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

$\frac{{^{20}C}_{10}×5^6}{6^{20}}$ $\frac{120×5^7}{6^{10}}$ $\frac{84×5^6}{6^{10}}$ None of these

$\frac{84×5^6}{6^{10}}$

Let success : getting '6' ; failure : no '6' P(required) = ${^9C}_3(\frac{1}{6})^3(\frac{5}{6})^6.\frac{1}{6}=\frac{84×5^6}{6^{10}}$