Probability Theory: Poisson Distribution

[mliuzzolino]

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) = λ^ke^-λ/k!

The Attempt at a Solution

P(X ≥ 4) = Ʃ_{k = 4}^∞ λ^ke^-λ/k!

= λ⁴e^-λ/4! + λ⁵e^-λ/5! + λ⁶e^-λ/6! + \cdots

= e^-λ [λ⁴/4! + λ⁵/5! + λ⁶/6! + \cdots]

= e^-λ Ʃ_{k = 4}^∞ λ^k/k!

Let n = k - 4

=e^-λ Ʃ_{n = 0}^∞ λⁿ⁺⁴/(n+4)!

plug in λ = 2

= e^-2 Ʃ_{n = 0}^∞ 2ⁿ⁺⁴/(n+4)!

= e^-2 Ʃ_{n = 0}^∞ 2ⁿ2⁴/(n+4)!

= 16e^-2 Ʃ_{n = 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, Ʃ_{n = 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!

 

[Ray Vickson]

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) = λ^ke^-λ/k!

The Attempt at a Solution

P(X ≥ 4) = Ʃ_{k = 4}^∞ λ^ke^-λ/k!

= λ⁴e^-λ/4! + λ⁵e^-λ/5! + λ⁶e^-λ/6! + \cdots

= e^-λ [λ⁴/4! + λ⁵/5! + λ⁶/6! + \cdots]

= e^-λ Ʃ_{k = 4}^∞ λ^k/k!

Let n = k - 4

=e^-λ Ʃ_{n = 0}^∞ λⁿ⁺⁴/(n+4)!

plug in λ = 2

= e^-2 Ʃ_{n = 0}^∞ 2ⁿ⁺⁴/(n+4)!

= e^-2 Ʃ_{n = 0}^∞ 2ⁿ2⁴/(n+4)!

= 16e^-2 Ʃ_{n = 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, Ʃ_{n = 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!

$P(X \geq 4) = 1-P(X \leq 3).$

 

[BruceW]

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.

 

[mliuzzolino]

Doh! That little fact completely slipped my mind. Thanks guys!