$A$ couple has two children. Find the probability that both children are females,if it is known that the elder child is a female.

For two events $A$ and $B$,$P(B) \neq 0$ and $P(A \mid B) = 1$,then . . . . . . .

Two events $E$ and $F$ are independent. If $P(E) = \frac{3}{5}$ and $P(F) = \frac{3}{10}$,then $P(E'/F) + P(F'/E) = \text{ . . . . . . }$

For a biased die,the probabilities for different faces to turn up are

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

The die is tossed and you are told that either face $4$ or face $5$ has turned up. The probability that it is face $4$ is

$E_1$ and $E_2$ are two independent events of a random experiment such that $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$. Match the items of List-$I$ with the items of List-$II$.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to