Probability Intersections

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$A$ and $B$ are two events from certain probability space $\Omega$.

Knowing that: $P(A)=0.6$, $P(B)=0.7 $ and $P(A\cup B)-P(A\cap B) = 0.2$

Determine $P(A\cap B)$

It says in this sheet that the result is $0.55$.

Studying on the internet said that when they're independent $P(A\cap B) = P(A)\times P(B)$

which results in $0.42$. Which is wrong supposedly. When they're dependent it's $P(A\cap B) = P(B|A)\times P(A) = P(B)\times P(A|B)$

What operation is $|$? $(B|A)$ - division?

I don't know if they're dependent or not but in a question below it states that "$A$ and $C$ are independent" (so if there's the need to say they're independent it means the others so far were dependent).

I'm asking for a fast answer with an explanation (reason why, formula) so I can apply it tomorrow in the exam.

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Hint:

In general: $$P(A\cup B)+P(A\cap B)=P(A)+P(B)$$

It follows directly from integrating both sides of the evident equation $$1_{A\cup B}+1_{A\cap B}=1_A+1_B$$ with respect to probability measure $P$.