can probability be negative I am solving a question that says:
Given that $$P(B)=P(A\cap B)=\frac{1}{2}$$ and $$P(A)=\frac{3}{8}$$
Find
$$P(A\cap {B}^{c})$$
My answer is 
$$P(A\cap {B}^{c})=P(A)-P(A\cap B)=(-)\frac{1}{8}$$

It's because the two conditions contradict each other.  The first says that $P(B)$ is $\frac{1}{2}$, and also that whenever $B$ happens $A$ happens, so already we know that $P(A) \ge \frac{1}{2}$.  But the second says that $P(A) \lt \frac{1}{2}$.  The derivation you have carried out is another way to say the same thing (as a proof by contradiction), because negative probability violates Kolmogorov's first probability axiom.
On the other hand, negative "probability" can have meaning if we expand our scope somewhat.
