If $$A$$ and $$B$$ are two events such that $$P(B) \neq 0$$ and $$P(B) \neq 1$$, then $$P(\bar{A} \mid \bar{B})$$ is

Two brothers $$X$$ and $$Y$$ appeared for an exam. Let $$A$$ be the event that $$X$$ has passed the exam and $$B$$ is the event that $$Y$$ has passed. The probability of $$A$$ is $$\frac{1}{7}$$ and of $$B$$ is $$\frac{2}{9}$$. Then, the probability that both of them pass the exam is

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red, is

In a Binomial distribution, if '$$n$$' is the number of trials and the mean and variance are 4 and 3 respectively, then $$2^{32} p\left(X=\frac{n}{2}\right)=$$