Two numbers are selected randomly from the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ without replacement one by one. The probability that the minimum of the two numbers is divisible by $3$ or the maximum of the two numbers is divisible by $4$ is:

If two numbers $p$ and $q$ are chosen randomly from the set $\{1, 2, 3, 4\}$,one by one,with replacement,then the probability of getting $p^2 > 4q$ is

Let $X$ and $Y$ be events such that $P(X \cup Y) = P(X \cap Y).$
Statement-$1$: $P(X \cap Y) = P(X' \cap Y') = 0$
Statement-$2$: $P(X) + P(Y) = 2P(X \cap Y).$

Three dice are thrown simultaneously and the sum of the numbers appeared on them is noted. If $A$ is the event of getting a sum greater than $14$ and $B$ is the event of getting a sum which is a multiple of $3$,then $P(A \cap \overline{B}) + P(\overline{A} \cap B) = $

Football teams $T_1$ and $T_2$ play two games against each other. The outcomes of the two games are independent. The probabilities of $T_1$ winning,drawing,and losing a game against $T_2$ are $\frac{1}{2}$,$\frac{1}{6}$,and $\frac{1}{3}$,respectively. Each team gets $3$ points for a win,$1$ point for a draw,and $0$ points for a loss. Let $X$ and $Y$ denote the total points scored by teams $T_1$ and $T_2$,respectively,after two games.
$(1)$ $P(X>Y)$ is
$(A)$ $\frac{1}{4}$ $(B)$ $\frac{5}{12}$ $(C)$ $\frac{1}{2}$ $(D)$ $\frac{7}{12}$
$(2)$ $P(X=Y)$ is
$(A)$ $\frac{11}{36}$ $(B)$ $\frac{1}{3}$ $(C)$ $\frac{13}{36}$ $(D)$ $\frac{1}{2}$

All face cards from a pack of $52$ playing cards are removed. From the remaining $40$ cards,two are drawn randomly without replacement. The probability of drawing a pair (cards of the same denomination) is: