A chess game between two grandmasters X and Y is won by whoever first wins a total of two games. X's chances of winning or loosing any perticular game are a, b and c, respectively. The games are independent and a+b+c=1.
The probability that Y wins the match after the 4th game, is

Options

A. (a)`abc(2a+3b)`
B. (b)`bc^2(a+3b)`
C. (c)`2ac^2(b+c)`
D. (d)`3bc^2(2a+b)`

Answer

To solve the problem, we need to find the probability that player Y wins the chess match after the 4th game. The match is won by the first player to win 2 games. We denote the probabilities of player X winning, drawing, and losing as \( a \), \( b \), and \( c \) respectively, with the condition that \( a + b + c = 1 \).

### Step-by-Step Solution:

1. **Understanding Winning Conditions**:
Player Y can win the match after the 4th game if:
- Y wins 2 games and X wins 1 game.
- Y wins 2 games and there are 1 or more draws.

2. **Possible Outcomes**:
The possible sequences of outcomes for the first 4 games where Y wins after the 4th game can be:
- Y wins the 2nd game (Y wins 2, X wins 1).
- The draws can occur in any of the first three games.

3. **Calculating Probabilities**:
We can break down the probability into two cases:
- **Case 1**: Y wins 2 games, X wins 1 game, and there are draws.
- **Case 2**: Y wins 2 games, and X does not win any games (only draws).

4. **Case 1: Y wins 2 games, X wins 1 game**:
- The possible sequences can be represented as: Y, Y, X or Y, X, Y or X, Y, Y.
- The probability for each sequence is \( c \cdot a \cdot b \) (Y wins, X wins, and draws).
- The number of arrangements for 2 wins for Y and 1 win for X is \( \frac{3!}{2!1!} = 3 \).
- Therefore, the total probability for this case is:
\[ 3 \cdot c \cdot a \cdot b \]

5. **Case 2: Y wins 2 games, X wins 0 games**:
- The only sequence is Y, Y, and draws (D).
- The probability for this sequence is \( b \cdot b \cdot c \) (two draws and Y wins).
- The number of arrangements for 2 wins for Y and 2 draws is \( \frac{3!}{2!0!} = 3 \).
- Therefore, the total probability for this case is:
\[ 3 \cdot b^2 \cdot c \]

6. **Total Probability**:
The total probability that Y wins after the 4th game is the sum of the probabilities from both cases:
\[ P(Y \text{ wins after 4 games}) = 3abc + 3b^2c \]
Factoring out common terms, we get:
\[ P(Y \text{ wins after 4 games}) = 3bc(a + b) \]

### Final Result:
Thus, the probability that Y wins the match after the 4th game is:
\[ P(Y \text{ wins}) = 3bc(a + b) \]