Let $A$ and $B$ be two independent events such that $P(B) > P(A)$. If the probability that both $A$ and $B$ happen is $\frac{1}{12}$ and the probability that neither $A$ nor $B$ happens is $\frac{1}{2}$,then:

$A$ ship is fitted with three engines $E_1, E_2$,and $E_3$. The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4}$,and $\frac{1}{4}$. For the ship to be operational,at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X_1, X_2$,and $X_3$ denote respectively the events that the engines $E_1, E_2$,and $E_3$ are functioning. Which of the following is (are) true?
$(A) P(X_1^c \mid X) = \frac{3}{16}$
$(B) P(\text{Exactly two engines are functioning} \mid X) = \frac{7}{8}$
$(C) P(X \mid X_2) = \frac{5}{16}$
$(D) P(X \mid X_1) = \frac{7}{16}$

Consider the following statements:
Assertion $(A)$: If $P_1, P_2, P_3$ are probabilities of occurrence of three independent events, then the probability of occurrence of at least one of them is $1 - [(1 - P_1)(1 - P_2)(1 - P_3)]$.
Reason $(R)$: For any three independent events $A, B$, and $C$, $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C)$.
The correct option among the following is:

An urn contains marbles of four colours: red,white,blue,and green. When four marbles are drawn without replacement,the following events are equally likely:
$1.$ The selection of four red marbles.
$2.$ The selection of one white and three red marbles.
$3.$ The selection of one white,one blue,and two red marbles.
$4.$ The selection of one marble of each colour.
The smallest total number of marbles satisfying the given condition 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 ..........

If the numbers appearing on the two throws of a fair six-faced die are $\alpha$ and $\beta$,then the probability that $x^{2}+\alpha x+\beta > 0$ for all $x \in R$ is: