The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. Find the probability that a randomly selected passenger has a waiting time less than 0.75 minutes. Find the probability that a randomly selected passenger has a waiting time less than 0.75 minutes. (Simplify your answer. Round to three decimal places as needed.)

In a uniform distribution, the probability for a certain range can be calculated by considering the length of that range compared to the total length of the distribution. Here, the total waiting time is 6 minutes (ranging from 0 to 6 minutes). Thus, the probability that a passenger waits less than 0.75 minutes can be calculated as follows: \[ P(X < 0.75) = \frac{\text{Length of interval from 0 to 0.75}}{\text{Total length}} = \frac{0.75 - 0}{6 - 0} = \frac{0.75}{6} = 0.125 \] Rounding to three decimal places, the probability is approximately **0.125**. Now, if you want to reiterate this slightly more scientifically: Given that the waiting times are uniformly distributed, every moment in the range has an equal chance of being selected. That implies a 12.5% probability for any given passenger to experience a wait of less than 0.75 minutes!