Das Ereignis ${\displaystyle {}A}$ hat die Wahrscheinlichkeit ${\displaystyle {}{\frac {1}{1024}}}$. Für das Ereignis ${\displaystyle {}B}$ haben wir die (groben) Abschätzungen

${\displaystyle {}{\begin{aligned}P(B)&=\sum _{k=0}^{10}{\frac {\binom {100}{k}}{2^{100}}}\\&\leq 11\cdot {\frac {\binom {100}{10}}{2^{100}}}\\&=11\cdot {\frac {\,\,{\frac {100\cdot 99\cdots 91}{10\cdot 9\cdots 1}}\,\,}{2^{100}}}\\&\leq {\frac {11\cdot 100^{10}}{2^{100}}}\\&\leq {\frac {100^{11}}{2^{100}}}\\&\leq {\frac {10^{22}}{{\left(2^{10}\right)}^{10}}}\\&\leq {\frac {10^{22}}{{\left(10^{3}\right)}^{10}}}\\&={\frac {10^{22}}{10^{30}}}\\&={\frac {1}{10^{8}}}.\end{aligned}}}$

${\displaystyle {}A}$