If A and B play a series of games in each of which probability that A wins is p and that B wins is q=1-p. Therefore the chance that A wins two games before B wins three is

Options

A. `p^(2)/(1+3q)`
B. `p^(2)/(1+3q^(2))`
C. `p^(2)(1+2q+3q^(2))`
D. None of these

Answer

To find the probability that player A wins 2 games before player B wins 3 games, we can break down the problem into manageable parts.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A wins with probability \( p \).
- B wins with probability \( q = 1 - p \).
- We need to find the probability that A wins 2 games before B wins 3 games.

2. **Setting Up the Cases**:
- Before A wins the second game, we need to consider how many games B can win. B can win 0, 1, or 2 games before A wins 2 games.
- We will denote the number of games won by A as \( W_A \) and by B as \( W_B \).

3. **Case Analysis**:
- **Case 1**: A wins 1 game, and B wins 0 games.
- Sequence: A, A (A wins the last game).
- Probability: \( P(A) = p \times p = p^2 \).

- **Case 2**: A wins 1 game, and B wins 1 game.
- Possible sequences: A, B, A (A wins the last game).
- Probability: \( P(A) = p \times q \times p = p^2q \).

- **Case 3**: A wins 1 game, and B wins 2 games.
- Possible sequences: A, B, B, A (A wins the last game).
- Probability: \( P(A) = p \times q \times q \times p = p^2q^2 \).

4. **Combining the Probabilities**:
- Now we sum the probabilities from all cases:
\[ P(A \text{ wins 2 before B wins 3}) = p^2 + p^2q + p^2q^2 \]

5. **Factoring Out Common Terms**:
- We can factor out \( p^2 \):
\[ P(A \text{ wins 2 before B wins 3}) = p^2(1 + q + q^2) \]

6. **Using the Formula for the Sum of a Geometric Series**:
- The expression \( 1 + q + q^2 \) can be simplified using the formula for the sum of a geometric series:
\[ 1 + q + q^2 = \frac{1 - q^3}{1 - q} = \frac{1 - (1 - p)^3}{p} \]
- Thus, we can substitute this back into our probability expression:
\[ P(A \text{ wins 2 before B wins 3}) = p^2 \cdot \frac{1 - (1 - p)^3}{p} = p(1 - (1 - p)^3) \]

### Final Answer:
The probability that A wins 2 games before B wins 3 games is:
\[ P(A \text{ wins 2 before B wins 3}) = p(1 - (1 - p)^3) \]