$P\left(A\mid B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$   - zauważmy, że aby wyznaczyć prawdopodobieństwo warunkowe, nie musimy znać prawdopodobieństwa P(A)

 

$p_2=\frac{1}{64}\cdot \left(\begin{array}{c}6\\ 2\end{array}\right)=\frac{1}{64}\cdot \frac{6!}{2!\cdot \left(6-2\right)!}=\frac{1}{64}\cdot \frac{4!\cdot 5\cdot 6}{2!\cdot 4!}=\frac{1}{64}\cdot \frac{30}{2}=\frac{15}{64}$  

 

$p_3=\frac{1}{64}\cdot \left(\begin{array}{c}6\\ 3\end{array}\right)=\frac{1}{64}\cdot \frac{6!}{32!\cdot \left(6-3\right)!}=\frac{1}{64}\cdot \frac{3!\cdot 4\cdot 5\cdot 6}{3!\cdot 3!}=\frac{1}{64}\cdot \frac{120}{6}=\frac{20}{64}$   

 

$p_4=\frac{1}{64}\cdot \left(\begin{array}{c}6\\ 4\end{array}\right)=\frac{1}{64}\cdot \frac{6!}{4!\cdot \left(4-2\right)!}=\frac{1}{64}\cdot \frac{4!\cdot 5\cdot 6}{4!\cdot 2!}=\frac{1}{64}\cdot \frac{30}{2}=\frac{15}{64}$  

 

$p_5=\frac{1}{64}\cdot \left(\begin{array}{c}6\\ 5\end{array}\right)=\frac{1}{64}\cdot \frac{6!}{5!\cdot \left(6-5\right)!}=\frac{1}{64}\cdot \frac{5!\cdot 6}{5!\cdot 1!}=\frac{1}{64}\cdot 6=\frac{6}{64}$  

 

$p_6=\frac{1}{64}\cdot \left(\begin{array}{c}6\\ 6\end{array}\right)=\frac{1}{64}\cdot 1=\frac{1}{64}$  

 

$P\left(B\right)=\frac{15}{64}+\frac{20}{64}+\frac{15}{64}+\frac{6}{64}+\frac{1}{64}=\frac{57}{64}$  

 

$A\cap B$   - zdarzenie polegające na wylosowani liczby 3 lub 5

  

$P\left(A\cap B\right)=p_3+p_5=\frac{20}{64}+\frac{6}{64}=\frac{26}{64}$  

 

 

$P\left(A\mid B\right)=\frac{\frac{26}{64}}{\frac{57}{64}}=\frac{26}{64}\cdot \frac{64}{57}=\underline{\frac{26}{57}}$