Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, is

The number of ways of arranging 10 balls in a row is 10! 

n(S)=10!

E= The number of ways of arranging Seven white balls and three black balls are such that no two black balls are placed adjacently

Seven white balls can be arranged in 7! ways .There are eight places in between the white balls to arrange the 3 black balls.They can be arranged in ${ }^8{p}_3$ ways

n(E)= ${ }^8{p}_3\times 7!$

 Probability  $P\left(E\right)=\frac{{ }^8{P}_3\cdot 7!}{10!}=\frac{8\cdot 7\cdot 6}{10\cdot 9\cdot 8}=\frac{7}{15}$