Answer: Option B
:
B
The number of ways to arrange 7 white an 3 black balls in a row
$\frac{10!}{7!.3!}=\frac{\mathrm{10.9.8}}{\mathrm{1.2.3}}=120$
Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8.
Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways.
${=}^8{C}_3=\frac{8\times 7\times 6}{1\times 2\times 3}=56$
So required probability = $\frac{56}{120}=\frac{7}{15}$