A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses required.
The probability that `X=3` equals

Options

A. `125//216`
B. `25//36`
C. `5//36`
D. `25//216`

Answer

D

Explanation

For `Xge6,` the probability is
`(5^(5))/(6^(6))+(5^(5))/(6^(7))+...oo=(5^(5))/(6^(6))((1)/(1-5//6))=((5)/(6))^(5)`
For `Xgt3,`
`(5^(3))/(6^(4))+(5^(4))/(6^(5))+(5^(5))/(6^(6))+...oo=((5)/(6))^(3)`
Hence, the conditional probability is
`((5//6)^(6))/((5//6)^(3))=25/36`