Question

A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that atleast one of the three marbles drawn be black, if the first marble is red?

Answer 1

Let $R$ and $B$ represent red marbles and black marbles respectively.

For atleast one of the three marbles drawn be black, if the first marble is red, the following three conditions will occur:
$\left(1\right)$ Second marble is black and third marble is red i.e., $RBR$
$\left(2\right)$ Second and third marble is black i.e., $RBB$
$\left(3\right)$ Second marble is red and third marble is black i.e., $RRB$

The required probability is given by,
$P\text{(atleast one black | first marble is red)}$
$=P\left(RBR\right)+P\left(RBB\right)+P\left(RRB\right)$
$=(\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6})+(\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6})+(\frac{5}{8}\times \frac{4}{7}\times \frac{3}{6})$
$=\frac{25}{56}$