd, odd$\times$even), but its complement "product is odd" is a single case: both spinners land on an odd number. Tool #16 (Change Focus / Count the Complement) swaps the messy event for the clean one, computes the probability of the complement, and subtracts from $1$. Tool #7 (Identify Subproblems) splits that complement into two independent subproblems — $P(A \text{ odd})$ and $P(B \text{ odd})$ — which multiply because the spinners are independent.
Execute — Answer: D
7.SP.C.8 Step 1 - Flip to the complement.
- The product $A \times B$ is even unless both $A$ and $B$ are odd, so $P(\text{even}) = 1 - P(\text{both odd})$.
- Solving for the complement is one case instead of three.
💡 Even $\times$ anything is even, so the only way to get an odd product is two odd factors. The complement is much smaller, so count it instead.
7.SP.C.7 Step 2 - Solve the spinner-$A$ subproblem.
- Spinner $A$ has $4$ equal regions; the odd ones are $1$ and $3$, so $2$ out of $4$ regions are odd.
💡 Equally likely regions means favorable $\div$ total. Two odd labels out of four gives one-half.
7.SP.C.7 Step 3 - Solve the spinner-$B$ subproblem.
- Spinner $B$ has $3$ equal regions; the odd ones are $1$ and $3$, so $2$ out of $3$ regions are odd.
💡 Same rule: two of the three labels are odd, so $P = 2/3$.
7.SP.C.8 Step 4 - Combine the two independent subproblems and subtract from $1$.
- Because the spinners are independent, multiply the two odd-probabilities to get $P(\text{both odd})$, then take the complement.
💡 Independent events multiply, so $1/2 \times 2/3 = 1/3$ for the odd-product side. The even-product probability is everything else, which is $2/3$.
7.SP.C.8 Flip to the complement. The product $A \times B$ is even unless both $A$ and $B$ 7.SP.C.7 Solve the spinner-$A$ subproblem. Spinner $A$ has $4$ equal regions; the odd one 7.SP.C.7 Solve the spinner-$B$ subproblem. Spinner $B$ has $3$ equal regions; the odd one 7.SP.C.8 Combine the two independent subproblems and subtract from $1$. Because the spinn Review
Reasonableness: Sanity check with the full $4 \times 3 = 12$ outcome grid. The products row by row are: $A=1 \to 1,2,3$; $A=2 \to 2,4,6$; $A=3 \to 3,6,9$; $A=4 \to 4,8,12$. Even products per row: $1 + 3 + 1 + 3 = 8$. Probability $= 8/12 = 2/3$, matching answer (D). The $2/3$ also feels right: spinner $A$ alone lands even half the time, which already forces the product to be even, so the answer must be at least $1/2$ — and $2/3$ comfortably exceeds it.
Alternative: Tool #2 (Make a Systematic List): build a $4 \times 3$ table of products. The $12$ products are $1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, 12$. Mark each as odd or even — the odd products are $1, 3, 3, 9$ (just $4$ of them), so the even products number $12 - 4 = 8$. Probability $= 8/12 = 2/3$, same answer (D), reached by brute force instead of the complement shortcut.
CCSS standards used (min grade 7)
4.OA.B.4Find all factor pairs, recognize multiples, and determine whether a whole number is prime or composite (Classifying each spinner label as odd or even and using the rule $\text{odd} \times \text{odd} = \text{odd}$ to identify when the product is even.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating each equal region as equally likely to compute $P(A \text{ odd}) = 1/2$ and $P(B \text{ odd}) = 2/3$ from favorable $\div$ total.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Multiplying the two independent odd-probabilities to get $P(\text{both odd}) = 1/3$ and subtracting from $1$ to land on $P(\text{product even}) = 2/3$.)
⭐ Even products have many sub-cases, but odd products only happen one way — both spinners must land on odd. Count that single easy case ($1/3$), then take the complement: $1 - 1/3 = 2/3$.
⭐ Even products have many sub-cases, but odd products only happen one way — both spinners must land on odd. Count that single easy case ($1/3$), then take the complement: $1 - 1/3 = 2/3$.
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