Calculating Probability using the Poisson Distribution

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The discussion focuses on calculating the probability of student arrivals using the Poisson distribution, specifically for the scenario where the time between two consecutive arrivals falls within a specified range. Participants clarify that the average arrival rate (λ) is 2 students per hour, and the original equation used was not appropriate for determining the time interval between arrivals. Instead, they suggest using the Exponential distribution to model the waiting time for the next arrival. The final calculation concludes that the probability of no arrivals in the first 10 minutes and at least one arrival in the subsequent 40 minutes is approximately 0.527. The thread emphasizes the importance of correctly interpreting the problem's constraints and applying the right statistical methods.
  • #31
sol59 said:
The probability of no arrivals in (9:10, 9:50) is P(Y=0)=e-4/3=0.264
Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?
 
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  • #32
haruspex said:
Right, so what is the probability of at least 1 arrival in (9:10, 9:50) ?

1-0.264=0.736
 
  • #33
sol59 said:
1-0.264=0.736
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)?
 
  • #34
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)?

0,736*0.716=0.527
 
  • #35
sol59 said:
0,736*0.716=0.527
You got there!
 
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  • #36
haruspex said:
You got there!

Thank you so much for your help! I've managed to solve all other problems but this one was too difficult for me:) Thanks again:)!
 

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