Met terugleggen

kansboom bij 49

${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$

$3\cdot {\left(\frac{1}{2}\right)}^3=\frac{3}{8}$

$\frac{1}{2}$; ja want de kans dat er meer mannetjes zijn als vrouwtjes is even groot.

$\frac{7}{16}\cdot \frac{6}{16}=\frac{42}{256}$

$\frac{9}{16}\cdot \frac{9}{16}=\frac{81}{256}$

$\frac{5}{25}\cdot \frac{20}{25}\cdot \frac{20}{25}+\frac{20}{25}\cdot \frac{5}{25}\cdot \frac{20}{25}+\frac{20}{25}\cdot \frac{20}{25}\cdot \frac{5}{25}=\frac{48}{125}$

${\left(\frac{4}{5}\right)}^4=\frac{256}{625}$

$\frac{5}{25}\cdot \frac{4}{24}=\frac{1}{30}$

$\frac{15}{25}\cdot \frac{14}{24}\cdot \frac{13}{23}\approx \mathrm{0,198}$

$3\cdot \frac{10}{25}\cdot \frac{15}{24}\cdot \frac{14}{23}\approx \mathrm{0,457}$

$\frac{10}{25}\cdot 5=2$

${\left(\mathrm{0,95}\right)}^5\approx \mathrm{0,774}$

${\left(\mathrm{0,9}\right)}^5\approx \mathrm{0,590}$

${\left(\mathrm{0,85}\right)}^5\approx \mathrm{0,444}$

$\frac{2}{7}$

$\frac{2}{7}\cdot \frac{1}{6}=\frac{1}{21}$

$\frac{5}{7}\cdot \frac{4}{6}=\frac{10}{21}$

$7$%

In de eerste auto $1$ en de tweede $3$, in beide auto's $2$ of in de eerste auto $3$ en in de tweede $1$.
De kans is: $\mathrm{0,58}\cdot \mathrm{0,13}+\mathrm{0,22}\cdot \mathrm{0,22}+\mathrm{0,13}\cdot \mathrm{0,58}=\mathrm{0,1992}$.

$\frac{5}{16}\cdot \frac{11}{15}+\frac{11}{16}\cdot \frac{5}{15}=\frac{11}{24}$

kansboom bij 56

Zonder terugleggen

De som van de drie kansen is $1$.

$1-\frac{11}{24}=\frac{13}{24}$

$3\cdot \mathrm{0,4}\cdot {\left(\mathrm{0,6}\right)}^2=\mathrm{0,432}$

$1-{\left(\mathrm{0,6}\right)}^3=\mathrm{0,784}$

$3\cdot \frac{6}{10}\cdot \frac{5}{9}\cdot \frac{4}{8}=\frac{1}{2}$

$1-\frac{6}{10}\cdot \frac{5}{9}\cdot \frac{4}{8}=\frac{5}{6}$