AMC 10 · 2007 · #13

Grade 7 probability
Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region
probability-basicgeometric-probabilityperiodic-function work-backwardscomplementary-counting ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
A traffic light cycles through three fixed durations forever. Someone starts watching at a random moment for a short fixed stretch. Find the probability the light changes while they watch.

Pick an answer.

(A)
$\frac{1}{63}$
(B)
$\frac{1}{21}$
(C)
$\frac{1}{10}$
(D)
$\frac{1}{7}$
(E)
$\frac{1}{3}$
How to solve
Strategy Work Backwards

The colors themselves are a distraction: what matters is three isolated instants at which the light changes. Use the repeat pattern (Tool #5) to shrink all of time down to one 63-second cycle, and draw that cycle as a number line (Tool #1) so the three change instants become three marks. Then work backwards from each mark (Tool #11) to the set of start times that would catch it. The one delicate point is whether those sets of start times overlap, which is a boundary question about the shortest phase (Tool #14). Finally count the opposite event, the starts that see nothing happen (Tool #16), as an independent check.

1STEP 1

Shrink all of time to one cycle

Repetition shrinks all of time to one cycle.

30+3+30=63 seconds per cycle, t∈[0,63)
2STEP 2

Mark the three change instants

There are exactly three change instants in it.

changes at t=30, t=33, t=63≡ 0
3STEP 3

Work back from a change to its start times

Each one turns into a stretch of winning start times.

t ＜ c ＜ t+3 ⇔ c-3 ＜ t ＜ c
4STEP 4

Prove the stretches do not overlap

The three stretches do not overlap, so their lengths add.

|(27,30)∪(30,33)∪(60,63)|=3+3+3=9, 9/63=1/7
5STEP 5

Cross-check by counting quiet starts

Counting the quiet starts confirms 1/7, choice (B).

1-(27+0+27)/63=1-54/63=9/63=1/7
Answer
1/7
The answer must lie between 0 and 1 and be small, since only 9 of the 63 possible start seconds work; 1/7≈ 0.14 fits. A third count confirms the size. One particular change is caught with probability 3/63=1/21, and there are three changes per cycle, so the expected number of changes Leah sees is 3×1/21=1/7. Because no two change instants are less than 3 seconds apart, she can never see two changes in one interval apart from the lone borderline start t=30, which has probability 0; so that expected count equals the probability of seeing at least one, namely 1/7. The wrong choices are the natural slips: 1/21 is catching one specific change and forgetting the other two, and 1/63 treats a single second of the cycle as the only chance.
💡Key takeaway

You catch a color change only if you start watching in the three seconds just before it, so three changes give three 3-second windows, and those windows just barely fail to overlap, leaving 9 good seconds out of 63.

  • Shrink all of time to one cycle
  • Mark the three change instants
  • Work back from a change to its start times
  • Prove the stretches do not overlap
  • Cross-check by counting quiet starts

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