Let $A$ and $B$ be two events such that $P(A \cap B) = \frac{1}{6}$,$P(A \cup B) = \frac{31}{45}$,and $P(\bar{B}) = \frac{7}{10}$. Then which of the following is true?

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 a value of $\frac{P(E)}{P(F)}$ 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 $E$)

An unbiased coin is tossed. If the result is a head,a pair of unbiased dice is rolled and the sum of the numbers on the two faces is noted. If the result is a tail,a card from a well-shuffled pack of eleven cards numbered $2, 3, 4, \dots, 12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$ is:

$A$ and $B$ are independent events. The probability that both $A$ and $B$ occur is $\frac{1}{20}$ and the probability that neither of them occurs is $\frac{3}{5}$. The probability of occurrence of $A$ 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: