Calculate the probability of drawing one white ball and one black ball in two draws from a bag containing 5 white and 3 black balls, without replacement.

There are 5 white balls and 3 black balls.
Total number of balls is $5 + 3 = 8$.

The probability of drawing a white ball first is $\frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{5}{8}$.
After drawing one white ball, there are 4 white balls and 3 black balls left, making a total of 7 balls.
The probability of drawing a black ball second is $\frac{\text{Number of black balls}}{\text{Total number of remaining balls}} = \frac{3}{7}$.
The probability of this sequence (White then Black) is:

The probability of drawing a black ball first is $\frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{3}{8}$.
After drawing one black ball, there are 5 white balls and 2 black balls left, making a total of 7 balls.
The probability of drawing a white ball second is $\frac{\text{Number of white balls}}{\text{Total number of remaining balls}} = \frac{5}{7}$.
The probability of this sequence (Black then White) is:

To find the probability of drawing one ball of each color, add the probabilities of the two possible sequences (White then Black, or Black then White).

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2.

The probability of drawing one ball of each color is $\frac{15}{28}$.