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

${ }^8 C_3 \frac{5^5}{6^8}$ $\frac{{ }^7 C_2 . 5^5}{6^8}$ $\frac{{ }^7 C_2 . 5^5}{6^7}$ none of these

$\frac{{ }^7 C_2 . 5^5}{6^8}$

Third six occurs on 8th trial. It means that in first 7 trials we must exactly 2 sixes and 8th trial must result in a six. ⇒ Required probability = ${ }^7 C_2 . (1 / 6)^2 . (5 / 6)^5 . (1 / 6)=\frac{{ }^7 C_2 . 5^5}{6^8}$