A fair die is tossed eight times. Probability that on the eighth throw a third six is observed is

$\frac{{^7C}_2.5^5}{6^8}$

Third six occurs on 8th trial. It means that in first 7 trials we must have exactly 2 sixes and 8th trial must result in a six.

⇒ Required probability = ${^7C}_2 . (1/6)^2. (5/6)^5. (1/6) =\frac{{^7C}_2.5^5}{6^8}$