余事象の独立性の証明

（i）事象 $A$ と事象 $B$ の余事象 $$\begin{array}{rl}P(A\cap B^c)& =P\left(A\right)-P(A\cap B)\\ & =P\left(A\right)-P\left(A\right)\cdot P\left(B\right)\\ & =P\left(A\right)\{1-P\left(B\right)\}\end{array}$$ 確率の基本性質 $P\left(A^C\right)=1-P\left(A\right)$ より、 $$\begin{array}{r}P(A\cap B^c)=P\left(A\right)\cdot P\left(B^c\right)\end{array}$$

（ii）事象 $A$ の余事象と事象 $B$ $$\begin{array}{rl}P(A^c\cap B)& =P\left(B\right)-P(A\cap B)\\ & =P\left(B\right)-P\left(A\right)\cdot P\left(B\right)\\ & =P\left(B\right)\{1-P\left(A\right)\}\end{array}$$ 確率の基本性質 $P\left(A^C\right)=1-P\left(A\right)$ より、 $$\begin{array}{r}P(A^c\cap B)=P\left(A^c\right)\cdot P\left(B\right)\end{array}$$

（iii）事象 $A$ の余事象と事象 $B$ の余事象
ド・モルガンの法則 ${(A\cup B)}^C=A^C\cap B^C$ より、 $$\begin{array}{rl}P(A^c\cap B^c)& =1-P(A\cup B)\\ & =1-P\left(A\right)-P\left(B\right)+P(A\cap B)\\ & =1-P\left(A\right)-P\left(B\right)+P\left(A\right)\cdot P\left(B\right)\\ & =\{1-P\left(A\right)\}-P\left(B\right)\{1-P\left(A\right)\}\\ & =\{1-P\left(A\right)\}\{1-P\left(B\right)\}\end{array}$$ 確率の基本性質 $P\left(A^C\right)=1-P\left(A\right)$ より、 $$\begin{array}{r}P(A^c\cap B^c)=P\left(A^C\right)\cdot P\left(B^c\right)\end{array}$$ $◼$