AMC 8 · 2006 · #17
Grade 7 probabilityProblem
Jeff rotates spinners P, Q and R and adds the resulting numbers. What is the probability that his sum is an odd number?
Pick an answer.
AMC 8 2006 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: Jeff spins three spinners $P$, $Q$, $R$ and adds the three numbers. Spinner $P$ has $\{1, 2, 3\}$, spinner $Q$ has $\{2, 4, 6, 8\}$, and spinner $R$ has $\{1, 3, 5, 7, 9, 11\}$. Each region on a given spinner is equally likely. What is the probability that the sum $P + Q + R$ is odd?
Givens: Spinner $P$: $\{1, 2, 3\}$, three equal regions; Spinner $Q$: $\{2, 4, 6, 8\}$, four equal regions; Spinner $R$: $\{1, 3, 5, 7, 9, 11\}$, six equal regions; The three 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 $P + Q + R$ is an odd number
Understand
Restated: Jeff spins three spinners $P$, $Q$, $R$ and adds the three numbers. Spinner $P$ has $\{1, 2, 3\}$, spinner $Q$ has $\{2, 4, 6, 8\}$, and spinner $R$ has $\{1, 3, 5, 7, 9, 11\}$. Each region on a given spinner is equally likely. What is the probability that the sum $P + Q + R$ is odd?
Givens: Spinner $P$: $\{1, 2, 3\}$, three equal regions; Spinner $Q$: $\{2, 4, 6, 8\}$, four equal regions; Spinner $R$: $\{1, 3, 5, 7, 9, 11\}$, six equal regions; The three 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: #11 Find an Invariant
Secondary: #7 Identify Subproblems
The actual values do not matter — only their parities. Looking at the three spinners, two of them have parities that never change: every number on $Q$ is even, and every number on $R$ is odd. Tool #11 (Find an Invariant) names that. With $Q$ fixed even and $R$ fixed odd, the parity of the sum depends on only one thing: the parity of $P$. Tool #7 (Identify Subproblems) reduces the three-spinner question to a single subproblem — the probability that spinner $P$ lands on an even number.
Execute — Answer: B
4.OA.B.4 Step 1 - Record the parity of each spinner.
- Read off the numbers on each spinner and label them odd or even.
💡 Spotting that every $Q$-value is even and every $R$-value is odd is the invariant — the parity on those two spinners is locked in before Jeff even spins.
4.OA.B.4 Step 2 - Use the invariants to reduce the problem.
- Since $Q$ contributes "even" and $R$ contributes "odd" no matter what, the sum's parity is $(\text{parity of } P) + \text{even} + \text{odd} = (\text{parity of } P) + \text{odd}$.
- So the sum is odd exactly when $P$ is even.
💡 Adding an odd number flips parity, adding an even number keeps it. So adding one even and one odd flips the parity of $P$ exactly once: $P$ even $\to$ sum odd; $P$ odd $\to$ sum even.
7.SP.C.7 Step 3 - Compute the only probability that matters.
- Spinner $P$ has three equal regions, and only one of them ($2$) is even.
💡 Equal regions means each value is equally likely, so favorable $/$ total $= 1/3$.
7.SP.C.8 Step 4 - Confirm with the full independent-events product.
- Multiplying the three required probabilities — $P$ even, $Q$ even (certain), $R$ odd (certain) — gives the same answer.
💡 Independent spinners multiply, and two of the three factors are $1$, so the answer is just the $P$-side probability.
4.OA.B.4 Record the parity of each spinner. Read off the numbers on each spinner and labe 4.OA.B.4 Use the invariants to reduce the problem. Since $Q$ contributes "even" and $R$ c 7.SP.C.7 Compute the only probability that matters. Spinner $P$ has three equal regions, 7.SP.C.8 Confirm with the full independent-events product. Multiplying the three required Review
Reasonableness: Cross-check by trying every parity case for $P$. If $P$ is odd (prob $2/3$), the sum is odd $+$ even $+$ odd $=$ even. If $P$ is even (prob $1/3$), the sum is even $+$ even $+$ odd $=$ odd. The two cases cover all outcomes and their probabilities add to $2/3 + 1/3 = 1$, so the probability of an odd sum is exactly $1/3$, matching answer (B). The number $1/3$ also makes sense: $Q$ and $R$ are both "stuck" on a fixed parity, so the answer must be a probability that already lives on spinner $P$ — and the only $P$-fractions in sight are $1/3$ and $2/3$.
Alternative: Tool #2 (Make a Systematic List) on parities: write each outcome as a triple (parity of $P$, parity of $Q$, parity of $R$). Only two patterns exist — (odd, even, odd) and (even, even, odd) — with probabilities $2/3$ and $1/3$. The first sums to even, the second to odd, so the odd-sum probability is $1/3$. Same answer (B), reached by a brute-force parity table instead of the invariant 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's numbers as odd or even (divisibility by $2$) and using parity rules to reason about the sum.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating each region of a spinner as equally likely to compute $P(\text{P is even}) = 1/3$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Combining the three independent spinners by multiplying probabilities to get $P(\text{sum is odd}) = \tfrac{1}{3} \times 1 \times 1 = \tfrac{1}{3}$.)
⭐ Spinner Q is always even and spinner R is always odd — those two parities are locked in, so the only spinner that decides the sum's parity is P. The answer is just the probability that P lands even: $1/3$.
⭐ Spinner Q is always even and spinner R is always odd — those two parities are locked in, so the only spinner that decides the sum's parity is P. The answer is just the probability that P lands even: $1/3$.
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