a)

A - wypadła co najmniej raz szóstka.

$A=\left\{\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right),\left(1,6\right),\left(2,6\right),\left(3,6\right),\left(4,6\right),\left(5,6\right)\right\}$  

$\overline{\overline{A}}=11$

 

B - suma oczek, które wypadły to liczba nieparzysta.

$B=\{\left(1,2\right),\left(1,4\right),\left(1,6\right),\left(2,1\right),\left(2,3\right),\left(2,5\right),\left(3,2\right),\left(3,4\right),\left(3,6\right),$   

$\left(4,1\right),\left(4,3\right),\left(4,5\right),\left(5,2\right),\left(5,4\right),\left(5,6\right),\left(6,1\right),\left(6,3\right),\left(6,5\right)\}$  

$\overline{\overline{B}}=18$  

 

$A\cap B=\left\{\left(1,6\right),\left(3,6\right),\left(5,6\right),\left(6,1\right),\left(6,3\right),\left(6,5\right)\right\}$  

$\overline{\overline{A\cap B}}=6$  

 

$P\left(A\mid B\right)=\frac{\overline{\overline{A\cap B}}}{\overline{\overline{B}}}=\frac{6}{18}=\frac{1}{3}$  

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b)

C - suma oczek, które wypadły to liczba parzysta.

$C=\{\left(1,1\right),\left(1,3\right),\left(1,5\right),\left(2,2\right),\left(2,4\right),\left(2,6\right),\left(3,1\right),\left(3,3\right),\left(3,5\right),$  

$\left(4,2\right),\left(4,4\right),\left(4,6\right),\left(5,1\right),\left(5,3\right),\left(5,5\right),\left(6,2\right),\left(6,4\right),\left(6,6\right)\}$  

$\overline{\overline{C}}=18$  

 

$A\cap C=\left\{\left(2,6\right),\left(4,6\right),\left(6,2\right),\left(6,4\right),\left(6,6\right)\right\}$  

$\overline{\overline{A\cap C}}=5$  

 

$P\left(A\mid C\right)=\frac{\overline{\overline{A\cap C}}}{\overline{\overline{C}}}=\frac{5}{18}$  

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c)

D - w pierwszym rzucie wypadło więcej oczek niż w drugim

$D=\left\{\left(2,1\right),\left(3,1\right),\left(3,2\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right)\right\}$  

$\overline{\overline{D}}=15$  

 

$A\cap D=\left\{\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right)\right\}$  

$\overline{\overline{A\cap D}}=5$  

 

$P\left(A\mid D\right)=\frac{\overline{\overline{A\cap D}}}{\overline{\overline{D}}}=\frac{5}{15}=\frac{1}{3}$  

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d)

E - w drugim rzucie wypadła szóstka

$E=\left\{\left(1,6\right),\left(2,6\right),\left(3,6\right),\left(4,6\right),\left(5,6\right),\left(6,6\right)\right\}$  

$\overline{\overline{E}}=6$  

 

$A\cap E=\left\{\left(1,6\right),\left(2,6\right),\left(3,6\right),\left(4,6\right),\left(5,6\right),\left(6,6\right)\right\}$  

$\overline{\overline{A\cap E}}=6$  

 

$P\left(A\mid E\right)=\frac{\overline{\overline{A\cap E}}}{\overline{\overline{E}}}=\frac{6}{6}=1$