Two players $A$ and $B$ alternatively toss $3$ coins simultaneously. The player who gets $2$ heads and $1$ tail first, wins the game. If the game continues until someone wins and if $A$ begins the game, the probability that $B$ wins the game is

$A$ biased die is marked with numbers $2, 4, 8, 16, 32, 32$ on its faces. The probability of getting a face with mark $n$ is $\frac{1}{n}$. If the die is thrown thrice,then the probability that the sum of the numbers obtained is $48$ 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:

If a year is chosen at random from the years having $53$ Sundays and $53$ Mondays,what is the probability that it is a leap year?

For the three events $A, B$ and $C$,$P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs) = $P$ (exactly one of the events $C$ or $A$ occurs) = $p$ and $P$ (all the three events occur simultaneously) = $p^2$,where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is

Let $A, B, C$ be three pairwise independent events of a random experiment. If $P(\bar{B} \cup \bar{C}) = \frac{1}{2}$, $P(A) > 0$, $P(B) = b$, and $P(C) = c$, then $P((\bar{B} \cap \bar{C}) \mid A) = $