Let $a$ and $b$ be given positive real numbers, with $a<b.$ If two points are selected at random from a straight line segment of length $b,$ what is the probability that the distance between them is at least $a?$
1. **Define the problem geometrically**: We need to find the probability that the distance between two randomly chosen points on a line segment of length \( b \) is at least \( a \). We can represent this problem in a coordinate system where the points are \((x, y)\) with \( 0 \leq x, y \leq b \).

2. **Identify the acceptable region**: The condition \(|x - y| \geq a\) can be split into two cases:
   - \( x - y \geq a \)
   - \( y - x \geq a \)

3. **Graph the inequalities**:
   - For \( x - y \geq a \), this translates to \( y \leq x - a \).
   - For \( y - x \geq a \), this translates to \( y \geq x + a \).

4. **Visualize the regions**: Plot these inequalities on the coordinate plane:
   - The line \( y = x - a \) starts at \((a, 0)\) and ends at \((b, b-a)\).
   - The line \( y = x + a \) starts at \((0, a)\) and ends at \((b-a, b)\).

5. **Determine the area of the acceptable region**: The acceptable region consists of two right-angled triangles:
   - Triangle 1: Vertices at \((0, a)\), \((0, b)\), and \((b-a, b)\).
   - Triangle 2: Vertices at \((a, 0)\), \((b, 0)\), and \((b, b-a)\).

6. **Calculate the area of each triangle**:
   - The base and height of each triangle are both \( b - a \).
   - The area of one triangle is \(\frac{1}{2} \times (b - a) \times (b - a) = \frac{1}{2} (b - a)^2\).

7. **Sum the areas of the two triangles**:
   \[
   \text{Total area of acceptable region} = 2 \times \frac{1}{2} (b - a)^2 = (b - a)^2
   \]

8. **Calculate the total area of the square**:
   \[
   \text{Total area of the square} = b^2
   \]

9. **Find the probability**:
   \[
   \text{Probability} = \frac{\text{Area of acceptable region}}{\text{Total area of the square}} = \frac{(b - a)^2}{b^2}
   \]

Conclusion:
\[
\boxed{\frac{(b - a)^2}{b^2}}
\]