There are three evens `E_(1),E_(2)`and `E_(3)` one of which must, and only one can happen. The odds are 7 to 4 against `E_(1)` and 5 to 3 against `E_(2)`. Find odds against `E_(3)`.

Answer

To solve the problem, we need to find the odds against event \( E_3 \) given the odds against events \( E_1 \) and \( E_2 \).

### Step-by-Step Solution:

1. **Understand the Odds Against Events**:
- The odds against \( E_1 \) are given as 7 to 4. This means for every 7 unfavorable outcomes, there are 4 favorable outcomes.
- The odds against \( E_2 \) are given as 5 to 3. This means for every 5 unfavorable outcomes, there are 3 favorable outcomes.

2. **Convert Odds to Probabilities**:
- For \( E_1 \):
\[ \text{Probability of } E_1 = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{4}{4 + 7} = \frac{4}{11} \]
- For \( E_2 \):
\[ \text{Probability of } E_2 = \frac{3}{3 + 5} = \frac{3}{8} \]

3. **Use the Property of Exhaustive Events**:
- Since \( E_1, E_2, \) and \( E_3 \) are mutually exclusive and exhaustive events, we know:
\[ P(E_1) + P(E_2) + P(E_3) = 1 \]
- Substituting the known probabilities:
\[ \frac{4}{11} + \frac{3}{8} + P(E_3) = 1 \]

4. **Find a Common Denominator**:
- The least common multiple of 11 and 8 is 88. We convert the fractions:
\[ \frac{4}{11} = \frac{4 \times 8}{11 \times 8} = \frac{32}{88} \]
\[ \frac{3}{8} = \frac{3 \times 11}{8 \times 11} = \frac{33}{88} \]

5. **Substitute and Solve for \( P(E_3) \)**:
- Now we can substitute these values into the equation:
\[ \frac{32}{88} + \frac{33}{88} + P(E_3) = 1 \]
\[ \frac{65}{88} + P(E_3) = 1 \]
\[ P(E_3) = 1 - \frac{65}{88} = \frac{88 - 65}{88} = \frac{23}{88} \]

6. **Calculate Odds Against \( E_3 \)**:
- The probability of \( E_3 \) is \( \frac{23}{88} \). The probability against \( E_3 \) is:
\[ P(\text{not } E_3) = 1 - P(E_3) = 1 - \frac{23}{88} = \frac{65}{88} \]
- The odds against \( E_3 \) are given by the ratio of unfavorable outcomes to favorable outcomes:
\[ \text{Odds against } E_3 = \frac{P(\text{not } E_3)}{P(E_3)} = \frac{\frac{65}{88}}{\frac{23}{88}} = \frac{65}{23} \]

### Final Answer:
The odds against \( E_3 \) are \( 65 \) to \( 23 \).