Let's break down this problem step-by-step:
1. First, let's calculate the probability of drawing a white ball first and then a black ball:
   - The probability of drawing a white ball first is 7/15 (since there are 7 white balls out of 15 total balls).
   - After drawing a white ball, there are now 14 balls left in the bag, including 6 black balls. So, the probability of drawing a black ball next is 6/14.
   - Therefore, the probability of drawing a white ball first and then a black ball is (7/15) * (6/14).
2. Next, let's calculate the probability of drawing a black ball first and then a white ball:
   - The probability of drawing a black ball first is 6/15 (since there are 6 black balls out of 15 total balls).
   - After drawing a black ball, there are now 14 balls left in the bag, including 7 white balls. So, the probability of drawing a white ball next is 7/14.
   - Therefore, the probability of drawing a black ball first and then a white ball is (6/15) * (7/14).
3. Since we want to find the probability of either drawing a white ball first and then a black ball or drawing a black ball first and then a white ball, we can add these two probabilities together:
   - P(one white ball and one black ball) = [(7/15) * (6/14)] + [(6/15) * (7/14)]

The final answer is that the probability of drawing one white ball and one black ball is [(7/15) * (6/14)] + [(6/15) * (7/14)], which simplifies to 7/15 or approximately 46.67%.
A bag contains 7 white balls, 6 black balls, and 2 yellow balls. You draw two balls from the bag without replacement. What is the probability that you draw one white ball and one black ball?