If $A$ and $B$ are two independent events,then $A$ and $\bar{B}$ are

If $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$,then $P(A \mid B) = $ . . . . . . .

Suppose that Box-$I$ contains $8$ red,$3$ blue and $5$ green balls,Box-$II$ contains $24$ red,$9$ blue and $15$ green balls,Box-$III$ contains $1$ blue,$12$ green and $3$ yellow balls,and Box-$IV$ contains $10$ green,$16$ orange and $6$ white balls. $A$ ball is chosen randomly from Box-$I$; call this ball $b$. If $b$ is red,then a ball is chosen randomly from Box-$II$; if $b$ is blue,then a ball is chosen randomly from Box-$III$; and if $b$ is green,then a ball is chosen randomly from Box-$IV$. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened,is equal to:

Let $A$ and $B$ be two events such that $P(B \mid A) = \frac{2}{5}$,$P(A \mid B) = \frac{1}{7}$ and $P(A \cap B) = \frac{1}{9}$. Consider:
$(S1) P(A' \cup B) = \frac{5}{6}$
$(S2) P(A' \cap B') = \frac{1}{18}$.
Then:

If $E_1$ and $E_2$ are two events of a sample space such that $P(E_1) = \frac{1}{4}$,$P(E_2 \mid E_1) = \frac{1}{2}$,and $P(E_1 \mid E_2) = \frac{1}{4}$,then $P(\bar{E}_1 \mid E_2) = $

Two dice $A$ and $B$ are rolled. If it is known that the number on $B$ is $5$, then the probability that the sum of the numbers on the two dice will be greater than $9$ is