A person throws a die until the side with two spots appears. The probability of throwing the die exactly n times before throwing a "2" is $\left( \frac { 5 } { 6 } \right) ^ { n }$ $\left( \frac { 1 } { 6 } \right)$ , n ≥ 0. What is the probability that the number of throws before throwing a "2" is even? Enter just a reduced fraction.

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The Answer of A person throws a die until the side with two spots appears. The probability of throwing the die exactly n times before throwing a "2" is $\left( \frac { 5 } { 6 } \right) ^ { n }$ $\left( \frac { 1 } { 6 } \right)$ , n ≥ 0. What is the probability that the number of throws before throwing a "2" is even? Enter just a reduced fraction.

A Person Throws a Die Until the Side with Two (56)n\left( \frac { 5 } { 6 } \right) ^ { n }(65​)n

A Person Throws a Die Until the Side with Two $\left( \frac { 5 } { 6 } \right) ^ { n }$