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

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

$A$ biased die is tossed and the respective probabilities for various faces to turn up are given below:

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

If an even face has turned up,then the probability that it is face $2$ or face $4$ is:

Let $A$ and $B$ be two events such that $P(A) = \frac{5}{11}$, $P(B) = \frac{2}{11}$, and $P(A \cup B) = \frac{3}{11}$, then $P(A'|B')$ is . . . . . . .

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) = $

Find $P(E | F)$ for a coin tossed three times,where $E$ is the event 'at least two heads' and $F$ is the event 'at most two heads'.