$A$ and $B$ throw a pair of dice alternately and they note the sum of the numbers appearing on the dice. $A$ wins if he throws $6$ before $B$ throws $7$,and $B$ wins if he throws $7$ before $A$ throws $6$. If $A$ begins,the probability of $A$ winning is:

$A$ and $B$ are independent events with $P(A)=\frac{3}{10}$ and $P(B)=\frac{2}{5}$. Then,the value of $P(A^{\prime} \cup B)$ is:

$A$ draws two cards with replacement from a pack of $52$ cards and $B$ throws a pair of dice. What is the chance that $A$ gets both cards of the same suit and $B$ gets a total of $6$?

The probability of occurrence of an event is $\frac{2}{5}$ and the probability of non-occurrence of another event is $\frac{3}{10}$. If these events are independent,then the probability that only one of the two events occur is

$A$ player $X$ has a biased coin whose probability of showing heads is $p$ and a player $Y$ has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If $X$ starts the game,and the probability of winning the game by both the players is equal,then the value of $p$ is

$A$ boy throws an unbiased die. Whenever he gets $1$ on the die, he has a further chance to throw it once again immediately. The probability that the boy gets a score of $7$ in this process is