If $A$ and $B$ are two events,then which of the following is $NOT$ true?

Three distinct numbers are selected randomly from the set $\{1, 2, 3, \ldots, 40\}$. If the probability that the selected numbers are in an increasing $G.P.$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to . . . . . . .

Cards are drawn one after the other without replacement from a well-shuffled pack of cards until an ace card appears. If the probability that exactly $5$ cards are drawn before the first ace card appears is $\frac{4}{49}\left(\frac{p_1 \cdot p_2 \cdot p_3}{p_4 \cdot p_5 \cdot p_6}\right)$, where $p_i$ is prime for $i=1, 2, 3, 4, 5, 6$, then $(\max \{p_i\} - \min \{p_i\}) = $

If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B) = \frac{1}{6}$ and $P(\bar{A} \cap \bar{B}) = \frac{1}{3}$,then $P(A)$ is equal to (Here,$\bar{E}$ is the complement of the event $E$)

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:

Three numbers are selected at random from the set ${1, 2, 3, \dots, 8}$ without replacement. Given that the minimum of the selected numbers is $3$ and the maximum is $6$, what is the probability that the third number is $4$ or $5$?