$A$ box contains $3$ white and $2$ red balls. One ball is drawn at random,and then a second ball is drawn without replacement. What is the probability that the second ball is red?

If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$,then $P(A^{\prime} \cap B^{\prime})$ is

One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

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

If $A$ and $B$ are two events such that $P(A) = \frac{3}{8}$,$P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$,then $P(A|B) = $