A dice is thrown. The probability that the first time 1 occurs at the even throw is

Options

A. `1//6`
B. `5//11`
C. `6//11`
D. `5//36`

Answer

B

Explanation

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) `