Maximum and minimum values of probability with two events Question:

Consider two events $a$ and $b$ such that $P(a) = \alpha$ and $P(b) = \beta$. Given only that knowledge, what is the maximum and minimum values of the probability of the events $(a \cap b$), and $(a \cup b)$. Can you characterize the situations in which each of these extreme values occur?

My take:

If $a$ and $b$ are disjoint, $(a \cup b)$ will have the maximum value of $1$ and $(a \cap b)$ will have the minimum value of $0$.
if $a$ and $b$ are identical, ($a \cap b$) will have the maximum value of $1$ and $(a \cup b)$ will have the minimum value of $0$.

I am not sure if my explanation for this question is correct or not, if it is wrong please guide me towards the answer. Thanks!
 A: Your answer is partly right and partly wrong. Here's an easy take on it:
We know that: $0\leq P(A)\leq1$ and $P(A)=n(A)/n(S)$.

$$0\leq P(A\cap B)\leq1$$
Min possible: $P(A\cap B)=0$
$\Rightarrow n(A\cap B)=0$
$\Rightarrow A\cap B=\phi\quad$ [Disjoint sets]
Min$(P(A\cap B))=0$ when $A$ and $B$ are disjoint sets.
Example: Act: Die rolled; A: Prime number; B: Composite number
Max possible: $P(A\cap B)=1$
$\Rightarrow n(A\cap B)=n(S)$
$\Rightarrow A=B=S\quad$ [Both are Universal events]
Max$(P(A\cap B))=1$ when $A$ and $B$ are both universal events.
Example: Act: Die rolled; A: Number$>0$; B: Number$<7$

$$0\leq P(A\cup B)\leq1$$
Min possible: $P(A\cup B)=0$
$\Rightarrow n(A\cup B)=0$
$\Rightarrow A=B=\phi\quad$ [Impossible events]
Min$(P(A\cup B))=0$ when $A$ and $B$ are impossible events.
Example: Act: Die rolled; A: Number$<1$; B: Number$>6$
Max possible: $P(A\cup B)=1$
$\Rightarrow n(A\cup B)=n(S)$
$\Rightarrow A\cup B=S\quad$ [Mutually exhaustive events]
Max$(P(A\cup B))=1$ when $A$ and $B$ are mutually exhaustive events.
Example: Act: Die rolled; A: Even number; B: Odd number

For given events, $A$ and $B$, we also know that:
$$P(A\cup B)\geq P(A\cap B)$$
So, the extremes we may have together are:
$\rightarrow P(A\cup B) = P(A\cap B)=1$
$\Rightarrow$ (Mutually Exhaustive events $\cap$ Both Universal events) $=$ Both Universal events
Example: Act: Die rolled; A: Number$>0$; B: Number$<7$
$\rightarrow P(A\cup B) = 1 \text{ and }P(A\cap B)=0$
$\Rightarrow$ (Mutually Exhaustive events $\cap$ Disjoint Sets) $=$ Mutually exclusive and exhaustive events
Example: Act: Die rolled; A: Even number; B: Odd number
$\rightarrow P(A\cup B) = P(A\cap B)=0$
$\Rightarrow$ (Impossible events $\cap$ Disjoint sets) $=$ Impossible events
Example: Act: Die rolled; A: Number$<1$; B: Number$>6$
A: Without further info and, w.l.o.g., take $\mathbb{P}(A)\ge \mathbb{P}(B)$,
$$
\mathbb{P}(A\cup B)\ge \mathbb{P}(A)
$$
with equality if $B \subset A$. Also,
$$
1\ge \mathbb{P}(A\backslash(A\cap B)\cup B\backslash(A\cap B)\cup (A\cap B) )\ge \mathbb{P}(A)+\mathbb{P}(B)-2\mathbb{P}(A\cap B)+\mathbb{P}(A\cap B).
$$
Thus,
$$
\mathbb{P}(A\cap B)\ge \max\{\mathbb{P}(A)+\mathbb{P}(B)-1, 0\}.
$$
The equality can be attained too.
A: Your reasoning is correct. However, you can say a bit more about your research question by using the Frechet inequalities. Note in particular that you can formulate you probabilistic question by using logical conjunction $(\wedge)$, which would coincide with an "and" and the logical disjunction ( $\lor$ ), which would coincide with an "or".
If $A_{i}$ are logical propositions or events, the Fréchet inequalities are

*

*Probability of a logical conjunction $(\wedge)$
$$
\max \left(0, \sum_{k=1}^{n} \mathbb{P}\left(A_{k}\right)-(n-1)\right) \leq \mathbb{P}\left(\bigwedge_{k=1}^{n} A_{k}\right) \leq \min _{k}\left\{\mathbb{P}\left(A_{k}\right)\right\}
$$

*Probability of a logical disjunction $(\vee)$
$$
\max _{k}\left\{\mathbb{P}\left(A_{k}\right)\right\} \leq \mathbb{P}\left(\bigvee_{k=1}^{n} A_{k}\right) \leq \min \left(1, \sum_{k=1}^{n} \mathbb{P}\left(A_{k}\right)\right)
$$
where $\mathbb{P}( )$ denotes the probability of an event or proposition. In the case where there are only two events, say $A$ and $B$, the inequalities reduce to

*Probability of a logical conjunction $(\wedge)$
$$
\max (0, \mathbb{P}(A)+\mathbb{P}(B)-1) \leq \mathbb{P}(A \wedge B) \leq \min (\mathbb{P}(A), \mathbb{P}(B))
$$

*Probability of a logical disjunction ( $\lor$ )
$$
\max (\mathbb{P}(A), \mathbb{P}(B)) \leq \mathbb{P}(A \vee B) \leq \min (1, \mathbb{P}(A)+\mathbb{P}(B))
$$
This comes from: https://en.wikipedia.org/wiki/Fr%C3%A9chet_inequalities
