The odds in favour of A winning a game of chess against B are 5:2. if three games are to be played, then the odds in favour of A's winning at least one game are 355:8 .If this is true give answer 1 else 0.

Options

A. 1
B. 0
C. both
D. none

Answer

To solve the problem, we need to determine if the odds in favor of A winning at least one game out of three games played is indeed 355:8, given that the odds in favor of A winning a single game against B are 5:2.

### Step-by-Step Solution:

1. **Understanding Odds**:
The odds in favor of A winning are given as 5:2. This means:
- For every 5 wins by A, there are 2 wins by B.
- Total outcomes = 5 + 2 = 7.
- Probability of A winning a single game (P(A)) = Number of favorable outcomes / Total outcomes = 5/7.

2. **Calculating Probability of A Losing**:
- The probability of A losing a single game (P(A')) = 1 - P(A) = 1 - 5/7 = 2/7.

3. **Calculating Probability of A Losing All Three Games**:
- The probability of A losing all three games is given by:
\[ P(A' \text{ in 3 games}) = P(A')^3 = \left(\frac{2}{7}\right)^3 = \frac{8}{343}. \]

4. **Calculating Probability of A Winning At Least One Game**:
- The probability of A winning at least one game is:
\[ P(\text{at least 1 win}) = 1 - P(A' \text{ in 3 games}) = 1 - \frac{8}{343} = \frac{343 - 8}{343} = \frac{335}{343}. \]

5. **Calculating Odds in Favor of A Winning At Least One Game**:
- The odds in favor of A winning at least one game can be calculated as:
\[ \text{Odds in favor} = \frac{P(\text{A wins at least 1})}{P(\text{A loses all})} = \frac{335/343}{8/343} = \frac{335}{8}. \]

6. **Comparing with Given Odds**:
- The given odds in favor of A winning at least one game are 355:8.
- We calculated the odds as 335:8.

7. **Conclusion**:
Since 335:8 is not equal to 355:8, the statement provided in the question is false.

### Final Answer:
The answer is **0**.