If $P(A) = 0.65$ and $P(B) = 0.15$,then $P(\bar{A}) + P(\bar{B}) = $

Consider the experiment of rolling a die. Let $A$ be the event 'getting a prime number' and $B$ be the event 'getting an odd number'. Write the set representing the event 'not $A$'.

For the two events $A$ and $B$,$P(A) = 0.38$ and $P(B) = 0.41$. Find the value of $P(\text{not } A)$.

If $P(A) = 0.25, P(B) = 0.50$ and $P(A \cap B) = 0.14,$ then $P(A \cap \bar{B})$ is equal to

$A$ and $B$ are events such that $P(A)=0.42$,$P(B)=0.48$ and $P(A \cap B)=0.16$. Determine $P(\text{not } B)$.

If $P(A) = 0.25$,$P(B) = 0.50$,and $P(A \cap B) = 0.14$,then what is the value of $P(A \cap \overline{B})$?