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Find $P(1 < X \le 4)$, where $X$ is a Poisson variable with parameter $\lambda = 0.1$.
Hint: For a Poisson distribution, $P(X = r) = \mathrm{e}^{-\lambda} \dfrac{\lambda^r}{r!}$.
$P(X = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c0}}\hspace{35px}}~}) = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c2}}\hspace{100px}}~}}{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c3}}\hspace{100px}}~}} = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c4R}}\hspace{100px}}~}$ $P(X = [2]) = [0.90483742] \times \dfrac{[0.01]}{[2]} = [0.00452419]$
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c2}}\hspace{100px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c3}}\hspace{100px}}~}} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c4R}}\hspace{100px}}~}$ $P(X = [3]) = [0.90483742] \times \dfrac{[0.001]}{[6]} = [0.00015081]$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c2}}\hspace{100px}}~}}{\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c3}}\hspace{100px}}~}} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c4R}}\hspace{100px}}~}$ $P(X = [4]) = [0.90483742] \times \dfrac{[0.0001]}{[24]} = [0.00000377]$
Therefore, Hint: Add together the individual probabilities.
$P(1 < X \le 4) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{100px}}~}$ $P(1 < X \le 4) = [0.00467876]$