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Probability Theory: Poisson Distribution

  • #1

Homework Statement



A random variable has a Poisson distribution with parameter λ = 2. Compute the following probabilities, giving an exact answer and a decimal approximation.

P(X ≥ 4)


Homework Equations



P(X = k) = λke/k!

The Attempt at a Solution



P(X ≥ 4) = Ʃk = 4 λke/k!

= λ4e/4! + λ5e/5! + λ6e/6! + [itex]\cdots[/itex]

= e4/4! + λ5/5! + λ6/6! + [itex]\cdots[/itex]]

= e Ʃk = 4 λk/k!

Let n = k - 4

=e Ʃn = 0 λn+4/(n+4)!

plug in λ = 2

= e-2 Ʃn = 0 2n+4/(n+4)!

= e-2 Ʃn = 0 2n24/(n+4)!

= 16e-2 Ʃn = 0 2n/(n+4)!

This is as far as I have gotten, but I'm not sure I'm on the correct track. I used wolfram alpha to reduce the summation term, Ʃn = 0 2n/(n+4)!, to [1/48(3e2-19)], but I'm at a loss as to how to get there myself.

Anyone have any suggestions? It would be greatly appreciated!
 

Answers and Replies

  • #2
Ray Vickson
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Homework Helper
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Homework Statement



A random variable has a Poisson distribution with parameter λ = 2. Compute the following probabilities, giving an exact answer and a decimal approximation.

P(X ≥ 4)


Homework Equations



P(X = k) = λke/k!

The Attempt at a Solution



P(X ≥ 4) = Ʃk = 4 λke/k!

= λ4e/4! + λ5e/5! + λ6e/6! + [itex]\cdots[/itex]

= e4/4! + λ5/5! + λ6/6! + [itex]\cdots[/itex]]

= e Ʃk = 4 λk/k!

Let n = k - 4

=e Ʃn = 0 λn+4/(n+4)!

plug in λ = 2

= e-2 Ʃn = 0 2n+4/(n+4)!

= e-2 Ʃn = 0 2n24/(n+4)!

= 16e-2 Ʃn = 0 2n/(n+4)!

This is as far as I have gotten, but I'm not sure I'm on the correct track. I used wolfram alpha to reduce the summation term, Ʃn = 0 2n/(n+4)!, to [1/48(3e2-19)], but I'm at a loss as to how to get there myself.

Anyone have any suggestions? It would be greatly appreciated!
##P(X \geq 4) = 1-P(X \leq 3).##
 
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  • #3
BruceW
Homework Helper
3,611
119
Ray's 'clue' is what you must use for this problem. generally, this would be a difficult problem. But luckily, 4 is not a large number.
 
  • #4
Doh! That little fact completely slipped my mind. Thanks guys!
 

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