An urn contains 'a' green and 'b'pink balls `k(lt, a,b)` balls are drawn and laid a side, their colour being ignored. Then , one more ball is drawn. Find the probability that it is green.

Answer

To solve the problem, we need to find the probability that the last ball drawn from an urn containing 'a' green balls and 'b' pink balls is green, after 'k' balls have been drawn and set aside, ignoring their colors.

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- The urn contains 'a' green balls and 'b' pink balls.
- A total of 'k' balls are drawn and set aside, and their colors are ignored.
- We need to find the probability that the next ball drawn is green.

2. **Total Balls in the Urn**:
- Initially, the total number of balls in the urn is \( a + b \).

3. **Balls Remaining After Drawing k Balls**:
- After drawing 'k' balls, the total number of balls left in the urn is \( (a + b - k) \).
- The number of green balls remaining can vary depending on how many green balls were drawn in the first 'k' draws.

4. **Define the Random Variable**:
- Let \( X \) be the number of green balls drawn in the first 'k' draws. The possible values of \( X \) can range from \( 0 \) to \( \min(k, a) \).

5. **Probability of Drawing i Green Balls**:
- The probability of drawing exactly \( i \) green balls (and hence \( k - i \) pink balls) can be computed using the hypergeometric distribution:
\[ P(X = i) = \frac{\binom{a}{i} \binom{b}{k-i}}{\binom{a+b}{k}} \]
- This is valid for \( i = 0, 1, 2, \ldots, \min(k, a) \).

6. **Remaining Green Balls**:
- If \( i \) green balls are drawn, the number of green balls remaining in the urn is \( a - i \).
- The total number of balls remaining is \( a + b - k \).

7. **Probability of Drawing a Green Ball**:
- The probability that the next ball drawn is green, given that \( i \) green balls were drawn, is:
\[ P(\text{Next ball is green} | X = i) = \frac{a - i}{a + b - k} \]

8. **Total Probability**:
- We can use the law of total probability to find the overall probability that the next ball drawn is green:
\[ P(\text{Next ball is green}) = \sum_{i=0}^{\min(k, a)} P(\text{Next ball is green} | X = i) P(X = i) \]
- Substituting the expressions we have:
\[ P(\text{Next ball is green}) = \sum_{i=0}^{\min(k, a)} \frac{a - i}{a + b - k} \cdot \frac{\binom{a}{i} \binom{b}{k-i}}{\binom{a+b}{k}} \]

9. **Simplifying the Expression**:
- Factor out \( \frac{1}{a + b - k} \):
\[ P(\text{Next ball is green}) = \frac{1}{a + b - k} \sum_{i=0}^{\min(k, a)} (a - i) \cdot \frac{\binom{a}{i} \binom{b}{k-i}}{\binom{a+b}{k}} \]
- The sum \( \sum_{i=0}^{\min(k, a)} (a - i) \cdot \frac{\binom{a}{i} \binom{b}{k-i}}{\binom{a+b}{k}} \) can be computed using combinatorial identities.

10. **Final Result**:
- After simplification, we find that:
\[ P(\text{Next ball is green}) = \frac{a}{a + b} \]

### Conclusion:
The probability that the next ball drawn is green is \( \frac{a}{a + b} \).