Prawdopodobieństwo w zbiorach

[liquido]

Wiadomo, że \(\displaystyle{ P(A) = \frac{3}{25} , P(B') = \frac{7}{10} , P(A \cup B) = \frac{4}{10}}\). Oblicz \(\displaystyle{ P(A \cap B), P(A \backslash B), P(A' \cap B)}\).

Z góry bardzo dziękuje za wszelką pomoc

[bstq]

\(\displaystyle{ P(A)=\frac{3}{25}\Rightarrow P\left(A^{\prime}\right)=P\left(\Omega\backslash A\right)=1-P(A)=\frac{22}{25}}\)

\(\displaystyle{ P\left(B^{\prime}\right)=\frac{7}{10}\Rightarrow P\left(B\right)=P\left(\Omega\backslash B^{\prime}\right)=1-P(B^{\prime})=\frac{3}{10}}\)

\(\displaystyle{ P\left(A\cup B\right)=\frac{4}{10}}\)
\(\displaystyle{ P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac{22}{25}+\frac{3}{10}-\frac{4}{10}=0.78}\)

\(\displaystyle{ P(A\backslash B)=P(A)-P(A\cap B)=\frac{22}{25}-0.78=0.1}\)

\(\displaystyle{ P\left(A^{\prime}\cap B\right)=P\left(\left(\Omega\backslash A\right)\cap B\right)=P\left(\left(\Omega\cap B\right)\backslash\left(A\cap B\right)\right)=P\left(\Omega\backslash\left(A\cap B\right)\right)=1-P\left(\Omega\cap\left(A\cap B\right)\right)=1-P\left(A\cap B\right)=1-0.78=0.22}\)

[PES34]

\(\displaystyle{ P\left(A^{\prime}\cap B\right)=P\left(\left(\Omega\backslash A\right)\cap B\right)=P\left(\left(\Omega\cap B\right)\backslash\left(A\cap B\right)\right)=P\left(\Omega\backslash\left(A\cap B\right)\right)=1-P\left(\Omega\cap\left(A\cap B\right)\right)=1-P\left(A\cap B\right)=1-0.78=0.22}\)

może mi ktoś wytłumaczyć skąd się to wzięło? z góry dziękuję

[Maciej87]

Raczej
\(\displaystyle{ \mathbb{P}\left(A^{\prime}\cap B\right)=\mathbb{P}\left(B\backslash\left(A\cap B\right)\right)}\)