Refer to question 74 above. If the probability that exactly two of the three balls were red, then the first ball being red, is

Let ${E_1} =$ Event that first ball being red
and ${E_2} =$ Event that exactly two of the three balls being red

$\therefore$ $P\left( {{E_1}} \right) = {P_R} \cdot {P_R} \cdot {P_R} + {P_R} \cdot {P_R} \cdot {P_{\bar R}} + {P_R} \cdot {P_{\bar R}} \cdot {P_R} + {P_R} \cdot {P_{\bar R}} \cdot {P_{\bar R}}$

$= \frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} + \frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} + \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6} + \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}$
$= \frac{{60 + 60 + 60 + 30}}{{336}} = \frac{{210}}{{336}}$

$P\left( {{E_1} \cap {E_2}} \right) = {P_R} \cdot {P_{\bar R}} \cdot {P_R} + {P_R} \cdot {P_R} \cdot {P_{\bar R}}$

$= \frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6} + \frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} = \frac{{120}}{{336}}$

$\therefore$ $P\left( {{E_2}/{E_1}} \right) = \frac{{P\left( {{E_1} \cap {E_2}} \right)}}{{P\left( {{E_1}} \right)}} = \frac{{120/336}}{{210/336}} = \frac{4}{7}$