A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears, is

$\frac{84×5^6}{6^{10}}$

In the first 9 throws we should have three sixes and six non-sixes; and a six in the 10th throw, and thereafter whatever face appears no matter.

∴ Required probability

$={^9C}_3 \left(\frac{1}{6}\right)^3×\left(\frac{5}{6}\right)^6×\frac{1}{6}×1×1×1×1×1×1×1×1×1×1=\frac{84×5^6}{6^{10}}$