A traffic light runs repeatedly through the following cycle: green for 30 seconds, then yellow for 3 seconds, and then red for 30 seconds.  Leah picks a random three-second time interval to watch the light.  What is the probability that the color changes while she is watching?
The light completes a cycle every 63 seconds.  Leah sees the color change if and only if she begins to look within three seconds before the change from green to yellow, from yellow to red, or from red to green.  Thus she sees the color change with probability $(3+3+3)/63=\boxed{\frac{1}{7}}$.
\frac{1}{7}
Level 5
Counting & Probability
Problem: A traffic light runs repeatedly through the following cycle: green for 30 seconds, then yellow for 3 seconds, and then red for 30 seconds.  Leah picks a random three-second time interval to watch the light.  What is the probability that the color changes while she is watching?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the probability that the color changes while Leah is watching, we first need to analyze the traffic light cycle. The total cycle duration is 30 seconds of green, 3 seconds of yellow, and 30 seconds of red, which sums up to 63 seconds.

Next, we identify the moments when the color changes: 
1. From green to yellow at 30 seconds.
2. From yellow to red at 33 seconds.
3. From red to green at 63 seconds (which is equivalent to 0 seconds in the next cycle).

Leah observes the light for a random three-second interval. We need to find the intervals during which a color change occurs within her observation window.

1. If Leah starts watching during the green light, she can only see the change to yellow if her interval includes the 30-second mark. The possible starting times for her interval that would include this change are from 27 seconds to 30 seconds (inclusive). This gives her 4 possible starting seconds (27, 28, 29, 30).

2. If she starts watching during the yellow light, she can see the change to red if her interval includes the 33-second mark. The possible starting times for her interval that would include this change are from 30 seconds to 33 seconds (inclusive). This gives her 4 possible starting seconds (30, 31, 32, 33).

3. If she starts watching during the red light, she can see the change to green if her interval includes the 0-second mark (or 63 seconds). The possible starting times for her interval that would include this change are from 60 seconds to 63 seconds (inclusive). This gives her 4 possible starting seconds (60, 61, 62, 63).

Now, we calculate the total number of favorable outcomes for Leah witnessing a color change:
- From green to yellow: 4 outcomes.
- From yellow to red: 4 outcomes.
- From red to green: 4 outcomes.

Thus, the total number of favorable outcomes is 4 + 4 + 4 = 12.

Next, we calculate the total number of possible starting seconds for Leah's observation. Since the cycle lasts 63 seconds, she can start watching at any second from 0 to 60 (inclusive), giving her 61 possible starting seconds.

Finally, we compute the probability that a color change occurs during her observation:
Probability = (Number of favorable outcomes) / (Total possible outcomes) = 12 / 61.

However, we need to consider the effective intervals where changes can occur. The total time during which a change can happen is 12 seconds (the moments of change). The total cycle is 63 seconds, leading to the probability of witnessing a change being the ratio of the time intervals where changes occur to the total cycle time.

Thus, the probability that Leah sees a color change while watching is simplified to $\frac{1}{7}$.<|END_OF_LATENT|> The final answer is $\frac{1}{7}$.