Two persons A and B play a game of throwing a pair of dice until one of them wins. A will win if sum of numbers on dice appear to be 6 and B will win. If sum is 7. What is the probability that A wins the game if A starts the game.

Options

A. `31/61`
B. `30/61`
C. `29/61`
D. `32/61`

Answer

To solve the problem, we need to calculate the probability that player A wins the game when they start. Let's break it down step by step.

### Step 1: Calculate the probabilities of winning for A and B
When two dice are thrown, the possible sums range from 2 to 12. We need to find the probability of getting a sum of 6 (for A) and a sum of 7 (for B).

- **Total outcomes when throwing two dice**: There are \(6 \times 6 = 36\) total outcomes.

- **Outcomes for A (sum = 6)**: The combinations that yield a sum of 6 are:
- (1, 5)
- (2, 4)
- (3, 3)
- (4, 2)
- (5, 1)

Thus, there are 5 outcomes where A wins.

- **Outcomes for B (sum = 7)**: The combinations that yield a sum of 7 are:
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)

Thus, there are 6 outcomes where B wins.

### Step 2: Calculate the probabilities
- Probability that A wins on a single throw:
\[ P(A \text{ wins}) = \frac{5}{36} \]

- Probability that B wins on a single throw:
\[ P(B \text{ wins}) = \frac{6}{36} = \frac{1}{6} \]

- Probability that neither A nor B wins (sum is neither 6 nor 7):
\[ P(\text{neither wins}) = 1 - P(A \text{ wins}) - P(B \text{ wins}) = 1 - \frac{5}{36} - \frac{6}{36} = \frac{25}{36} \]

### Step 3: Set up the equation for A's winning probability
Let \(P_A\) be the probability that A wins the game. A can win in two scenarios:
1. A wins on the first throw.
2. A does not win on the first throw (neither A nor B wins), and the game restarts with the same probabilities.

Thus, we can write the equation:
\[ P_A = P(A \text{ wins}) + P(\text{neither wins}) \cdot P_A \]
Substituting the probabilities we calculated:
\[ P_A = \frac{5}{36} + \frac{25}{36} P_A \]

### Step 4: Solve for \(P_A\)
Rearranging the equation:
\[ P_A - \frac{25}{36} P_A = \frac{5}{36} \]
\[ \left(1 - \frac{25}{36}\right) P_A = \frac{5}{36} \]
\[ \frac{11}{36} P_A = \frac{5}{36} \]
\[ P_A = \frac{5}{11} \]

### Conclusion
The probability that A wins the game is \(\frac{5}{11}\).