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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)/(216)`

C

`(5)/(36)`

D

`(25)/(36)`

Text Solution

Verified by Experts

The correct Answer is:
D
`P({(Xge6)//(Xgt3)}=(P{Xgt3)//(Xge6)}*P(Xge6))/(P(Xgt3))`
`=(1*[((5)/(6))^(5)*((1)/(6))+((5)/(6))^(6)*(1)/(6)+...oo])/([((5)/(6))^(3)*(1)/(6)+((5)/(6))^(4)*(1)/(6)+...oo])=(25)/(36)`
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