If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3}, P(B)=\frac{1}{2}$$ and $$P(A \cap B)=\frac{1}{6}$$, then $$P\left(A^{\prime} / B\right)$$ is

Events $$E_1$$ and $$E_2$$ from a partition of the sample space $$S$$. $$A$$ is any event such that $$P\left(E_1\right)=P\left(\dot{E}_2\right)=\frac{1}{2}, P\left(E_2 / A\right)=\frac{1}{2}$$ and $$P\left(A / E_2\right)=\frac{2}{3}$$, then $$P\left(E_1 / A\right)$$ is

The probability of solving a problem by three persons $$A, B$$ and $$C$$ independently is $$\frac{1}{2}, \frac{1}{4}$$ and $$\frac{1}{3}$$ respectively. Then the probability of the problem is solved by any two of them is

If $$A, B, C$$ are three mutually exclusive and exhaustive events of an experiment such that $$P(A)=2 P(B)=3 P(C)$$, then $$P(B)$$ is equal to