Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then

Two friends $A$ and $B$ meet every weekend either at a party or at a Sports Club. The probability that they meet at the Sports Club is $\frac{4}{9}$. The probability that they dine together at a party and at the Club are respectively $\frac{1}{3}$ and $\frac{2}{5}$. On a certain weekend, the probability that they disperse without dining together is:

$A$ 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?
$(A) P(X_1^c \mid X) = \frac{3}{16}$
$(B) P(\text{Exactly two engines are functioning} \mid X) = \frac{7}{8}$
$(C) P(X \mid X_2) = \frac{5}{16}$
$(D) P(X \mid X_1) = \frac{7}{16}$

$A$ box contains $15$ tickets numbered $1, 2, \dots, 15$. Seven tickets are drawn at random one after the other with replacement. The probability that the greatest number on a drawn ticket is $9$ 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 $B_{i} (i=1, 2, 3)$ 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 ..........