In a horse race the odds in favour of three horses are 1 : 2, 1 : 3 and 1 : 4. The probability that one of the horse will win the race is

Options

A. `(37)/(60)`
B. `(47)/(60)`
C. `(1)/(4)`
D. `(3)/(4)`

Answer

To solve the problem, we need to find the probability that at least one of the three horses wins the race based on the given odds.

### Step-by-Step Solution:

1. **Understand the Odds**:
The odds in favor of the horses A, B, and C are given as:
- Horse A: 1 : 2
- Horse B: 1 : 3
- Horse C: 1 : 4

2. **Convert Odds to Probability**:
The probability of winning for each horse can be calculated using the formula:
\[ P(\text{Horse}) = \frac{\text{Odds of winning}}{\text{Total Odds}} = \frac{1}{\text{Odds of winning} + \text{Odds against winning}} \]

- For Horse A:
\[ P(A) = \frac{1}{1 + 2} = \frac{1}{3} \]
- For Horse B:
\[ P(B) = \frac{1}{1 + 3} = \frac{1}{4} \]
- For Horse C:
\[ P(C) = \frac{1}{1 + 4} = \frac{1}{5} \]

3. **Calculate Total Probability**:
Since the events are mutually exclusive (only one horse can win), we can add the probabilities directly:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) \]
Substituting the values we found:
\[ P(A \cup B \cup C) = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \]

4. **Find a Common Denominator**:
The least common multiple (LCM) of 3, 4, and 5 is 60. We convert each fraction:
- For Horse A:
\[ P(A) = \frac{1}{3} = \frac{20}{60} \]
- For Horse B:
\[ P(B) = \frac{1}{4} = \frac{15}{60} \]
- For Horse C:
\[ P(C) = \frac{1}{5} = \frac{12}{60} \]

5. **Add the Probabilities**:
Now we can add the converted probabilities:
\[ P(A \cup B \cup C) = \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{47}{60} \]

6. **Conclusion**:
The probability that at least one of the horses will win the race is:
\[ \boxed{\frac{47}{60}} \]