$P\left(A\right)=3\cdot P\left(B\right)$  

$P\left(A{}^{\prime }\text{}B\right)=\frac{1}{9}$  

$P\left(A\cap B\right)=0$  

 

$P\left(A\right)=?$  

$P\left(B\right)=?$  

 

 

Wykonajmy rysunek pomocniczy:

Zauważmy, że skoro zdarzenia A i B są rozłączne, to:

$B\subset A{}^{\prime }$  , a więc:

$A{}^{\prime }\text{}B=\Omega \text{}A\text{}B$  , a więc:

$P\left(A{}^{\prime }\text{}B\right)=P\left(\Omega \right)-P\left(A\right)-P\left(B\right)=1-P\left(A\right)-P\left(B\right)=1-3\cdot P\left(B\right)-P\left(B\right)=1-4\cdot P\left(B\right)=\frac{1}{9}$  

 

$1-4\cdot P\left(B\right)=\frac{1}{9}$  

 

$-4\cdot P\left(B\right)=-\frac{8}{9}\mid :\left(-4\right)$  

 

$P\left(B\right)=\frac{2}{9}$  

 

$P\left(A\right)=3\cdot \frac{2}{9}=\frac{2}{3}$