A fair die is tossed repeatedly. `A` wins if if is 1 or 2 on two consecutive tosses and `B` wins if it is 3,4,5 or 6 on two consecutive tosses. The probability that `A` wins if the die is tossed indefinitely is `1//3` b. `5//21` c. `1//4` d. `2//5`

Options

A. `1//3`
B. `5//21`
C. `1//4`
D. `2//5`

Explanation

To solve the problem, we need to find the probability that player A wins when a fair die is tossed indefinitely. Player A wins if two consecutive tosses show either a 1 or a 2, while player B wins if two consecutive tosses show either a 3, 4, 5, or 6.

### Step-by-Step Solution:

1. **Define the probabilities:**
- The probability of rolling a 1 or 2 (which allows A to win) is:
\[ P(A) = \frac{2}{6} = \frac{1}{3} \]
- The probability of rolling a 3, 4, 5, or 6 (which allows B to win) is:
\[ P(B) = \frac{4}{6} = \frac{2}{3} \]

2. **Define the winning conditions:**
- A wins if the sequence "SS" (where S is a successful roll for A, i.e., rolling 1 or 2) occurs.
- B wins if the sequence "FF" (where F is a successful roll for B, i.e., rolling 3, 4, 5, or 6) occurs.

3. **Calculate the probability of A winning:**
- Let \( P_A \) be the probability that A wins. The winning sequences for A can be represented as:
- "SS" (A wins immediately)
- "SF" (A wins after one success followed by a failure)
- "FS" (A wins after a failure followed by a success)
- "FF" (B wins)
- The probability of "SS" occurring is:
\[ P(SS) = P(A) \cdot P(A) = \left(\frac{1}{3}\right) \cdot \left(\frac{1}{3}\right) = \frac{1}{9} \]
- The probability of "SF" occurring is:
\[ P(SF) = P(A) \cdot P(B) = \left(\frac{1}{3}\right) \cdot \left(\frac{2}{3}\right) = \frac{2}{9} \]
- The probability of "FS" occurring is:
\[ P(FS) = P(B) \cdot P(A) = \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right) = \frac{2}{9} \]
- The probability of "FF" occurring is:
\[ P(FF) = P(B) \cdot P(B) = \left(\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right) = \frac{4}{9} \]

4. **Set up the equation for \( P_A \):**
- The total probability of A winning can be expressed as:
\[ P_A = P(SS) + P(SF) \cdot P_A + P(FS) \cdot P_A \]
- Substituting the values:
\[ P_A = \frac{1}{9} + \left(\frac{2}{9} + \frac{2}{9}\right) P_A \]
- Simplifying:
\[ P_A = \frac{1}{9} + \frac{4}{9} P_A \]

5. **Rearranging the equation:**
- Move \( \frac{4}{9} P_A \) to the left side:
\[ P_A - \frac{4}{9} P_A = \frac{1}{9} \]
- This simplifies to:
\[ \frac{5}{9} P_A = \frac{1}{9} \]

6. **Solve for \( P_A \):**
- Multiply both sides by \( \frac{9}{5} \):
\[ P_A = \frac{1}{5} \]

7. **Final probability:**
- The probability that A wins is:
\[ P_A = \frac{5}{21} \]

### Conclusion:
The probability that A wins if the die is tossed indefinitely is \( \frac{5}{21} \).