you need to sum up the probabilities of all the cases of the 3 ball transfer

$P[\text{2 balls from bag 2 were different color}] = \sum \limits_{\vec{t}}P[\text{2 balls from bag 2 were different color} | \vec{t}]P[\vec{t}]\\

\text{where }\vec{t} \text{ is the combination of balls that were transferred from bag 1}$

for example for $\vec{t}=(3,0) =\text{ (3 white, 0 black)}$

$P[\text{2 different colors from bag 2}|(3,0)]P[(3,0)] = \dfrac{\dbinom{5}{1}\dbinom{9}{1}}{\dbinom{14}{2}} \cdot \dfrac{\dbinom{6}{0}\dbinom{5}{3}}{\dbinom{11}{3}}$

repeat this for all the combinations of the 3 black and white balls and sum them all up.