Given P(A) and P(B), what would be the minimum probability of the intersection? Like if I was given a P(A) of .5 and a P(B) value of .4 how would I get the minimum of the P(A∩B)?
 A: Quickly generalising from cardinal's comment, we have:
$$\mathbb P(A \cap B) = \mathbb P(A) + \mathbb P(B) - \mathbb P( A \cup B )$$
Now the unknown term on the right must be between $0$ and $1$.  So a loose bound is given by using these limits.
$$\mathbb P(A) + \mathbb P(B) \geq \mathbb P(A \cap B) \geq \mathbb P(A) + \mathbb P(B) - 1$$
However, we can do a little better than this.  Firstly, the intersection cannot be larger than the smallest marginal probability, as we have 
$$\mathbb P(A \cap B)=\mathbb P(A)\mathbb P(B|A)\leq \mathbb P(A)$$
$$\mathbb P(A \cap B)=\mathbb P(B)\mathbb P(A|B)\leq \mathbb P(B)$$
Also, the probability cannot be negative.  Hence we have:
$$min\left[\mathbb P(A),\mathbb P(B)\right] \geq \mathbb P(A \cap B) \geq max\left[\mathbb P(A) + \mathbb P(B) - 1,0\right]$$
A: $A$ and $B$ could be disjoint, so the minimum possible value of $P(A \cap B)$ is zero.
Example: Suppose a number is chosen from {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, with each number equally likely to be chosen. Let $A$ be the event that the chosen number is less than or equal to 5; let $B$ be the event that the chosen number is greater than or equal to 7. Then $P(A)=0.5$, $P(B)=0.4$, and $A \cap B$ is empty, so $P(A \cap B)=0$.
