$A$ family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

If $P(A / B) = \frac{3}{10}$, $P(B / A) = \frac{4}{5}$ and $P(A \cup B) = K P(B)$, then $\frac{1}{K} =$

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:
$A) P(X^{\prime} \mid Y)=\frac{1}{2}$
$B) P(X \cap Y)=\frac{1}{5}$
$C) P(X \cup Y)=\frac{2}{5}$
$D) P(Y)=\frac{4}{15}$

$A$ 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) = $

$A$ and $B$ are mutually exclusive events of a random experiment and $P(B) \neq 1$,then $P(A \mid B^c) =$

$A, B$ are the events in a random experiment. If $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}, P(A \cap B)=\frac{1}{4}$, then $P\left(\frac{A^{c}}{B^{c}}\right)+P\left(\frac{A}{B}\right)=$