An urn contains n white and m black balls. You draw repeatedly at random and without replacement. What is the probability that the first black ball comes in the k-th draw, k = 1, 2, . . . , n + 1 ?

To solve this problem, let's break it down:

### Problem Restatement:
We have an urn containing:
- \( n \) white balls
- \( m \) black balls

We are drawing balls **without replacement**, and we want to find the probability that the **first black ball** comes on the \( k \)-th draw, where \( k = 1, 2, \dots, n + 1 \).

### Solution:

1. **Total number of balls**: There are \( n + m \) balls in total.

2. **Number of ways to arrange the balls**: The total number of possible ways to draw all \( n + m \) balls (in any order) is:
   \[
   \binom{n + m}{n}
   \]
   This represents the number of ways to choose \( n \) white balls and \( m \) black balls from a total of \( n + m \).

3. **Number of favorable outcomes**: For the first black ball to appear in the \( k \)-th draw, the first \( k - 1 \) draws must all be white balls, and the \( k \)-th draw must be a black ball. After that, we are free to choose any of the remaining \( n + m - k \) balls.

   - The number of ways to choose \( k - 1 \) white balls from the \( n \) available white balls is \( \binom{n}{k - 1} \).
   - The remaining \( n - (k - 1) \) white balls and \( m - 1 \) black balls can be arranged in any order, which has \( \binom{n + m - k}{n - (k - 1)} \) ways.

4. **Probability**: The probability that the first black ball appears in the \( k \)-th draw is the ratio of favorable outcomes to the total number of outcomes.

   Thus, the probability \( P(\text{first black ball on } k\text{-th draw}) \) is:
   \[
   P_k = \frac{\binom{n}{k - 1} \binom{m}{1} \binom{n + m - k}{n - (k - 1)}}{\binom{n + m}{n}}
   \]
   Simplifying further:
   \[
   P_k = \frac{\binom{n}{k - 1} \binom{n + m - k}{m - 1}}{\binom{n + m}{m}}
   \]

### Conclusion:
The probability that the first black ball appears on the \( k \)-th draw is:
\[
P_k = \frac{\binom{n}{k - 1} \binom{n + m - k}{m - 1}}{\binom{n + m}{m}}
\]

Would you like further details on any part of the solution?

### Related Questions:
1. How does the probability change if we add more black balls?
2. What is the probability of drawing exactly two black balls in the first three draws?
3. Can we use this approach for drawing with replacement instead of without replacement?
4. What is the expected number of draws until the first black ball?
5. How does the probability change if we start with equal numbers of black and white balls?

### Tip:
When calculating probabilities without replacement, understanding combinations is key, as it reflects the number of ways to arrange objects under specific constraints.

An urn contains n white and m black balls. You draw repeatedly at random and without replacement. What is the probability that the first black ball comes in the k-th draw, k = 1, 2, . . . , n + 1?

Explore how to calculate the probability that the first black ball is drawn on the k-th attempt from an urn containing n white and m black balls, using combinatorial methods and binomial coefficients. This detailed solution is ideal for advanced high school and undergraduate students studying probability.

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