AMC 8 · 1999 · #10
Grade 7 probabilityProblem
A complete cycle of a traffic light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly chosen time, what is the probability that the light will NOT be green?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Total cycle length $= 60$ seconds; Green lasts $25$ s, yellow lasts $5$ s, red lasts $30$ s; The chosen moment is uniformly random within the cycle; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{5}{12}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{7}{12}$
Unknowns: The probability that the light is not green at the random moment
Understand
Restated: 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?
Givens: Total cycle length $= 60$ seconds; Green lasts $25$ s, yellow lasts $5$ s, red lasts $30$ s; The chosen moment is uniformly random within the cycle; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{5}{12}$, (D) $\tfrac{1}{2}$, (E) $\tfrac{7}{12}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #16 Change Focus / Count the Complement
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.
Execute — Answer: E
4.MD.A.2 Step 1 - Draw the cycle as a bar of length $60$ seconds, split into three colored pieces: a green piece of length $25$, a yellow piece of length $5$, and a red piece of length $30$.
- Check that the three lengths add to the whole cycle.
💡 Treating one cycle as a $60$-second bar turns the time question into a length question, which is the Grade 4 way of measuring "how much" of a whole.
4.OA.A.3 Step 2 - Add the lengths of the pieces that are NOT green.
- The yellow piece is $5$ seconds long and the red piece is $30$ seconds long, so together they cover $35$ seconds of the bar.
💡 Once the bar is drawn, the "not green" region is just two adjacent pieces — combine their lengths by adding.
7.SP.C.7 Step 3 - Because the random moment is uniform on the cycle, the probability equals favorable length $\div$ total length.
- Plug in $35$ and $60$.
💡 With equally likely moments, probability is just the fraction of the timeline that is shaded — exactly the Grade 7 uniform probability model.
4.NF.A.1 Step 4 - Reduce the fraction.
- The greatest common factor of $35$ and $60$ is $5$, so divide top and bottom by $5$.
💡 Dividing numerator and denominator by the same nonzero number does not change the fraction's value — Grade 4 equivalent fractions.
4.MD.A.2 Draw the cycle as a bar of length $60$ seconds, split into three colored pieces: 4.OA.A.3 Add the lengths of the pieces that are NOT green. The yellow piece is $5$ second 7.SP.C.7 Because the random moment is uniform on the cycle, the probability equals favora 4.NF.A.1 Reduce the fraction. The greatest common factor of $35$ and $60$ is $5$, so divi Review
Reasonableness: Cross-check with the complement: green covers $25$ of the $60$ seconds, so $P(\text{green}) = \tfrac{25}{60} = \tfrac{5}{12}$, and $P(\text{not green}) = 1 - \tfrac{5}{12} = \tfrac{7}{12}$ — matches answer (E). Size check: green is less than half the cycle, so "not green" should be more than half, and $\tfrac{7}{12} > \tfrac{1}{2}$ fits. Choices (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{5}{12}$ are all $\le \tfrac{1}{2}$, so they fail this sanity check; (D) $\tfrac{1}{2}$ would mean green and not-green were equal, which the $25$ vs. $35$ split rules out.
Alternative: Tool #16 (Count the Complement) directly: instead of adding yellow $+$ red, subtract green from the whole. $P(\text{green}) = \tfrac{25}{60} = \tfrac{5}{12}$, so $P(\text{not green}) = 1 - \tfrac{5}{12} = \tfrac{12}{12} - \tfrac{5}{12} = \tfrac{7}{12}$. Same answer, with only one fraction to reduce instead of one fraction to build from two pieces.
CCSS standards used (min grade 7)
4.MD.A.2Use the four operations to solve word problems involving distances, intervals of time, and other measurements (Modeling the $60$-second cycle as a measured bar split into $25$, $5$, and $30$ second intervals so the question becomes a length-on-a-bar question.)4.OA.A.3Solve multistep word problems posed with whole numbers (Adding the yellow and red durations to find the total $35$ seconds during which the light is not green.)4.NF.A.1Explain why a fraction $a/b$ is equivalent to $(n \times a)/(n \times b)$ (Reducing $\tfrac{35}{60}$ to $\tfrac{7}{12}$ by dividing numerator and denominator by the common factor $5$.)7.SP.C.7Develop a uniform probability model and use it to find probabilities of events (Treating every moment in the cycle as equally likely so the probability of "not green" equals favorable seconds ($35$) over total seconds ($60$).)
⭐ 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.
⭐ 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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