1.4.1 problem 43

(a)

Inequality can be demonstrated using the first property of probabilities, $$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$$

and the first axiom of probabilities,

$$P\left(S\right)=1.$$

$P\left(A\right)+P\left(B\right)-P(A\cap B)\le 1\Rightarrow P\left(A\right)+P\left(B\right)-1\le P(A\cap B)$.

Strict equality holds if and only if $A\cup B=S$ where $S$ is the sample space.

(b)

Since $A\cap B\subseteq A\cup B$, $P(A\cap B)\le P(A\cup B)$ by the second property of probabilites.

Strict equality holds if and only if $A=B.$

(c)

Inequality follows directly from the first property of probabilities with strict equality if and only if $P(A\cap B)=0.$