Lower bound of intersection negative? Let A and B be events where $P(A) = 1/3$ and $P(B) = 1/4$. Show bounds for $P(A \cup B)$ and $P(A \cap B)$.
For the union I got $1/3 \le P(A \cup B) \le 7/12$. For the intersection I got $1/4$ for the upper bound. When calculating the lower bound I'm using $P(A) + P(B) - 1$ but I'm getting $-5/12$. Am I doing something wrong?
 A: We know that $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.
the left hand side is $\frac{7}{12}$, so it follows that $0 \leq P(A \cup B),P(A \cap B) \leq \frac 7{12}$.
However, $P(A \cup B)$ is at least as large as $P(A)$, so we get $\frac 13 \leq P(A \cup B) \leq \frac 7{12}$.
On the other hand, $P(A \cap B)$ must be smaller than $P(B)$, hence $0 \leq P(A \cap B) \leq \frac 14$.
You are doing nothing wrong. It's just that $P(A) + P(B) - 1$ is a negative quantity, but since  the probability can never go below zero, the lower bound is automatically zero. However, if $P(A) + P(B)-1$ is not a negative quantity, then that quantity would be the lower bound, rather than $0$.
A: Seems like you have the right idea! Really, the lower bound formula should be $\min(P(A) + P(B) - 1, 0)$ since the intersection clearly must be greater than $0$, and as you're aware, the intersection must be greater than $P(A) + P(B) - 1$ also.
A: Lower bound for intersection should be $0$ which would occur, for example, if $A \cap B = \emptyset$. The fact that you are getting a value below zero means you should cut off the probability at $0$.
If you get a value above $0$ that means that $A$ and $B$ have to intersect. For instance if $P(A) = 3/4$ and $P(B) = 4/5$ then
$$ P(A \cap B) \ge \frac{11}{20}. $$
A: We have: $0 \le P(X) \le 1$ for any event $X$, thus a lower bound for $X = A \cap B$ is clearly $0$, for upper bound, note that $ X = A \cap B \subseteq B$. Thus $P(A \cap B) \le P(B) = \dfrac{1}{4}$ which is the upper bound .
A: $$P(A\cup B)=\frac{7}{12}-P(A\cap B).$$
Thus, $$\frac{1}{3}\leq P(A\cup B)\leq\frac {7}{12}$$ and since 
$$P(A\cap B)=\frac{7}{12}-P(A\cup B),$$ we obtain:
$$0\leq P(A\cap B\leq\frac{1}{4}$$
