first case
- probability of white ball in first replacement P(w)=4/10
- probability of black ball in first replacement P(b)=6/10
- probability of white ball in second replacement P'(w)=3/9
- probability of black ball in second replacement P'(b)=5/9
Explanation:
now for ,
the probability that the balls removed were the same colour =P(w)×P'(w)+P'(b)×P(b)
=7/15