A bag contains 3 red and 3 green balls and a person draws out 3 at random. Then, he drops 3 blue balls into the bag and again draws out 3 at random. The chance that the 3 later balls being all of different colours, is

$\text{ There are 3 red and }3\text{ green balls in a bag}$

$\mathrm{and} a \mathrm{person} \mathrm{draws} 3\text{ balls. The possibilities are: }$

$\text{2Red, }1\text{ Green then }1R+2G+3\mathrm{Blue} \mathrm{balls} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{bag}$

$1\text{ Red, }2\text{ Green then }2R+1G+3Blue balls are in the bag$

$\text{ Probability of withdrawing }2R,IG=\frac{3C_2·3C_1}{6C_3}=\frac{9}{20}$

$$\begin{gathered} \text{ Probabilityof withdrawing }\mathrm{IR},\mathrm{IG},\mathrm{IB} \\ =\frac{2c_1\times 1C_1\times 3C_1}{6c_3}=\frac{6}{20} \end{gathered}$$

Two cases are arise, 1R, 2G, 3B, 2R, 1G, 3B Required probability
$=2\left(\frac{{ }^3{\mathrm{C}}_2{\cdot }^3{\mathrm{C}}_1}{{ }^6{\mathrm{C}}_3}\cdot \frac{{ }^1{\mathrm{C}}_1{\cdot }^2{\mathrm{C}}_1{\cdot }^3{\mathrm{C}}_1}{{ }^6{\mathrm{C}}_3}\right)$=$\frac{27}{100}$