Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #25
Grade 7 probabilitygeometry-2d
Pick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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(odd) and P(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 1s and 2s 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.
Find the region areas
Sketch the board: each inner region has area 3π while each outer-ring region has area 9π — three times as large.
Drawing the dartboard makes it clear that an outer-ring region (area 9π) is three times as big as an inner region (area 3π), so they are not equally likely targets.
7.G.B.4Draw A DiagramFind the odd probability
Add every region scoring 1 — one inner (3π) plus two outer (18π) = 21π out of 36π — so P(odd) = .
When outcomes are not equally likely, the Grade 7 probability model says add favorable areas and divide by the total area.
7.SP.C.7Identify SubproblemsFind the even probability
Add every region scoring 2 — two inner (6π) plus one outer (9π) = 15π — so P(even) = , and P(odd) + P(even) = 1.
Computing P(even) directly (rather than 1 - P(odd)) gives an instant arithmetic check that the two probabilities sum to 1.
7.SP.C.7Identify SubproblemsList the odd-sum cases
List the four parity cases OO, OE, EO, EE; only the mixed ones, where exactly one dart is odd, sum to odd.
A clean Grade 7 sample-space list of compound events isolates the two cases that matter without missing any.
7.SP.C.8Make A Systematic ListMultiply and add the cases
Multiply within each case and add the two: 2 · ()() = , choice (B).
Independent throws multiply; mutually exclusive cases add. Subproblems 1 and 2 plug straight into the formula from Subproblem 3.
The chance that the two darts' scores add to an odd total equals two times the single-dart chance of an odd score multiplied by the single-dart chance of an even score.
▸ Why?
An odd total happens exactly when one dart scores odd and the other scores even, and that pairing can arrive in two separate orders — odd-then-even or even-then-odd — whose chances add together.
▸ Why?
Adding an odd score to an even score always lands on an odd total: the even score splits into two equal whole groups, and the odd score is one past such a split, so the combined pile is still two equal groups with a single leftover — one past a multiple of two, which is exactly what makes a whole number odd.
▸ Why?
The odd-then-even order and the even-then-odd order are separate outcomes that never overlap, so the chance of getting one order or the other is the two orders' chances added up.
▸ Why?
Because the two throws do not affect each other, the chance of a fixed order — say odd first, then even — is the even-chance taken out of the odd-chance slice, and taking a part of a part is carried out by multiplying the two chances.
Split a scary two-dart probability into three small subproblems — find each region's area, get the single-dart P(odd) and P(even), then list the four parity cases. The mixed ones give the odd sum, and the answer drops out as .
- Find the region areas
- Find the odd probability
- Find the even probability
- List the odd-sum cases
- Multiply and add the cases
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