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How do I go about answering the following question.
Given two events A,B with (A) = 3/4 and P(B) = 1/3, what is the smallest possible value of P(A $\cap$ B)? The largest? That is, a and b such that, $$a \leq P(A \cap B) \leq b,$$ holds and any value in the closed interval [a,b] is possible.
a = ? b = ?
Below is a image of what I think is the shaded area of interest. Is this correct? If so, then is the probability 1/4 x 2/3 = 1/6. In which case, a = 1/4 and b = 2/3?
A venn diagram might be helpful.
In this case,
$\max P(A\cap B)=\min(P(A),P(B))=\frac13$
$\min P(A\cap B)=P(A)+P(B)-1=\frac1{12}$ (the inclusion-exclusion principle)
Concerning the largest value we have $P(A\cap B)\leq\min(P(A),P(B))$ and taking $A\subseteq B$ or $B\subseteq A$ (at least one of the options is open) we find that: $$\min(P(A),P(B))$$ is indeed possible as value of $P(A\cap B)$, so is the largest value.
Concerning the smallest value we have $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ so it will be achieved if $P(A\cup B)$ is maximal. That leads to value:$$P(A)+P(B)-\min(1,P(A)+P(B))$$as smallest value.
Since $P(A \cap B) \le P(A) = \frac34$ and $P(A \cap B) \le P(B)=\frac13$ you must have $P(A \cap B) \le \frac13$
Meanwhile $P(A \cup B) \le 1$ and $P(A)+P(B)=\frac{13}{12}$ so you must have $P(A \cap B) \ge \frac1{12}$
If you want to visualise this, try something like this
