Wir berechnen die Wahrscheinlichkeiten für ${\displaystyle {}n\leq 5}$ und vergleichen sie mit den oben gefundenen. Es ist

${\displaystyle {}\sum _{k=0}^{1}(-1)^{k}{\frac {1}{k!}}=1-1=0\,,}$

${\displaystyle {}\sum _{k=0}^{2}(-1)^{k}{\frac {1}{k!}}=1-1+{\frac {1}{2}}={\frac {1}{2}}\,,}$

${\displaystyle {}\sum _{k=0}^{3}(-1)^{k}{\frac {1}{k!}}=1-1+{\frac {1}{2}}-{\frac {1}{6}}={\frac {1}{3}}\,,}$

${\displaystyle {}\sum _{k=0}^{4}(-1)^{k}{\frac {1}{k!}}=1-1+{\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{24}}={\frac {9}{24}}\,,}$

${\displaystyle {}\sum _{k=0}^{5}(-1)^{k}{\frac {1}{k!}}=1-1+{\frac {1}{2}}-{\frac {1}{6}}+{\frac {1}{24}}-{\frac {1}{120}}={\frac {44}{120}}\,.}$