If $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then $P(B/A) = $

If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B \mid A) = 1$,then . . . . . . .

Three numbers are chosen at random without replacement from $\{1, 2, 3, 4, 5, 6, 7, 8\}$. The probability that their minimum is $3$,given that their maximum is $6$,is:

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P(E_{1} \mid E_{2}) = \frac{1}{2}$,$P(E_{2} \mid E_{1}) = \frac{3}{4}$ and $P(E_{1} \cap E_{2}) = \frac{1}{8}$. Then:

If $P(AB) = P(A)P(B)$,$P(A/B) = 1/4$,and $P(B/A) = 1/3$,then which of the following is true?

Let $E^c$ denote the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G)>0$ and $P(E \cap F \cap G) = 0$. Then $P(E^c \cap F^c \mid G)$ equals