1. **Problem statement:** We have a biased coin that comes up heads with probability $p$. The coin is flipped repeatedly. We want to find the probability that the first tail appears on an even-numbered flip. 2. To get the first tail on the $2n$-th flip (where $n=1,2,3,\dots$), the first $2n-1$ flips must be heads and the $2n$-th flip must be a tail. 3. The probability of heads is $p$, so the probability of $2n-1$ heads in a row is $p^{2n-1}$. The probability of tail is $1-p$. 4. Hence, the probability that the first tail appears on the $2n$-th flip is: $$p^{2n-1}(1-p)$$ 5. We want the total probability of first tail on any even flip, which is the sum over all $n \geq 1$: $$\sum_{n=1}^\infty p^{2n-1}(1-p) = (1-p)\sum_{n=1}^\infty p^{2n-1}$$ 6. The sum $\sum_{n=1}^\infty p^{2n-1}$ is a geometric series with first term $p$ and ratio $p^2$: $$\sum_{n=1}^\infty p^{2n-1} = p + p^3 + p^5 + \cdots = \frac{p}{1-p^2}$$ 7. Substitute back to get: $$\text{Probability} = (1-p) \cdot \frac{p}{1-p^2} = \frac{(1-p)p}{(1-p)(1+p)}$$ 8. Simplify numerator and denominator: $$\frac{p}{1+p}$$ 9. **Final answer:** The probability that the first tail appears on an even-numbered flip is: $$\boxed{\frac{p}{1+p}}$$ The learner can interpret this as the bias $p$ playing directly into the chance of getting the first tail on an even flip, with the denominator accounting for the infinite sum of outcomes.