Certainly! Let's break down the process step by step:

1. **Starting Point:**
   - Red Balls (RB): 50
   - Blue Balls (BB): 50
   - White Balls (WB): 0
   - Total Balls: 100

2. **Drawing Process:**
   - Whenever you draw a blue ball, two white balls are added to the container.

Let's proceed through the drawing process:

- **First Draw:**
  - If you draw a blue ball: BB = 49, WB = 2, Total Balls = 101
  - If you draw a red ball: RB = 49, Total Balls = 99

If we follow the pattern, every time you draw a blue ball, the total number of balls in the container increases by 1 (since you remove one blue ball but add two white balls), and every time you draw a red ball, the total number of balls in the container decreases by 1.

We need to find a scenario where you've drawn 100 balls, so the total number of draws will be 100. Let's denote the number of blue balls drawn as \( x \) and the number of red balls drawn as \( y \).

From the drawing process, we have two equations:
\[ x + y = 100 \] (since you've drawn 100 balls)
\[ 50 - x + 2x = 50 + x \] (since each blue ball drawn adds 1 to the total count of balls in the container)

Now, solving these equations will give us the values of \( x \) and \( y \), the number of blue and red balls drawn respectively.

Solving the second equation gives:
\[ 50 - x + 2x = 50 + x \]
\[ x = 0 \]

Substituting \( x = 0 \) into the first equation gives:
\[ 0 + y = 100 \]
\[ y = 100 \]

So in this scenario, you've drawn all 50 red balls and none of the blue balls.

- **Ending Scenario:**
  - Red Balls (RB): 0 (all have been drawn)
  - Blue Balls (BB): 50 (none have been drawn)
  - White Balls (WB): 0 (since no blue balls were drawn, no white balls were added)
  - Total Balls: 50

This is one possible scenario. However, there could be other scenarios based on the order in which balls are drawn. But in any case, the same logic can be applied to find the number of each color of ball left in the container.