$\backslash qquad\backslash \; \backslash rule\{11cm\}\{0.5mm\}$

$P={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}\backslash \; P=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}+\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}+\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}+\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}+\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}+\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}$

$P={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}\cdot (1+1+1+1+1+1)\backslash \; P=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot(1+1+1+1+1+1)$

$P={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{4}}\cdot 6\Rightarrow P={\displaystyle \frac{6}{32}}\Rightarrow P={\displaystyle \frac{3}{16}}\Rightarrow P=\mathrm{0,1875}\backslash \; P=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot\backslash dfrac\{1\}\{4\}\backslash cdot6\backslash \; \backslash Rightarrow\backslash \; P=\backslash dfrac\{6\}\{32\}\backslash \; \backslash Rightarrow\backslash \; P=\backslash dfrac\{3\}\{16\}\backslash \; \backslash Rightarrow\backslash \; P=0\{,\}1875$

$P=\mathrm{18,75}\mathrm{\%}\backslash \; P=18\{,\}75\backslash \%$

Die Wahrscheinlichkeit, dass jeder Buchstabe bei dreimaligem Drehen einmal erzielt wird, beträgt,$\mathrm{18,75}\mathrm{\%}\backslash \; 18\{,\}75\backslash \%$.