$A$ and $B$ are two independent events of a random experiment and $P(A) > P(B)$. If the probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$,then the probability of the occurrence of $B$ is

Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game, then the probability that $B$ wins the game is:

$A$ box contains $100$ balls,numbered from $1$ to $100$. If $3$ balls are selected one after the other at random with replacement from the box,then the probability that the sum of the three numbers on the balls selected is an odd number,is

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} = $

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

In a tournament,there are $12$ players $P_1, P_2, P_3, \dots, P_{12}$ divided into $6$ pairs at random. From each game,a winner is decided based on the game played between the two players of the pair. Assuming each player is of equal strength,what is the probability that exactly one out of $P_1$ and $P_2$ is among the losers?