5. A bus arrives at a station every day at a random time between 1:00 P.M. and 1:30 P.M.
A person arrives at this station at 1:00 and waits for the bus. If at 1:15 the bus has
not yet arrived, what is the probability that the person will have to wait at least an
additional 5 minutes?

We are given that the bus arrives at a random time between 1:00 P.M. and 1:30 P.M. (uniformly distributed), and a person arrives at 1:00 P.M. If the bus hasn’t arrived by 1:15 P.M., we are asked to find the probability that the person will have to wait at least an additional 5 minutes.

### Step-by-step solution:

1. **Time interval:**  
   The bus can arrive at any time between 1:00 P.M. and 1:30 P.M., meaning the bus arrives uniformly on the interval \([0, 30]\) minutes (where 0 corresponds to 1:00 P.M. and 30 corresponds to 1:30 P.M.).

2. **Conditional time range:**  
   At 1:15 P.M., we know that the bus has not arrived yet. This means the bus will arrive sometime between 1:15 P.M. and 1:30 P.M., corresponding to the interval \([15, 30]\) minutes.

3. **Required event:**  
   We are asked to find the probability that the person will have to wait at least 5 more minutes from 1:15 P.M., i.e., the bus arrives no earlier than 1:20 P.M. This corresponds to the bus arriving in the interval \([20, 30]\) minutes.

4. **Uniform distribution:**
   Since the bus arrives uniformly in the interval \([15, 30]\), the probability that it arrives after 1:20 P.M. (i.e., in the interval \([20, 30]\)) is given by the ratio of the length of the interval \([20, 30]\) to the length of the interval \([15, 30]\).

   - The length of the interval \([20, 30]\) is \(30 - 20 = 10\) minutes.
   - The length of the interval \([15, 30]\) is \(30 - 15 = 15\) minutes.

5. **Final calculation:**
   The probability is:

   \[
   \text{Probability} = \frac{\text{Length of the interval [20, 30]}}{\text{Length of the interval [15, 30]}} = \frac{10}{15} = \frac{2}{3}.
   \]

Thus, the probability that the person will have to wait at least an additional 5 minutes after 1:15 P.M. is \(\boxed{\frac{2}{3}}\).

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Would you like further details on any step or have additional questions?

Here are five related questions to explore:

1. What is the probability that the bus arrives between 1:10 P.M. and 1:20 P.M.?
2. How would the probability change if the bus were to arrive between 1:00 P.M. and 1:45 P.M. instead?
3. If the person arrives at 1:05 P.M., what is the probability that they will wait less than 10 minutes?
4. What is the expected waiting time for the person if they arrive at 1:00 P.M.?
5. If the person arrives randomly between 1:00 P.M. and 1:30 P.M., what is the probability that they will have to wait more than 10 minutes?

**Tip:** When dealing with uniform probability distributions, the probability of an event is often simply the ratio of the favorable interval length to the total possible interval length.

A bus arrives at a station every day at a random time between 1:00 P.M. and 1:30 P.M. A person arrives at this station at 1:00 and waits for the bus. If at 1:15 the bus has not yet arrived, what is the probability that the person will have to wait at least an additional 5 minutes?

This problem explores the probability of waiting for a bus that arrives randomly between 1:00 P.M. and 1:30 P.M., given that it has not arrived by 1:15 P.M. Learn how to calculate conditional probabilities using uniform distributions. Suitable for students in Grades 9-11.

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