This problem involves a uniform distribution, where the waiting time $X$ is uniformly distributed between 0 and 99 minutes. That means:

$X \sim U(0,99)$

For a uniform distribution over an interval $[a, b]$, the probability density function (PDF) is:

$f(x) = \frac{1}{b-a}, \quad a \leq x \leq b$

In this case:

$f(x) = \frac{1}{99 - 0} = \frac{1}{99}$

To find $P(X < 3.75)$, we use the cumulative distribution function (CDF):

$P(X < 3.75) = \frac{3.75 - 0}{99 - 0} = \frac{3.75}{99}$

Calculation:

$\frac{3.75}{99} \approx 0.038$

Final Answer:

$\mathbf{0.038}$

(Rounded to three decimal places)

---

Would you like a more detailed explanation or have any other questions?

Here are some related questions you might find useful:

1. What is the probability that a passenger waits more than 50 minutes?
2. Find the expected value and variance of the waiting time.
3. How does the probability change if the waiting time is uniformly distributed between 0 and 120 minutes?
4. What is the median waiting time in this scenario?
5. If a passenger arrives every 10 minutes, what is the probability that their wait is more than 20 minutes?

Tip: In a uniform distribution, the probability of any interval is simply the length of the interval divided by the total range.