$A$ coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which the first throw is a head,then $P(E|F) = $

$A$ couple has two children. If at least one of them is a boy,then the probability that the other is also a boy is:

If $A$ and $B$ are events such that $P(A) > 0$ and $P(B) \neq 1$,then $P(A \mid B^{\prime}) = $ . . . . . . .

Find $P(E | F)$ when two coins are tossed once,where $E$ is the event that no tail appears and $F$ is the event that no head appears.

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

Let $X$ and $Y$ be two events such that $P(X)=\frac{1}{3}$,$P(X \mid Y)=\frac{1}{2}$ and $P(Y \mid X)=\frac{2}{5}$. Then:
$A) P(X^{\prime} \mid Y)=\frac{1}{2}$
$B) P(X \cap Y)=\frac{1}{5}$
$C) P(X \cup Y)=\frac{2}{5}$
$D) P(Y)=\frac{4}{15}$