A container holds $3$ green marbles and $5$ yellow marbles. One marble is randomly drawn and discarded. Then a second marble is drawn. Find the probability that the second marble is yellow, given that the first marble was green.
Total number of marbles $= 8$
The first marble can be selected in $8$ ways; the second marble can be selected in $7$ ways.
Let $A$ be the event of drawing the first marble (green).
Let $B$ be the event of drawing the second marble (yellow).
$\therefore \;$ P(selecting a green marble) $= P \left(A\right) = \dfrac{3}{8}$
$1$ yellow marble can be selected from $5$ yellow marbles in $5$ ways
$\therefore \;$ P(selecting a yellow marble) $= \dfrac{5}{7}$
$\left(A \cap B\right) = \;$ event of selecting a green \textbf{AND} a yellow marble
$\therefore \;$ $P \left(A \cap B\right) = \dfrac{3}{8} \times \dfrac{5}{7}$
Now, P(selecting a yellow marble given that the first marble is green)
$= P \left(B|A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{\dfrac{3}{8} \times \dfrac{5}{7}}{\dfrac{3}{8}} = \dfrac{5}{7}$