Let's break down the problem step by step:

Given:

• A bag contains xxx red balls and yyy blue balls.
  
  

• If zzz red balls are removed, the probability of drawing a red ball becomes 12\frac{1}{2}21​.
  
  

• The ratio of red balls to blue balls is x:y=3:2x : y = 3 : 2x:y=3:2, meaning xy=32\frac{x}{y} = \frac{3}{2}yx​=23​, so x=32yx = \frac{3}{2} yx=23​y.
  
  

• z=y2z = \frac{y}{2}z=2y​, meaning zzz is half the number of blue balls.
  
  

Step 1: Expression for the probability of drawing a red ball after removal

The total number of balls initially is x+yx + yx+y, and after removing zzz red balls, the total number of balls becomes:

Total number of balls after removal=(x+y)−z\text{Total number of balls after removal} = (x + y) - zTotal number of balls after removal=(x+y)−z

The number of red balls remaining after removal is:

Remaining red balls=x−z\text{Remaining red balls} = x - zRemaining red balls=x−z

The probability of drawing a red ball after removing zzz red balls is:

P(red after removal)=x−zx+y−zP(\text{red after removal}) = \frac{x - z}{x + y - z}P(red after removal)=x+y−zx−z​

According to the problem, this probability is 12\frac{1}{2}21​, so:

x−zx+y−z=12\frac{x - z}{x + y - z} = \frac{1}{2}x+y−zx−z​=21​

Step 2: Substitute z=y2z = \frac{y}{2}z=2y​ into the equation

Substitute z=y2z = \frac{y}{2}z=2y​ into the equation:

x−y2x+y−y2=12\frac{x - \frac{y}{2}}{x + y - \frac{y}{2}} = \frac{1}{2}x+y−2y​x−2y​​=21​

Simplifying the denominator:

x+y−y2=x+y2x + y - \frac{y}{2} = x + \frac{y}{2}x+y−2y​=x+2y​

So the equation becomes:

x−y2x+y2=12\frac{x - \frac{y}{2}}{x + \frac{y}{2}} = \frac{1}{2}x+2y​x−2y​​=21​

Step 3: Cross-multiply to simplify

Now, cross-multiply to eliminate the fraction:

2(x−y2)=x+y22 \left( x - \frac{y}{2} \right) = x + \frac{y}{2}2(x−2y​)=x+2y​

Expanding both sides:

2x−y=x+y22x - y = x + \frac{y}{2}2x−y=x+2y​

Step 4: Solve for xxx and yyy

Now, move all terms involving xxx to one side and all terms involving yyy to the other side:

2x−x=y2+y2x - x = \frac{y}{2} + y2x−x=2y​+y

Simplifying:

x=y2+yx = \frac{y}{2} + yx=2y​+y x=3y2x = \frac{3y}{2}x=23y​

Step 5: Use the ratio x=32yx = \frac{3}{2} yx=23​y

We already know that x=32yx = \frac{3}{2} yx=23​y, so this matches the equation derived above. Therefore, the equation is satisfied, and we don’t need to do any further manipulation.

Step 6: Conclusion

Thus, the original number of red balls is x=32yx = \frac{3}{2} yx=23​y. Since we do not have additional information about the exact number of blue balls yyy, this is the general relationship between xxx and yyy. To find the actual number of red balls, we would need the value of yyy.

Let’s carefully solve this step by step.

Step 1: Express what we know

• Initially, the bag contains xx red balls and yy blue balls.

• The ratio x:y=3:2x : y = 3 : 2, so:

x=32yx = \frac{3}{2}y

• After removing z=y2z = \frac{y}{2} red balls, the number of remaining red balls becomes x−z=x−y2x - z = x - \frac{y}{2}.

