If $5$ distinct balls are placed at random into $5$ cells, then the probability that exactly one cell remains empty is (in $/ 125$)

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

If $S$ is the sample space of a random experiment $\xi$ and $P$ is a probability function defined on the power set $\mathcal{P}(S)$ of $S$,then which one of the following is not satisfied by $P$?
$(i)$ $P(\phi) = 0$
(ii) If $E^c$ is the complementary event of $E$,then $P(E^c) = 1 - P(E)$
(iii) $0 \leq P(E) \leq 1, \forall E \subseteq S$
(iv) If $E_1 \subseteq E_2$,then $P(E_2) \leq P(E_1)$

Suppose four balls labelled $1, 2, 3, 4$ are randomly placed in boxes $B_1, B_2, B_3, B_4$. The probability that exactly one box is empty is

For independent events $A$ and $B$,$P(A \cup B) =$ . . . . . . .

Of the three independent events $E_1, E_2$ and $E_3$,the probability that only $E_1$ occurs is $\alpha$,only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let the probability $p$ that none of events $E_1, E_2$ or $E_3$ occurs satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$. All the given probabilities are assumed to lie in the interval $(0, 1)$. Then $\frac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3} = $