A fair die is thrown 20 times. The probability that on the 10th throw, the fourth six appears is

Options

A. `(.^20C_(10)xx5^5)/(6^20)`
B. `(120xx5^7)/(6^10)`
C. `(84xx5^6)/(6^10)`
D. none of these

Answer

C

Explanation

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.
`therefore` Required Probability
`=.^9C_(3)((1)/(6))^3xx((5)/(6))^6xx(1)/(6)xx^(1xx1xx1....xx1)10 -times =(84xx5^6)/(6^10)`.