Hello I'm Presented with the following Poisson distribution question

[tex]P(X = x) = \frac{e^{-\lambda} \cdot \lambda^{x}}{x!}[/tex]

where [tex]x \in (1,2,3,\ldots)[/tex] and [tex]\lambda > 0[/tex]

Then I'm suppose to show that the above can be re-written if

[tex]P(X \leq 1) = 1 - e^{- \lambda}[/tex]

Any idears on how I do that?

I'm told [tex]\sum_{x=0} ^ {\infty} p(x) = \frac{e^{- \lambda} \cdot \lambda^{x}}{x!} = e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!}[/tex]

if [tex] \lambda ^{x} = 1[/tex]

then [tex] p(x) = e^{- \lambda} [/tex]

This must give [tex]P(X \leq 1) = 1- e^{- \lambda}[/tex]

Can anybody please tell me if I'm on the right track here?

Sincerley Bob

[tex]P(X = x) = \frac{e^{-\lambda} \cdot \lambda^{x}}{x!}[/tex]

where [tex]x \in (1,2,3,\ldots)[/tex] and [tex]\lambda > 0[/tex]

Then I'm suppose to show that the above can be re-written if

[tex]P(X \leq 1) = 1 - e^{- \lambda}[/tex]

Any idears on how I do that?

I'm told [tex]\sum_{x=0} ^ {\infty} p(x) = \frac{e^{- \lambda} \cdot \lambda^{x}}{x!} = e^{- \lambda} \sum _{x=0} ^{\infty} \frac{\lambda ^{x}}{x!}[/tex]

if [tex] \lambda ^{x} = 1[/tex]

then [tex] p(x) = e^{- \lambda} [/tex]

This must give [tex]P(X \leq 1) = 1- e^{- \lambda}[/tex]

Can anybody please tell me if I'm on the right track here?

Sincerley Bob

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