Study the following information carefully and answer the question.

I. Bag X contains a total of 30 balls, which are either red or green. The probability of selecting one red ball and one green ball from the bag is given as $\frac{40}{87}$. The number of red balls in the bag is $A$.

II. A certain number $B$ of green balls are taken out, painted red, and then placed back into Bag X.

III. After this, the probability of selecting two green balls from Bag X is $\frac{7}{29}$.

IV. All balls from Bag X are then transferred to Bag Y. From Bag Y, $C$ green balls are removed, and $(C-5)$ red balls are added. The probability of selecting two red balls from Bag Y is $\frac{2}{5}$.

**Question:** Find the value of $C$.

Question

Study the following information carefully and answer the question.

I. Bag X contains a total of 30 balls, which are either red or green. The probability of selecting one red ball and one green ball from the bag is given as $\frac{40}{87}$. The number of red balls in the bag is $A$.

II. A certain number $B$ of green balls are taken out, painted red, and then placed back into Bag X.

III. After this, the probability of selecting two green balls from Bag X is $\frac{7}{29}$.

IV. All balls from Bag X are then transferred to Bag Y. From Bag Y, $C$ green balls are removed, and $(C-5)$ red balls are added. The probability of selecting two red balls from Bag Y is $\frac{2}{5}$.

**Question:** Find the value of $C$.

Accepted Answer

**Solution:**

1. Let the number of red balls in Bag X be $A$ and green balls be $30 - A$.

2. The probability of choosing one red and one green ball from Bag X is:
   $$P = \frac{A 	imes (30 - A)}{\binom{30}{2}} = \frac{40}{87}$$

3. Calculate total number of ways to choose 2 balls from 30:
   $$\binom{30}{2} = \frac{30 	imes 29}{2} = 435$$

4. So,
   $$\frac{A (30 - A)}{435} = \frac{40}{87} \implies A (30 - A) = \frac{40}{87} 	imes 435 = 200$$

5. Simplify the quadratic:
   $$A (30 - A) = 200 \implies 30A - A^2 = 200 \implies A^2 - 30A + 200 = 0$$

6. Solve the quadratic equation:
   $$A = \frac{30 \pm \sqrt{30^2 - 4 	imes 200}}{2} = \frac{30 \pm \sqrt{900 - 800}}{2} = \frac{30 \pm 10}{2}$$

7. Possible values:
   - $A = \frac{30 + 10}{2} = 20$
   - $A = \frac{30 - 10}{2} = 10$

8. So, red balls = 10 or 20, green balls = 20 or 10 respectively.

9. Next, $B$ green balls are taken out, painted red, and placed back. So, new red balls = $A + B$, green balls = $30 - A - B$.

10. Probability of choosing two green balls now is given as $\frac{7}{29}$:
    $$\frac{\binom{30 - A - B}{2}}{\binom{30}{2}} = \frac{7}{29}$$

11. Using $\binom{n}{2} = \frac{n(n-1)}{2}$ and total combinations $435$:
    $$\frac{(30 - A - B)(29 - A - B)/2}{435} = \frac{7}{29}$$

12. Multiply both sides by 435:
    $$(30 - A - B)(29 - A - B) = \frac{7}{29} 	imes 435 	imes 2$$

Calculate right side:
    $$\frac{7}{29} 	imes 435 	imes 2 = 7 	imes 15 	imes 2 = 210$$

13. So,
    $$(30 - A - B)(29 - A - B) = 210$$

14. Check for $A=10$:
    $$(30 - 10 - B)(29 - 10 - B) = (20 - B)(19 - B) = 210$$

Expand:
    $$380 - 39B + B^2 = 210 \implies B^2 - 39B + 170 = 0$$

15. Solve quadratic for $B$:
    $$B = \frac{39 \pm \sqrt{39^2 - 4 	imes 170}}{2} = \frac{39 \pm \sqrt{1521 - 680}}{2} = \frac{39 \pm \sqrt{841}}{2} = \frac{39 \pm 29}{2}$$

16. Possible $B$ values:
    - $B = \frac{39 + 29}{2} = 34$ (not possible, since max green balls are 20)
    - $B = \frac{39 - 29}{2} = 5$

17. So, $B=5$ is valid.

18. Now, all balls are transferred to Bag Y. Number of balls in Bag Y:
    - Red balls = $A + B = 10 + 5 = 15$
    - Green balls = $30 - A - B = 30 - 10 - 5 = 15$

19. From Bag Y, $C$ green balls are removed, and $(C-5)$ red balls are added.

New counts:
    - Red balls = $15 + (C - 5) = C + 10$
    - Green balls = $15 - C$

20. Total balls in Bag Y now:
    $$ (C + 10) + (15 - C) = 25$$

21. Probability of choosing two red balls from Bag Y is $\frac{2}{5}$:
    $$\frac{\binom{C + 10}{2}}{\binom{25}{2}} = \frac{2}{5}$$

22. Calculate total combinations:
    $$\binom{25}{2} = \frac{25 	imes 24}{2} = 300$$

23. So,
    $$\frac{(C + 10)(C + 9)/2}{300} = \frac{2}{5} \implies \frac{(C + 10)(C + 9)}{600} = \frac{2}{5}$$

24. Cross-multiplied:
    $$(C + 10)(C + 9) = \frac{2}{5} 	imes 600 = 240$$

25. Expand:
    $$C^2 + 19C + 90 = 240 \implies C^2 + 19C - 150 = 0$$

26. Solve quadratic:
    $$C = \frac{-19 \pm \sqrt{19^2 + 600}}{2} = \frac{-19 \pm \sqrt{361 + 600}}{2} = \frac{-19 \pm \sqrt{961}}{2} = \frac{-19 \pm 31}{2}$$

27. Possible values:
    - $C = \frac{-19 + 31}{2} = 6$
    - $C = \frac{-19 - 31}{2} = -25$ (not possible)

28. Therefore, the value of $C$ is $6$.

**Answer:** $6$

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