Prawa de Morgana ${\left(A\cup B\right)}^{{}^{\prime }}=A^{{}^{\prime }}\cap B^{{}^{\prime }}$ ${\left(A\cap B\right)}^{{}^{\prime }}=A^{{}^{\prime }}\cup B^{{}^{\prime }}$

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a)

Wiemy, że:

$P\left(A\cap B\right)=\frac{1}{3},P\left(A\right)=\frac{1}{2},P\left(B^{{}^{\prime }}\right)=\frac{1}{6}$

Wiemy, że: 

$A^{{}^{\prime }}\cap B^{{}^{\prime }}={\left(A\cup B\right)}^{{}^{\prime }}$  

Zatem:

$P\left(A^{{}^{\prime }}\cap B^{{}^{\prime }}\right)=P\left({\left(A\cup B\right)}^{{}^{\prime }}\right)=1-P\left(A\cup B\right)=$

$=1-\left(P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\right)=$  

$=1-\left(P\left(A\right)+1-\left(P\left(B^{{}^{\prime }}\right)\right)-P\left(A\cap B\right)\right)=$

$=1-\left(\frac{1}{2}+\left(1-\frac{1}{6}\right)-\frac{1}{3}\right)=$  

$=1-\left(\frac{1}{2}+\frac{5}{6}-\frac{1}{3}\right)=$    

$=1-\left(\frac{3}{6}+\frac{5}{6}-\frac{2}{6}\right)=$  

$=1-1=0$  

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b)

Wiemy, że:

$A\subset B,P\left(B\right)=\frac{3}{4},P\left(B\text{}A\right)=\frac{1}{3}$

Wiemy, że:

$A^{{}^{\prime }}\cup B^{{}^{\prime }}={\left(A\cap B\right)}^{{}^{\prime }}$  

Zatem:

$P\left(A^{{}^{\prime }}\cup B^{{}^{\prime }}\right)=P\left({\left(A\cap B\right)}^{{}^{\prime }}\right)=1-P\left(A\cap B\right)=$

$=1-\left(P\left(B\right)-P\left(B\text{}A\right)\right)=$  

$=1-\left(\frac{3}{4}-\frac{1}{3}\right)=$

$=1-\left(\frac{9}{12}-\frac{4}{12}\right)=$  

$=1-\frac{5}{12}=\frac{7}{12}$