$A$ and $B$ are two independent events. $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$. Match the following List-$I$ with List-$II$.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

$A$ pack of cards contains $4$ aces,$4$ kings,$4$ queens,and $4$ jacks. Two cards are drawn at random. The probability that at least one of these is an ace is:

For three events $A$, $B$, and $C$ of a sample space, $P(\text{exactly one of } A \text{ or } B \text{ occurs}) = P(\text{exactly one of } B \text{ or } C \text{ occurs}) = P(\text{exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$. If the probability of all the three events occurring simultaneously is $\frac{1}{16}$, then the probability that at least one of the events occurs is:

$A$ and $B$ are two independent events such that $P(A \cup B) = 0.8$ and $P(A) = 0.3$. The value of $P(B)$ is:

If $A$ and $B$ are independent events and $P(A) = p, P(B) = 2p$,and $P(\text{exactly one from } A \text{ and } B) = \frac{5}{9}$,then find the value of $p$.

$A$ coin is tossed $8$ times. If the probability that exactly $4$ heads appear in the first six tosses and exactly $3$ heads appear in the last five tosses is $p$,then $96p$ is equal to ————