A - co najmniej 1 szóstka
B - taka sama liczba na każdej kostce
[tex]$$\begin{aligned}&P(A \mid B)=\frac{P(A \cap B)}{P(B)}\\&|\Omega|=6^3\\&\begin{aligned}& |A \cap B|=1 \\& P(A \cap B)=\frac{1}{6^3}\end{aligned}\\&\begin{aligned}& |B|=6 \\& P(B)=\frac{6}{6^3}=\frac{1}{6^2}\end{aligned}\\&P(A \mid B)=\frac{\frac{1}{6^3}}{\frac{1}{6^2}}=\frac{1}{6^3} \cdot 6^2=\frac{1}{6}\end{aligned}$$[/tex]
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A - co najmniej 1 szóstka
B - taka sama liczba na każdej kostce
[tex]$$\begin{aligned}&P(A \mid B)=\frac{P(A \cap B)}{P(B)}\\&|\Omega|=6^3\\&\begin{aligned}& |A \cap B|=1 \\& P(A \cap B)=\frac{1}{6^3}\end{aligned}\\&\begin{aligned}& |B|=6 \\& P(B)=\frac{6}{6^3}=\frac{1}{6^2}\end{aligned}\\&P(A \mid B)=\frac{\frac{1}{6^3}}{\frac{1}{6^2}}=\frac{1}{6^3} \cdot 6^2=\frac{1}{6}\end{aligned}$$[/tex]