Final answer:
The probability of two individuals meeting within an hour given one waits 5 minutes and another waits 10 minutes cannot be determined with the information provided. We can, however, solve similar problems based on uniform distribution and exponential distribution, given more context and appropriate data.
Step-by-step explanation:
The question asks about the probability of two persons meeting within an hour if one waits for 5 minutes and the other for 10 minutes before they can meet. This problem involves uniform probability distribution and can be visualized as a geometric shape over the time frame of one hour. As we don't have specific details on the requirements for them to meet (do they arrive at random times, do they just need to be at the location at the same time, etc.), we cannot provide a precise answer to the question.
We can, however, look at similar problems involving waiting times and probabilities provided in the reference information. For example, if the amount of time for a bus to arrive is uniformly distributed between 0 and 15 minutes, the probability of waiting fewer than 12.5 minutes is the ratio of the interval length (12.5 minutes) to the total length of the time interval (15 minutes). Calculating it would give us 12.5/15 which simplifies to 5/6 or approximately 0.8333.
To answer a similar question listed in the reference, such as finding the probability of the amount of time between two successive visits to an urgent care being more than 15 minutes, given that visits are exponentially distributed with an average rate of one visit per every seven minutes, we can use the exponential distribution formula. The probability of the inter-arrival time being more than a certain number of minutes can be given as e^(-lambda * x), where lambda is the rate of arrivals (1/7 per minute) and x is the time threshold (15 minutes).