AMC 8 · 2007 · #25
Grade 7 probabilitygeometry-2dProblem
On the dart board shown in the figure below, the outer circle has radius 6 and the inner circle has radius 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?
Pick an answer.
AMC 8 2007 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: A dart board has an outer circle of radius $6$ and an inner circle of radius $3$. Three radii cut each circle into three congruent regions, labeled with point values $1$ or $2$. The inner circle has scores $1, 2, 2$ (one $1$ and two $2$s). The outer ring has scores $2, 1, 1$ (one $2$ and two $1$s). The chance a dart lands in a region is proportional to that region's area. Two independent darts are thrown. What is the probability that the sum of the two scores is odd?
Givens: Outer circle radius $R = 6$; Inner circle radius $r = 3$; Three congruent regions in the inner disk with scores $1, 2, 2$; Three congruent regions in the outer ring with scores $2, 1, 1$; Probability of landing in a region is proportional to its area; Two darts are thrown independently; Answer choices: (A) $\tfrac{17}{36}$, (B) $\tfrac{35}{72}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{37}{72}$, (E) $\tfrac{19}{36}$
Unknowns: The probability that the sum of the two dart scores is odd
Understand
Restated: A dart board has an outer circle of radius $6$ and an inner circle of radius $3$. Three radii cut each circle into three congruent regions, labeled with point values $1$ or $2$. The inner circle has scores $1, 2, 2$ (one $1$ and two $2$s). The outer ring has scores $2, 1, 1$ (one $2$ and two $1$s). The chance a dart lands in a region is proportional to that region's area. Two independent darts are thrown. What is the probability that the sum of the two scores is odd?
Givens: Outer circle radius $R = 6$; Inner circle radius $r = 3$; Three congruent regions in the inner disk with scores $1, 2, 2$; Three congruent regions in the outer ring with scores $2, 1, 1$; Probability of landing in a region is proportional to its area; Two darts are thrown independently; Answer choices: (A) $\tfrac{17}{36}$, (B) $\tfrac{35}{72}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{37}{72}$, (E) $\tfrac{19}{36}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #2 Make a Systematic List
The two-dart probability is intimidating, but it breaks cleanly into smaller pieces. Tool #7 (Identify Subproblems) splits the work into three stages: (1) find the area of each region; (2) collapse those areas into a single-dart probability $P(\text{odd})$ and $P(\text{even})$; (3) combine the two darts. Tool #1 (Draw a Diagram) keeps the six regions and their scores straight — the inner disk and outer ring carry different amounts of area, so the $1$s and $2$s are not equally likely. Tool #2 (Systematic List) handles step (3): the two darts have only four parity outcomes (OO, OE, EO, EE), and exactly the mixed ones give an odd sum.
Execute — Answer: B
7.G.B.4 Step 1 - Sketch the board and label the six regions.
- The outer disk has area $\pi R^2 = 36\pi$.
- The inner disk has area $\pi r^2 = 9\pi$.
- The outer ring (annulus) is the difference: $36\pi - 9\pi = 27\pi$.
- Each of the three inner regions has area $\tfrac{9\pi}{3} = 3\pi$.
- Each of the three outer-ring regions has area $\tfrac{27\pi}{3} = 9\pi$.
💡 Drawing the dartboard makes it clear that an outer-ring region (area $9\pi$) is three times as big as an inner region (area $3\pi$), so they are not equally likely targets.
7.SP.C.7 Step 2 - Subproblem 1: find the probability that one dart scores odd.
- Collect every region labeled $1$: one inner region (area $3\pi$) and two outer-ring regions ($2 \times 9\pi = 18\pi$).
- The total area for an odd score is $3\pi + 18\pi = 21\pi$.
- Divide by the total board area $36\pi$.
💡 When outcomes are not equally likely, the Grade 7 probability model says add favorable areas and divide by the total area.
7.SP.C.7 Step 3 - Subproblem 2: find the probability that one dart scores even.
