Two events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{4}$ and $P(B|A)=\frac{1}{2}$. Consider the following statements:
$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$
$(II) A$ and $B$ are mutually exclusive
$(III) P(A|B)+P(A|\bar{B})=1$
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) = $

An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is $0.9$ and that of the second unit is $0.8$. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and the second unit is functioning is $p$,then $98p$ is equal to ..... .

$A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B \mid A)$ if: $(i)$ $A \subset B$ (ii) $A \cap B = \phi$.

If two events $A$ and $B$ are such that $P(\overline{A}) = 0.3$,$P(B) = 0.4$,and $P(A \cap \overline{B}) = 0.5$,then $P(B | (A \cup \overline{B})) = $

If $A$ and $B$ are two events of a sample space $S$ such that $P(A)=0.2$,$P(B)=0.6$ and $P(A \mid B)=0.5$,then $P(A^{\prime} \mid B) = $