Find $P(E | F)$ if a mother,father,and son line up at random for a family picture,where $E$ is the event that the son is on one end and $F$ is the event that the father is in the middle.

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

Consider the following statements.
Statement $(I)$: If $E$ and $F$ are two independent events,then $E^{\prime}$ and $F^{\prime}$ are also independent.
Statement $(II)$: Two mutually exclusive events with non-zero probabilities of occurrence cannot be independent.
Which of the following is correct?

If $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$,and $P(A \cup B) = \frac{7}{11}$,find $P(B | A)$.

In a random experiment,events $A$ and $B$ are such that $P(A) = \frac{1}{4}$,$P(A \mid B) = \frac{1}{2}$,and $P(B \mid A) = \frac{2}{3}$. Find $P(B)$.

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