Bag $A$ contains $3$ white and $4$ red balls,bag $B$ contains $4$ white and $5$ red balls,and bag $C$ contains $5$ white and $6$ red balls. If one ball is drawn at random from each of these three bags,then the probability of getting one white and two red balls is

$A$ bag contains $4$ red and $6$ black balls. $A$ ball is drawn at random from the bag,its colour is observed,and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag,then the probability that this drawn ball is red,is:

If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B)=\frac{1}{6}$ and $P(\bar{A} \cap \bar{B})=\frac{1}{3}$, then $P(A)$ is equal to (Here, $\bar{E}$ is the complement of the event $E$)

Let $p$ denote the probability that a man aged $x$ years will die in a year. The probability that out of $n$ men $A_1, A_2, A_3, ..., A_n$ each aged $x$,$A_1$ will die in a year and will be the first to die,is

Let $B_{i} (i=1, 2, 3)$ be three independent events in a sample space. The probability that only $B_{1}$ occurs is $\alpha$,only $B_{2}$ occurs is $\beta$,and only $B_{3}$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_{i}$ occurs,and these $4$ probabilities satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$ (All the probabilities are assumed to lie in the interval $(0, 1)$). Then $\frac{P(B_{1})}{P(B_{3})}$ is equal to ..........

Let $X$ and $Y$ be two events such that $P(X \cup Y) = P(X \cap Y)$.
Statement $1$: $P(X \cap Y') = P(X' \cap Y) = 0$.
Statement $2$: $P(X) + P(Y) = 2P(X \cap Y)$.