Total Possible cases of balls in the bag could be:
(2W,6B),(3W,5B),(4W,4B),(5W,3B),(6W,2B)
The favourable case is to choose 2W and 2B from the bag having (4W,4B)
So, the required probability is given by
P(E)=C28C22×C26+C28C23×C25+C28C24×C24+C28C25×C23+C28C26×C22C28C24×C24
⇒P(E)=15+3×10+36+30+156×6
⇒P(E)=12636
⇒P(E)=72