- Collect every region labeled $2$: two inner regions ($2 \times 3\pi = 6\pi$) and one outer-ring region ($9\pi$).
- Total even area is $15\pi$.
- As a check, $P(\text{odd}) + P(\text{even}) = \tfrac{7}{12} + \tfrac{5}{12} = 1$.
💡 Computing $P(\text{even})$ directly (rather than $1 - P(\text{odd})$) gives an instant arithmetic check that the two probabilities sum to $1$.
7.SP.C.8 Step 4 - Subproblem 3: list every two-dart parity outcome and keep the ones that sum to odd.
- With two independent throws, the parities of the two scores form four cases: OO, OE, EO, EE.
- Sum parity: O+O = even, E+E = even, O+E = odd, E+O = odd.
- Only the mixed cases give an odd sum.
💡 A clean Grade 7 sample-space list of compound events isolates the two cases that matter without missing any.
7.SP.C.8 Step 5 Multiply within each case (independence) and add across the two favorable cases.
💡 Independent throws multiply; mutually exclusive cases add. Subproblems 1 and 2 plug straight into the formula from Subproblem 3.
7.G.B.4 Sketch the board and label the six regions. The outer disk has area $\pi R^2 = 3 7.SP.C.7 Subproblem 1: find the probability that one dart scores odd. Collect every regio 7.SP.C.7 Subproblem 2: find the probability that one dart scores even. Collect every regi 7.SP.C.8 Subproblem 3: list every two-dart parity outcome and keep the ones that sum to o 7.SP.C.8 Multiply within each case (independence) and add across the two favorable cases. Review
Reasonableness: The answer $\tfrac{35}{72} \approx 0.486$ is just under $\tfrac{1}{2}$, which fits the setup: a slightly tilted coin should land odd-sum slightly less often than even-sum. Quick check via the complement: $P(\text{even sum}) = P(\text{O})^2 + P(\text{E})^2 = \tfrac{49}{144} + \tfrac{25}{144} = \tfrac{74}{144} = \tfrac{37}{72}$, and $\tfrac{35}{72} + \tfrac{37}{72} = \tfrac{72}{72} = 1$, exactly as required. Choice (C) $\tfrac{1}{2}$ would only hold if odd and even were equally likely on a single dart, which they aren't ($\tfrac{7}{12} \neq \tfrac{5}{12}$).
Alternative: Tool #2 (Make a Systematic List) at the region level: treat each of the six regions as a weighted outcome with weight equal to its area share. Single-dart weights are $\tfrac{3}{36}, \tfrac{3}{36}, \tfrac{3}{36}, \tfrac{9}{36}, \tfrac{9}{36}, \tfrac{9}{36}$ for the three inner and three outer-ring regions. Build a $6 \times 6$ outcome table for two darts, mark each cell odd or even by score sum, and add the area-products of the odd cells. The total simplifies to $\tfrac{35}{72}$, confirming (B).
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the area of the outer disk ($36\pi$), the inner disk ($9\pi$), the outer ring ($27\pi$), and each of the six congruent regions ($3\pi$ inner, $9\pi$ ring).)7.SP.C.7Develop a probability model and use it to find probabilities of events (Building a single-dart probability model where each region's probability equals its area divided by the total board area, giving $P(\text{odd}) = \tfrac{7}{12}$ and $P(\text{even}) = \tfrac{5}{12}$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the four parity outcomes (OO, OE, EO, EE) for two independent darts and combining the two favorable cases by multiplying within and adding across to get $\tfrac{35}{72}$.)
⭐ Split a scary two-dart probability into three small subproblems — find each region's area, get the single-dart $P(\text{odd})$ and $P(\text{even})$, then list the four parity cases. The mixed ones give the odd sum, and the answer drops out as $\tfrac{35}{72}$.
⭐ Split a scary two-dart probability into three small subproblems — find each region's area, get the single-dart $P(\text{odd})$ and $P(\text{even})$, then list the four parity cases. The mixed ones give the odd sum, and the answer drops out as $\tfrac{35}{72}$.
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