$A$ bag contains $3$ white and $2$ red balls. If the first ball drawn is not replaced,what is the probability that the second ball is red?

When a die is thrown twice,what is the probability that the sum of the numbers is $6$,given that at least one of the numbers is $4$?

It is given that $A$ and $B$ are such that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$. Then $P(B) = $?

Suppose that Box-$I$ contains $8$ red,$3$ blue and $5$ green balls,Box-$II$ contains $24$ red,$9$ blue and $15$ green balls,Box-$III$ contains $1$ blue,$12$ green and $3$ yellow balls,and Box-$IV$ contains $10$ green,$16$ orange and $6$ white balls. $A$ ball is chosen randomly from Box-$I$; call this ball $b$. If $b$ is red,then a ball is chosen randomly from Box-$II$; if $b$ is blue,then a ball is chosen randomly from Box-$III$; and if $b$ is green,then a ball is chosen randomly from Box-$IV$. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened,is equal to:

Two events $A$ and $B$ are such that $P(A)=\frac{1}{4}$, $P(A|B)=\frac{1}{4}$ and $P(B|A)=\frac{1}{2}$. Consider the following statements:
$(I) P(\bar{A}|\bar{B})=\frac{3}{4}$
$(II) A$ and $B$ are mutually exclusive
$(III) P(A|B)+P(A|\bar{B})=1$
Then,

For two events $A$ and $B$,if $P(A) = P(A|B) = \frac{1}{4}$ and $P(B|A) = \frac{1}{2}$,then: