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-$2$: $P(X) + P(Y) = 2P(X \cap Y).$

One die has two faces marked $1$,two faces marked $2$,one face marked $3$ and one face marked $4$. Another die has one face marked $1$,two faces marked $2$,two faces marked $3$ and one face marked $4$. The probability of getting the sum of numbers to be $4$ or $5$,when both the dice are thrown together,is

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then

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
$(A)$ $\frac{11}{36}$ $(B)$ $\frac{1}{3}$ $(C)$ $\frac{13}{36}$ $(D)$ $\frac{1}{2}$

Three screws are drawn at random from a lot of $50$ screws containing $5$ defective ones. The probability of the event that all $3$ screws drawn are non-defective,assuming that the drawing is $(a)$ with replacement and $(b)$ without replacement,is respectively:

Let $A$ be the set of all $4$-digit natural numbers whose exactly one digit is $7$. Then the probability that a randomly chosen element of $A$ leaves a remainder of $2$ when divided by $5$ is ..... .