A bag contains `a` white and `b` black balls. Two players, `Aa n dB` alternately draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game. `A` begins the game. If the probability of `A` winning the game is three times that of `B ,` then find the ratio `a : b`

Explanation

To solve the problem step by step, we need to find the ratio of white balls (a) to black balls (b) in the bag, given the probability conditions of players A and B.

### Step 1: Define Probabilities
Let:
- Probability of drawing a white ball (W) = \( \frac{a}{a + b} \)
- Probability of drawing a black ball (B) = \( \frac{b}{a + b} \)

### Step 2: Determine Probability of A Winning
Player A can win in the following scenarios:
1. A draws a white ball on the first draw.
2. A draws a black ball, then B draws a black ball, and then A draws a white ball.
3. A and B both draw black balls, and then A draws a white ball again, and this continues.

The probability of A winning can be expressed as:
\[ P(A \text{ wins}) = P(W) + P(B) \cdot P(B) \cdot P(W) + P(B)^2 \cdot P(B) \cdot P(W) + \ldots \]

This can be simplified to:
\[ P(A \text{ wins}) = P(W) \left( 1 + P(B) + P(B)^2 + P(B)^3 + \ldots \right) \]

### Step 3: Sum of Infinite Geometric Series
The series \( 1 + P(B) + P(B)^2 + P(B)^3 + \ldots \) is a geometric series with first term \( 1 \) and common ratio \( P(B) \):
\[ \text{Sum} = \frac{1}{1 - P(B)} = \frac{1}{1 - \frac{b}{a + b}} = \frac{a + b}{a} \]

### Step 4: Substitute Back into Probability of A Winning
Now substituting back, we get:
\[ P(A \text{ wins}) = P(W) \cdot \frac{a + b}{a} \]
\[ P(A \text{ wins}) = \frac{a}{a + b} \cdot \frac{a + b}{a} = \frac{a}{a + 2b} \]

### Step 5: Determine Probability of B Winning
Since the total probability must equal 1:
\[ P(B \text{ wins}) = 1 - P(A \text{ wins}) = 1 - \frac{a}{a + 2b} = \frac{b}{a + 2b} \]

### Step 6: Set Up the Given Condition
According to the problem, the probability of A winning is three times that of B:
\[ P(A \text{ wins}) = 3 \cdot P(B \text{ wins}) \]
Substituting the expressions we derived:
\[ \frac{a}{a + 2b} = 3 \cdot \frac{b}{a + 2b} \]

### Step 7: Cross-Multiply and Simplify
Cross-multiplying gives:
\[ a(a + 2b) = 3b(a + 2b) \]
Expanding both sides:
\[ a^2 + 2ab = 3ab + 6b^2 \]
Rearranging gives:
\[ a^2 - ab - 6b^2 = 0 \]

### Step 8: Solve the Quadratic Equation
This is a quadratic equation in terms of \( a \):
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 1, B = -b, C = -6b^2 \):
\[ a = \frac{b \pm \sqrt{(-b)^2 - 4 \cdot 1 \cdot (-6b^2)}}{2 \cdot 1} \]
\[ a = \frac{b \pm \sqrt{b^2 + 24b^2}}{2} \]
\[ a = \frac{b \pm 5b}{2} \]

### Step 9: Calculate Possible Values
This gives us two possible values:
1. \( a = \frac{6b}{2} = 3b \)
2. \( a = \frac{-4b}{2} = -2b \) (not possible since a cannot be negative)

### Step 10: Find the Ratio
Thus, we have \( a = 3b \). The ratio \( a : b = 3 : 1 \).

### Final Answer
The ratio of white balls to black balls is:
\[ \boxed{3 : 1} \]