Two numbers are randomly selected from interval $I = [0, 1]$. Given $\alpha \in I$, what is the probability that the smaller of the two numbers does not exceed $\alpha$?

Is the answer $(100 \alpha)$%, it just seems too easy. :|
1. Let \( X \) and \( Y \) be two numbers randomly selected from the interval \( I = [0, 1] \).
2. We need to find the probability that the smaller of the two numbers does not exceed \( \alpha \), where \( \alpha \in I \).

3. Define the event \( A \) as the event that the smaller of the two numbers does not exceed \( \alpha \). Mathematically, this can be written as:
   \[
   A = \{ \min(X, Y) \leq \alpha \}
   \]

4. To find the probability of \( A \), we first consider the complement event \( A^c \), which is the event that the smaller of the two numbers exceeds \( \alpha \). This can be written as:
   \[
   A^c = \{ \min(X, Y) > \alpha \}
   \]

5. For the event \( \min(X, Y) > \alpha \) to occur, both \( X \) and \( Y \) must be greater than \( \alpha \). The probability of this happening is:
   \[
   P(A^c) = P(X > \alpha \text{ and } Y > \alpha) = P(X > \alpha) \cdot P(Y > \alpha)
   \]
   Since \( X \) and \( Y \) are independently and uniformly distributed over \([0, 1]\), we have:
   \[
   P(X > \alpha) = 1 - \alpha \quad \text{and} \quad P(Y > \alpha) = 1 - \alpha
   \]
   Therefore,
   \[
   P(A^c) = (1 - \alpha)^2
   \]

6. The probability of the event \( A \) is the complement of \( P(A^c) \):
   \[
   P(A) = 1 - P(A^c) = 1 - (1 - \alpha)^2
   \]

7. Simplify the expression:
   \[
   P(A) = 1 - (1 - 2\alpha + \alpha^2) = 2\alpha - \alpha^2
   \]

Conclusion:
\[
\boxed{2\alpha - \alpha^2}
\]