- 42,799
- 10,495
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 focuses on calculating the probability of student arrivals using the Poisson distribution and the Exponential distribution. The average arrival rate is established at λ = 2 students per hour. The key challenge is determining the probability that the time between two consecutive arrivals falls within the interval of 10 to 50 minutes. The final probability calculated for the scenario is approximately 0.527, representing the likelihood of no arrivals in the first 10 minutes and at least one arrival in the subsequent 40 minutes.
PREREQUISITESStudents, educators, and data analysts interested in probability theory, particularly those working with arrival processes and statistical modeling in various fields such as operations research and queueing theory.
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!