A bag contains 3 red and 3 green balls and a person draws out 3 at random. He then 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 colors is x%. Then the value of x is______

(27) 
For drawing 3 balls of distinct colour in second draw, he can not draw 3 balls of same colour in first draw.
Case 1: first draw: 1 red, 2 green out of 3 red and 3 green
P(1R, 2G in first draw)  $=\frac{3_{C_1}\times 3_{C_2}}{6_{C_3}}=9/20$
second draw: 1 red, 1 green, 1 blue out of 2 red, 1 green and 3 blue balls.
P(1R, 1G, 1B in second draw)  $=\frac{2_{C_1}\times 1_{C_1}\times 3_{C_1}}{6_{C_3}}=6/20$
So, the P(drawing 3 distinct balls in second draw) =  $\frac{9}{20}\times \frac{6}{20}$
second draw: 1 red, 1 green, 1 blue out of 1 red, 2 green and 3 blue balls.
P(1R, 1G, 1B in second draw)  $=\frac{1_{C_1}\times 2_{C_1}\times 3_{C_1}}{6_{C_3}}=6/20$
So, the P(drawing 3 distinct balls in second draw) =  $\frac{9}{20}\times \frac{6}{20}$
therefore, total probability =  $2\times \frac{9}{20}\times \frac{6}{20}=27\%$