Total number of ways in which all the balls can be placed is
$ \large\frac{10!}{7!\: 3!} $$= 120$
If no two black balls are placed adjacently, the possible number of ways is
$ ^8C_3 = 56$
Hence the required chance is = $ \large\frac{56}{120}$
$ = \large\frac{7}{15}$
Ans : (B)