An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k balls of the same colour as that of the ball drawn. a ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.

Explanation

To solve the problem, we need to find the probability of drawing a white ball from the urn after a sequence of events. Let's break it down step by step.

### Step 1: Understand the Initial Setup
The urn contains:
- \( m \) white balls
- \( n \) black balls

Total number of balls in the urn = \( m + n \).

### Step 2: Calculate the Probability of Drawing a White or Black Ball
When a ball is drawn at random, the probabilities are:
- Probability of drawing a white ball (\( P(W_1) \)) = \( \frac{m}{m+n} \)
- Probability of drawing a black ball (\( P(B_1) \)) = \( \frac{n}{m+n} \)

### Step 3: Update the Urn After Drawing a Ball
When a ball is drawn, it is put back into the urn along with \( k \) additional balls of the same color. We will analyze both cases (drawing a white ball and drawing a black ball).

#### Case 1: Drawing a White Ball
If a white ball is drawn:
- The new composition of the urn will be:
- White balls = \( m + k \)
- Black balls = \( n \)

Total balls = \( (m + k) + n = m + n + k \).

#### Case 2: Drawing a Black Ball
If a black ball is drawn:
- The new composition of the urn will be:
- White balls = \( m \)
- Black balls = \( n + k \)

Total balls = \( m + (n + k) = m + n + k \).

### Step 4: Calculate the Probability of Drawing a White Ball in the Second Draw
Now, we need to find the probability of drawing a white ball in the second draw, \( P(W) \).

Using the law of total probability:
\[ P(W) = P(W | W_1) \cdot P(W_1) + P(W | B_1) \cdot P(B_1) \]

Where:
- \( P(W | W_1) \) is the probability of drawing a white ball given that a white ball was drawn first.
- \( P(W | B_1) \) is the probability of drawing a white ball given that a black ball was drawn first.

Calculating these probabilities:
1. If a white ball was drawn first:
\[ P(W | W_1) = \frac{m + k}{m + n + k} \]

2. If a black ball was drawn first:
\[ P(W | B_1) = \frac{m}{m + n + k} \]

### Step 5: Substitute into the Total Probability Formula
Substituting these probabilities into the total probability formula:
\[ P(W) = \left( \frac{m + k}{m + n + k} \cdot \frac{m}{m + n} \right) + \left( \frac{m}{m + n + k} \cdot \frac{n}{m + n} \right) \]

### Step 6: Simplifying the Expression
Now, we can simplify the expression:
\[ P(W) = \frac{m(m + k)}{(m + n)(m + n + k)} + \frac{mn}{(m + n)(m + n + k)} \]
Combining the terms:
\[ P(W) = \frac{m(m + k) + mn}{(m + n)(m + n + k)} = \frac{m^2 + mk + mn}{(m + n)(m + n + k)} \]

### Step 7: Final Probability Expression
Notice that the term \( k \) in the numerator and denominator cancels out when we consider the overall probability of drawing a white ball. Thus, the final expression simplifies to:
\[ P(W) = \frac{m}{m + n} \]

### Conclusion
The probability of drawing a white ball after the second draw does not depend on \( k \).