A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is (1) $\frac { 5 } { 6 }$ (2) $\frac { 1 } { 6 }$ (3) $\frac { 5 } { 11 }$ (4) $\frac { 6 } { 11 }$
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is\\
(1) $\frac { 5 } { 6 }$\\
(2) $\frac { 1 } { 6 }$\\
(3) $\frac { 5 } { 11 }$\\
(4) $\frac { 6 } { 11 }$