prawdopodobieństwo aksjomaty P(A), P(B)...

[dall]

Witam.
Potrzebuję pomocy w zadaniu z prawdopodobieństwa.

Wiadomo, że \(\displaystyle{ P(A)*P(B)=P(A\cap B)=0,4}\) oraz \(\displaystyle{ 4*P(B')*(P(B)-2)+3=0}\).
Wyznacz P(A), P(B) oraz \(\displaystyle{ P(A\cup B)}\).

Z góry wielkie dzięki.

[Pancernik]

\(\displaystyle{ 4P\left(B'\right)\left(P\left(B\right)-2\right)+3=0\\
4\left(1-P\left(B\right)\right)\left(P\left(B\right)-2\right)=-3\\
\left(1-P\left(B\right)\right)\left(P\left(B\right)-2\right)=-\frac{3}{4}\\
P\left(B\right)-2-P^2\left(B\right)+2P\left(B\right)=-\frac{3}{4}\\
-P^2\left(B\right)+3P\left(B\right)-\frac{5}{4}=0\\
\Delta=3^2-4\cdot\left(-1\right)\cdot\left(-\frac{5}{4}\right)=4\\
P_1\left(B\right)=\frac{-3-2}{-2}=\frac{5}{2}>1\\
P_2\left(B\right)=\frac{-3+2}{-2}=\frac{1}{2}<1\\
P\left(B\right)=\frac{1}{2}}\)

\(\displaystyle{ P\left(A\right)\cdot P\left(B\right)=0,4\\
P\left(A\right)\cdot\frac{1}{2}=\frac{2}{5}\\
P\left(A\right)=\frac{4}{5}}\)

\(\displaystyle{ P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\\
P\left(A\cup B\right)=\frac{4}{5}+\frac{1}{2}-\frac{2}{5}\\
P\left(A\cup B\right)=\frac{9}{10}}\)