Let $P(A)=0,3$. Only one of the following events has a probability less than $0.3$. Which one?
$A)A \cup B$
$B)\bar{A} \cup B$
$C)A \cap B$
$D)\bar{A} \cup \bar{B}$
Lets check one by one.
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ and then $\space 0 \leq P(A \cap B)\leq 0.3 \space$. This happen because, if $\space A \subset B \space$ then $\space P(A \cap B) =P(A) \space$ and so $\space P(B) \geq P(A) \space$. In other hand, if $\space P(A \cap B)=0 \space$ then $\space P(A \cup B)=P(A)+P(B) \space$, and so $\space P(A \cup B) \geq 0.3 \space$ because $\space A \subset B \space$.
If $\space B \subset A \space$, then$\space P(A \cap B)\leq 0.3 \space$, and so $\space P(A \cup B) \geq 0.3 \space$, because $\space P(A)=0.3 \space$ and $\space P(B)\leq P(A) \space$.
Now the second one:
$P(\bar{A} \cup B)=P(\bar{A})+P(B)-P(\bar{A} \cap B)$. It's known that $ \space P(\bar{A})=0.7 \space$, so $\space 0 \leq P(\bar{A} \cap B)\leq 0.7$. If $\bar{A} \subset B$ then $P(\bar{A} \cap B)=P(\bar{A})$ and so $P(\bar{A} \cup B)=P(B)$, wich is $\geq P(\bar{A}) \space$. If $P(\bar{A} \cap B)=0$ then $P(\bar{A} \cup B) \geq0.7$ because $P(\bar{A})=0.7$.
If $B \subset \bar{A}$ then $P(\bar{A} \cap B)=P(B)$, and so $P(\bar{A} \cup B)=P(\bar{A})$.
The third one:
If $A \subset B$ then $P(A \cap B)=P(A)$. But if $B \subset A$ then $P(A \cap B)=P(B)$, and in this case $P(B) \leq P(A)$. So $0 \leq P(A \cap B)\leq 0.3$.The third one would be the right one
This was a question test, without solutions.Can you see if my answer is the right one?Thanks