Calculating Probability using the Poisson Distribution

  • #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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