Zdarzenia A i B nazwiemy zdarzeniami niezależnymi, jeżeli zachodzi równość:  $\mathbf{P}\left(\mathbf{A}\cap \mathbf{B}\right)=\mathbf{P}\left(\mathbf{A}\right)\cdot \mathbf{P}\left(\mathbf{B}\right)$

Z treści zadania wnioskujemy, że

$A=\left\{3,6,9,12\right\}\Rightarrow P\left(A\right)=\frac{4}{12}=\frac{1}{3}$  

$B=\left\{1,3,5,7,9,11\right\}\Rightarrow P\left(B\right)=\frac{6}{12}=\frac{1}{2}$

$C=\left\{8,9,10,11,12\right\}\Rightarrow P\left(C\right)=\frac{5}{12}$  

Ponadto

$A{}^{\prime }=\left\{1,2,4,5,7,8,10,11\right\}\Rightarrow P\left(A{}^{\prime }\right)=\frac{2}{3}$  

$B{}^{\prime }=\left\{2,4,6,8,10,12\right\}\Rightarrow P\left(B{}^{\prime }\right)=\frac{1}{2}$

$C{}^{\prime }=\left\{1,2,3,4,5,6,7\right\}\Rightarrow P\left(C{}^{\prime }\right)=\frac{7}{12}$  

Mamy więc

$A{}^{\prime }\cap B{}^{\prime }=\left\{2,4,8,10\right\}\Rightarrow \mathbf{P}\left(\mathbf{A}{}^{\prime }\cap \mathbf{B}{}^{\prime }\right)=\frac{4}{12}=\frac{1}{3}=\frac{2}{3}\cdot \frac{1}{2}=\mathbf{P}\left(\mathbf{A}{}^{\prime }\right)\cdot \mathbf{P}\left(\mathbf{B}{}^{\prime }\right)$

$B\cap C=\left\{9,11\right\}\Rightarrow P\left(B\cap C\right)=\frac{2}{12}\ne P\left(B\right)\cdot P\left(C\right)$

$A\cap C{}^{\prime }=\left\{3,6\right\}\Rightarrow P\left(A\cap C{}^{\prime }\right)=\frac{2}{12}\ne P\left(A\right)\cdot P\left(C{}^{\prime }\right)$

$A\cap C=\left\{9,12\right\}\Rightarrow P\left(A\cap C\right)=\frac{2}{12}\ne P\left(A\right)\cdot P\left(C\right)$   

Zatem niezależne są zdarzenia A' i B'.

Odp.: A