AMC 8 · 1999 · #10

Grade 7 probability

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

probability-basiccomplementary-countingfraction-arithmetic complementary-countingidentify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 2 insights

Problem

A traffic light runs a 60-second cycle: green for 25 s, yellow for 5 s, red for 30 s. If you look at a random moment in the cycle, what is the probability the light is NOT green?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{1}{3}$

(C)

$\frac{5}{12}$

(D)

$\frac{1}{2}$

(E)

$\frac{7}{12}$

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The cycle is a single 60-second timeline broken into three colored chunks, so Tool #1 (Draw a Diagram) — a labeled bar of length 60 — makes the sample space and the favorable region visible at once. "Not green" covers two chunks (yellow + red), and reading their combined length straight off the bar gives the numerator. Tool #16 (Count the Complement) is also natural here because the word "NOT" flags it: instead of adding yellow + red, you can take 1 minus the green fraction. Both routes give the same answer; the diagram is shown first because it is the most concrete for a younger reader.

1STEP 1

Draw the cycle as one bar of 60 seconds, split into green 25, yellow 5, red 30 — the pieces fill the whole cycle.

25 + 5 + 30 = 60 seconds

2STEP 2

Add the not-green pieces — yellow 5 plus red 30 make 35 seconds of the bar.

not-green seconds = 5 + 30 = 35

3STEP 3

The moment is uniform, so probability = favorable ÷ total = $\frac{35}{60}$ .

P(not green) =

\frac{35}{60}

4STEP 4

Reduce by the GCF 5: $\frac{35}{60}$ = $\frac{7}{12}$ , which is choice (E).

\frac{35}{60}

\frac{35 \div 5}{60 \div 5}

\frac{7}{12}

→ (E)

Answer

\frac{7}{12}

Cross-check with the complement: green covers 25 of the 60 seconds, so P(green) =

\frac{25}{60}

\frac{5}{12}

, and P(not green) = 1 -

\frac{5}{12}

\frac{7}{12}

— matches answer (E). Size check: green is less than half the cycle, so "not green" should be more than half, and

\frac{7}{12}

＞

\frac{1}{2}

fits. Choices (A)

\frac{1}{4}

, (B)

\frac{1}{3}

, (C)

\frac{5}{12}

are all ≤

\frac{1}{2}

, so they fail this sanity check; (D)

\frac{1}{2}

would mean green and not-green were equal, which the 25 vs. 35 split rules out.

💡Key takeaway

Draw the cycle as a 60-second bar, add the seconds that are NOT green, and write the fraction — this AMC 8 problem only needs Grade 7 probability and Grade 4 fraction reducing.

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