$A$ and $B$ are independent events. The probability that both $A$ and $B$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $A$ is

Two squares are chosen one by one on a chessboard. The probability that they have a side in common is

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?

$A$ box $B_1$ contains $1$ white ball,$3$ red balls and $2$ black balls. Another box $B_2$ contains $2$ white balls,$3$ red balls and $4$ black balls. $A$ third box $B_3$ contains $3$ white balls,$4$ red balls and $5$ black balls.
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$,the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red,the probability that these $2$ balls are drawn from box $B_2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Choose the correct options for question $1$ and $2$.

Let there be 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 $p$ denote the probability that none of the events occur,which satisfies 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,$\frac{\text{Probability of occurrence of } E_{1}}{\text{Probability of occurrence of } E_{3}}$ is equal to ..........

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