The problem you uploaded asks to find the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes, given that the waiting times between subway departures are uniformly distributed between 0 and 6 minutes.

Approach:

This is a uniform distribution problem where the waiting time is uniformly distributed between 0 and 6 minutes. The probability density function (PDF) for a uniform distribution is:

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

where $a = 0$ and $b = 6$, as the waiting time is uniformly distributed between 0 and 6 minutes.

The probability that a randomly selected passenger waits more than 1.25 minutes is calculated as:

$P(X > 1.25) = \frac{\text{Length of interval where } X > 1.25}{\text{Total length of interval}} = \frac{6 - 1.25}{6}$

$P(X > 1.25) = \frac{4.75}{6} = 0.7917$

Thus, the probability that a passenger waits more than 1.25 minutes is approximately 0.7917 or 79.17%.

Would you like further explanation or details on this concept?

Here are some related questions to explore this topic further:

1. What is the general formula for finding probabilities in uniform distributions?
2. How would the probability change if the upper limit of the waiting time was extended to 10 minutes instead of 6?
3. How is the cumulative distribution function (CDF) related to uniform distributions?
4. What is the expected (average) waiting time for a passenger given this uniform distribution?
5. How does the uniform distribution differ from other distributions like the normal distribution?

Tip: In a uniform distribution, the probability of any specific event is directly proportional to the length of the interval, making it one of the simplest probability distributions to handle.