Correct Answer - B
Let E be the event of getting 1 on a die .
`implies P(E ) =(1)/(6) and P(bar (E )) =(5)/(6)`
`:.` P (first time 1 occurs at the even throw ) `=t_(2) or t_(4) or t_(6) or t(8) …… and ` so on
`={P(bar( E)) P(E ) }+{P(bar( E))P(bar(E )) P(bar(E )) }+ ... oo `
`((5)/(6))((1)/(6))+((5)/(3))^(3)((1)/(6))+((5)/(6))^(5)((1)/(6))+oo=((5)/(36))/(1-(25)/(36))=(5)/(11) `