• The total number of balls after removing zz red balls is:

(x−z)+y=(x−y2)+y=x+y2(x - z) + y = (x - \frac{y}{2}) + y = x + \frac{y}{2}

• The probability of drawing a red ball is then:

Number of Red BallsTotal Balls=x−y2x+y2=12\frac{\text{Number of Red Balls}}{\text{Total Balls}} = \frac{x - \frac{y}{2}}{x + \frac{y}{2}} = \frac{1}{2}

Step 2: Solve the probability equation

Set up the equation:

x−y2x+y2=12\frac{x - \frac{y}{2}}{x + \frac{y}{2}} = \frac{1}{2}

Cross-multiply to eliminate the fraction:

2(x−y2)=x+y22(x - \frac{y}{2}) = x + \frac{y}{2}

Expand both sides:

2x−y=x+y22x - y = x + \frac{y}{2}

Simplify:

2x−x=y+y22x - x = y + \frac{y}{2}

x=2y+y2x = \frac{2y + y}{2}

x=3y2x = \frac{3y}{2}

Step 3: Use the ratio x:y=3:2x : y = 3 : 2

From the ratio x:y=3:2x : y = 3 : 2, we already know that x=32yx = \frac{3}{2}y. Thus, this calculation aligns perfectly.

Step 4: Assign a numerical value for yy (optional)

If you'd like to find a numerical value for xx, assign a value to yy. For example:

• If y=20y = 20, then:

x=32y=32×20=30x = \frac{3}{2}y = \frac{3}{2} \times 20 = 30

Final Answer:

The original number of red balls is x=32yx = \frac{3}{2}y, or 30 if y=20y = 20. Let me know if you'd like further clarification or variations of the problem! 😊

Let's solve it step-by-step.

1. Set up the initial conditions:

• We have 'x' red balls and 'y' blue balls.

• The ratio x:y is 3:2, so we can write x = (3/2)y.

• 'z' red balls are removed, and z = y/2.

• After removing 'z' red balls, the probability of drawing a red ball is 1/2.

2. Express the new number of red balls and total balls:

• After removing 'z' red balls, the number of red balls becomes x - z.

• The total number of balls after removing 'z' red balls is (x - z) + y.

3. Set up the probability equation:

• The probability of drawing a red ball after removing 'z' balls is (x - z) / ((x - z) + y) = 1/2.

4. Substitute the given values:

• Substitute x = (3/2)y and z = y/2 into the probability equation:

  • ((3/2)y - y/2) / (((3/2)y - y/2) + y) = 1/2

5. Simplify and solve for y:

• Simplify the numerator: (3y/2 - y/2) = 2y/2 = y

• Simplify the denominator: (3y/2 - y/2 + y) = y + y = 2y

• Now the equation is: y / 2y = 1/2

• This equation is always true, which means that the ratio of red to total balls, after removing z red balls, will always result in 1/2, as long as z=y/2 and x/y=3/2. We must therefore use the information we have to solve for x and y.

• Since x = (3/2)y, we can rewrite y as y = (2/3)x.

• Also, since z = y/2, we can say z = (1/3)x.

6. Substitute z and x into the probability equation:

• (x - z) / (x - z + y) = 1/2

• (x - x/3) / (x - x/3 + 2x/3) = 1/2

• (2x/3) / (2x/3 + 2x/3) = 1/2

• (2x/3) / (4x/3) = 1/2

• (2x/3) * (3/4x) = 1/2

• 1/2 = 1/2.

• This equation confirms that our substitutions are correct, but it does not give us a specific value for x or y.

7. Use the information given to solve for y and then x:

• We know x = (3/2)y and z=y/2.

• We also know (x-z)/(x-z+y)=1/2.

• Substitute x and z into the probability equation:

  • ((3/2)y - y/2) / ((3/2)y - y/2 + y) = 1/2

  • y / (2y) = 1/2.

• This is true for any value of y.

• However, if we multiply both sides of the ratio x:y = 3:2 by a variable k, we get x = 3k and y = 2k.

• Also, z = y/2 = 2k/2 = k.

• After removing z red balls, the new number of red balls is x - z = 3k - k = 2k.

• The new total number of balls is x - z + y = 2k + 2k = 4k.

• The probability of drawing a red ball is 2k/4k = 1/2.

• This is consistent with the given information.

• However, we need a specific number for x and y.

• Without additional information, we cannot determine a specific value for x and y. We can only express x in terms of y, or vice versa, or express x and y in terms of k.

• Therefore, the problem is underdefined, and we cannot determine a single value for 'x'.

• However, if we add an additional constraint, we can solve for 'x' and 'y'. For instance, if we know the total number of balls originally, we can solve for 'x' and 'y'.

Conclusion:

Without more information, we cannot determine the original number of red balls 'x'.