Zhegan has 7 gray balls, 4 white balls, and 3 yellow balls in the bag. The question asks for the minimum number of balls that need to be removed to ensure that at least one ball of each color (gray, white, and yellow) is removed.

Strategy:

To minimize the number of balls removed while ensuring that one ball of each color is selected, Zhegan could potentially remove all balls of two colors before getting a ball of the third color.

Step-by-step Solution:

1. Worst case scenario: Suppose Zhegan removes all the gray and white balls before drawing any yellow balls. In that case, he would need to remove:

   • All 7 gray balls, and
   • All 4 white balls, which totals to $7 + 4 = 11$ balls.

2. At least one of the yellow balls: After removing 11 balls, the next ball must be yellow, as there are only 3 yellow balls left in the bag, and no gray or white balls remain.

Thus, Zhegan needs to remove 11 balls to ensure that at least one of each color is removed.

Final Answer:

The minimum number of balls that must be removed to ensure that at least one ball of each color is removed is 12 (11 balls to potentially remove all of one or two colors, plus 1 more to ensure the third color is drawn).

Would you like further details on this process or additional related problems?

Here are 5 questions to expand on this:

1. How would the strategy change if there were more colors of balls in the bag?
2. What is the probability of drawing at least one yellow ball in the first three draws?
3. What is the probability of drawing all gray balls in the first 5 draws?
4. How does the removal strategy change if you want to minimize the number of balls drawn to get only one specific color?
5. How would the problem change if the balls were drawn with replacement?

Tip: When dealing with probability or selection problems, it's useful to first consider the worst-case scenario to ensure the desired outcome.