Formulaic Maths Problems

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Find $P(X > 1)$, where $X$ is a Poisson variable with parameter $\lambda = 2.29$.

Hint: For a Poisson distribution, $P(X = r) = \mathrm{e}^{-\lambda} \dfrac{\lambda^r}{r!}$.

$P(X = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c0}}\hspace{35px}}~}) = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c2}}\hspace{100px}}~}}{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c3}}\hspace{100px}}~}} = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c4R}}\hspace{100px}}~}$ $P(X = [0]) = [0.10126646] \times \dfrac{[1]}{[1]} = [0.10126646]$

$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c2}}\hspace{100px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c3}}\hspace{100px}}~}} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c4R}}\hspace{100px}}~}$ $P(X = [1]) = [0.10126646] \times \dfrac{[2.29]}{[1]} = [0.2319002]$

Hint: Add together the individual probabilities.

$P(X \le 1) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{100px}}~}$ $P(X \le 1) = [0.33316666]$

Therefore,

$P(X > 1) = 1 - \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{C1}}\hspace{100px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$ $P(X > 1) = 1 - [0.33316666] = [0.66683334]$