7 white balls and 3 black balls are placed in a row at random. The probability that no two black balls are adjacent is

$\frac{7}{15}$

7 white and 3 black balls can be arranged in a row in $\frac{10!}{3!7!}$ ways.

When 7 white balls are arranged in a row there are 8 place in between. In these 8 places 3 black balls can be arranged in ${^8C}_3$ ways.

∴ Required probability =$\frac{^8C_3}{\frac{10!}{3!7!}}=\frac{7}{15}$