A fair die is tossed eight times. The probability that a third six is observed in eight throw is a. `(\ ^7C_2xx5^5)/(6^7)` b. `(\ ^7C_2xx5^5)/(6^8)` c. `(\ ^7C_2xx5^5)/(6^6)` d. none of these

Explanation

On the eighth throw third dice must appear as six, which implies that on first seven there must be two sixes.

Probability of getting a `6` is `1/6` and probability of not getting a six is `5/6`

We must get two six in seven throws,

`therefore` the Probability is `\ ^7 C_2 (1/6)^2 (5/6)^5`

Now Probability that the eight throw is six is `1/6`

`therefore` The final required probability is `\ ^7 C_2 (1/2)^2 (5/6)^5 1/6`

`=( \ ^7 C_2 * 5^5)/(6)^8`

`therefore` the correct answer is option `B`.