- 42,828
- 10,505
Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?sol59 said:The probability of no arrivals in (9:10, 9:50) is P(Y=0)=e-4/3=0.264
The discussion revolves around calculating probabilities using the Poisson distribution, specifically in the context of student arrivals in a classroom setting. The original poster presents a scenario where the average arrival rate is 2 students per hour and seeks to determine the probability that the time between two consecutive arrivals falls within a specified interval.
Participants are actively engaging with the problem, questioning assumptions and clarifying definitions. Some guidance has been provided regarding the use of the Exponential distribution and the implications of the Poisson process. The conversation reflects a productive exploration of the problem without reaching a consensus on the solution.
There is an ongoing discussion about the constraints of the problem, particularly regarding the interpretation of the time intervals and the implications of the average arrival rate. Participants are considering how to accurately apply the Poisson distribution to the scenario described.
Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?sol59 said:The probability of no arrivals in (9:10, 9:50) is P(Y=0)=e-4/3=0.264
haruspex said:Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?
Good. So put it all together. What is the probability of no arrivals in (9:00, 9:10) and at least one in (9:10, 9:50)?sol59 said:1-0.264=0.736
haruspex said:Good. So put it all together. What is the probability of no arrivals in (9:00, 9:10) and at least one in (9:10, 9:50)?
You got there!sol59 said:0,736*0.716=0.527
haruspex said:You got there!