In a city,$40\%$ of the people have brown hair,$25\%$ have brown eyes,and $15\%$ have both brown hair and brown eyes. If a person is selected at random from those having brown hair,what is the probability that they also have brown eyes?

$P(A / A \cap B) + P(B / A \cap B) =$

If the lengths of the sides of a triangle are decided by the three throws of a single fair die,then the probability that the triangle is of maximum area given that it is an isosceles triangle,is

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:

If $A$ and $B$ are any two events such that $P(A) = \frac{2}{5}$ and $P(A \cap B) = \frac{3}{20}$,then the conditional probability $P(A | A' \cup B')$,where $A'$ denotes the complement of $A$,is equal to:

If $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then $P(B/A) = $