Well..divided into 6/6 first to define a heavier group is certainly right. Randomly select 3/3 will result another heavier group.
' f& I5 Z( \1 ~) g6 I0 ]. \$ AThe problem is we don't know the weight of each individual ball, so it will not be accurate to figure out which ball is irregular by only weight them once.
& f( I2 s% V" t4 }: O. Ftvb now,tvbnow,bttvb |