Can P(A) be smaller than Pr(A∩B)?

Question

I saw this question: if $P(A)=0.4, P(B)=0.7, P(A∪B)=0.6$. What's $P(A|B)$?

As $P(A∪B)=P(A)+P(B)-P(A∩B), P(A∩B)=0.5$.

1. Then $P(A∩B)$ is greater than $P(A)$, is it possible? Is there something wrong with the question?

2. If there is nothing wrong with the question. The answer should be 5/7, right? Thank you.

Answer 1

You're given in the problem that $P(A\cup B)<P(B)$, which is impossible. So it's not strange that from this you can deduce $P(A\cap B)>P(A)$, which is also impossible.

Answer 2

$$P(A\cap B)=P(A)P(B|A)$$

Since $P(B|A)\le1$, we get

$$P(A\cap B)\le P(A)$$

The problem is incorrect.