Mamy dane:

$P\left(A\right)=\frac{1}{2}$  

$P\left(A\mid B\right)=\frac{3}{4}$  

$P\left(A\mid B{}^{\prime }\right)=\frac{1}{3}$  

Wobec tego:

$P\left(A\mid B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$  

$\frac{3}{4}=\frac{P\left(A\cap B\right)}{P\left(B\right)}\mid \cdot P\left(B\right)$  

$\frac{3P\left(B\right)}{4}=P\left(A\cap B\right)$  

oraz

$P\left(A\mid B\right)=\frac{P\left(A\cap B{}^{\prime }\right)}{P\left(B{}^{\prime }\right)}$  

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

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

Wykonajmy rysunek pomocniczy.

Zdarzenia A ∩ B' oraz A \ (A ∩ B) są tymi samymi zdarzeniami, więc:

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

Wobec tego:

$\frac{1}{3}=\frac{P\left(A\right)-P\left(A\cap B\right)}{1-P\left(B\right)}$  

$\frac{1}{3}=\frac{\frac{1}{2}-\frac{3P\left(B\right)}{4}}{1-P\left(B\right)}$  

$\frac{1}{3}=\frac{\frac{2}{4}-\frac{3P\left(B\right)}{4}}{1-P\left(B\right)}$  

$\frac{1}{3}=\frac{\frac{2-3P\left(B\right)}{4}}{1-P\left(B\right)}$  

$\frac{1}{3}=\frac{2-3P\left(B\right)}{4}\cdot \frac{1}{1-P\left(B\right)}$  

$\frac{1}{3}=\frac{2-3P\left(B\right)}{4\left(1-P\left(B\right)\right)}$  

$\frac{1}{3}=\frac{2-3P\left(B\right)}{4-4P\left(B\right)}$  

Korzystając z własności proporcji, otrzymujemy:

$4-4P\left(B\right)=3\left(2-3P\left(B\right)\right)$  

$4-4P\left(B\right)=6-9P\left(B\right)\mid +9P\left(B\right)$  

$4+5P\left(B\right)=6\mid -4$  

$5P\left(B\right)=2\mid :5$  

$P\left(B\right)=\frac{2}{5}$  

Wyznaczamy prawdopodobieństwo zdarzenia A ∩ B.

$P\left(A\cap B\right)=\frac{3P\left(B\right)}{4}=\frac{3\cdot \frac{2}{5}}{4}=\frac{\frac{6}{5}}{4}=\frac{6}{5}\cdot \frac{1}{4}=\frac{3}{10}$  

Wyznaczamy prawdopodobieństwo zdarzenia A ∪ B.

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=\frac{1}{2}+\frac{2}{5}-\frac{3}{10}=\frac{5}{10}+\frac{4}{10}-\frac{3}{10}=\frac{6}{10}=\frac{3}{5}$  

Teraz obliczymy prawdopodobieństwo warunkowe P(A'|B).

$P\left(A{}^{\prime }\mid B\right)=\frac{P\left(A{}^{\prime }\cap B\right)}{P\left(B\right)}$  

Wykonajmy rysunek pomocniczy.

Zdarzenia A' ∩ B oraz B \ (A ∩ B) są tymi samymi zdarzeniami, więc:

$P\left(A{}^{\prime }\cap B\right)=P\left(B\right)-P\left(A\cap B\right)=\frac{2}{5}-\frac{3}{10}=\frac{4}{10}-\frac{3}{10}=\frac{1}{10}$  

Wobec tego:

$P\left(A{}^{\prime }\mid B\right)=\frac{P\left(A{}^{\prime }\cap B\right)}{P\left(B\right)}=\frac{\frac{1}{10}}{\frac{2}{5}}=\frac{1}{10}\cdot \frac{5}{2}=\frac{1}{4}$