It is given that the events $$A$$ and $$B$$ are such that
$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is :

The solution of the differential equation

$${{dy} \over {dx}} = {{x + y} \over x}$$ satisfying the condition $$y(1)=1$$ is :

The area of the plane region bounded by the curves $$x + 2{y^2} = 0$$ and $$\,x + 3{y^2} = 1$$ is equal to :

The value of $$\sqrt 2 \int {{{\sin xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$ is