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A fair die is tossed repeatedly until a ...

A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses rerquired.
The conditional probability that `Xge6` given `Xgt3` equals

A

`125//216`

B

`25//36`

C

`5//36`

D

`25//216`

Text Solution

Verified by Experts

The correct Answer is:
D
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`
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