A fair die is thrown 20 times. The probability that on the 10th throw, the fourth six appears is
a.`(.^20 C_(10)xx5^6)/6^(20)`
b. `(120xx5^7)/6^(10)`
c. `(84xx5^6)/6^(10)`
d. none of these

Options

A. `""^(20)C_(10)xx5//""^(6)//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 nonsixes, and a six in the `10^(th)` throw, and thereafter it does not matter whatever face appears. So, the required probability is
`""^(9)C_(3)((1)/(6))^(3)xx((5)/(6))^(6)xx1/6xx1xx1xx1xx...xx1=(84xx5^(6))/(6^(10))`