# A bag contains 5 red and 3 blue balls.If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

$$P(R\;B\;B)\;=\Large\frac{5}{8}\;x\;\frac{3}{7}x\frac{2}{6}\;=\frac{5}{56}$$
$$P(B\;R\;B)\;=\;\Large\frac{3}{8}\;x\;\frac{5}{7}\;x\;\frac{2}{6}\;=\;\frac{5}{56}$$
$$P(B\;B\;R)\;=\;\Large\frac{3}{8}\;x\;\frac{2}{7}\;x\;\frac{5}{7}\;=\;\frac{5}{56}$$
$$(Exactly 1 Red)\;=\;\Large\frac{5}{56}\;+\;\frac{5}{56}\;+\;\frac{5}{56}$$
=$$\Large\frac{15}{56}$$
'C' option is correct

edited Mar 15, 2013