If $P(A') + P(B') P(A \cup B) = 0.7$,then $P(A') + P(B')$ is

$A$ 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 ————

If $A$ and $B$ are two independent events such that $P(A) > 0.5$,$P(B) > 0.5$,$P(A \cap \bar{B}) = \frac{3}{25}$,and $P(\bar{A} \cap B) = \frac{8}{25}$,then $P(A \cap B)$ is:

Let $A, B$ and $C$ be three events such that the probability that exactly one of $A$ and $B$ occurs is $(1-k)$,the probability that exactly one of $B$ and $C$ occurs is $(1-2k)$,the probability that exactly one of $C$ and $A$ occurs is $(1-k)$ and the probability that all $A, B$ and $C$ occur simultaneously is $k^2$,where $0 < k < 1$. Then the probability that at least one of $A, B$ and $C$ occurs is:

For three events $A, B$ and $C$,$P(\text{Exactly one of } A \text{ or } B \text{ occurs}) = P(\text{Exactly one of } B \text{ or } C \text{ occurs}) = P(\text{Exactly one of } C \text{ or } A \text{ occurs}) = \frac{1}{4}$ and $P(\text{All the three events occur simultaneously}) = \frac{1}{16}$. Then the probability that at least one of the events occurs is:

If a $3$-digit number is randomly chosen,what is the probability that either the number itself or some permutation of the number (which is a $3$-digit number) is divisible by $4$ and $5$?