The probabilities that three children can win a race are `1/3,1/4` and `1/5`. Find the probability that any one can win the race.

Explanation

To solve the problem of finding the probability that any one of the three children can win the race, we will follow these steps:

### Step 1: Define the probabilities
Let:
- \( P(A) = \frac{1}{3} \) (Probability that child A wins)
- \( P(B) = \frac{1}{4} \) (Probability that child B wins)
- \( P(C) = \frac{1}{5} \) (Probability that child C wins)

### Step 2: Identify the formula
Since the events are mutually exclusive (only one child can win), the probability that any one of them can win is given by:
\[ P(A \text{ or } B \text{ or } C) = P(A) + P(B) + P(C) \]

### Step 3: Substitute the values
Now, we substitute the values of the probabilities into the formula:
\[ P(A \text{ or } B \text{ or } C) = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \]

### Step 4: Find a common denominator
To add these fractions, we need a common denominator. The least common multiple (LCM) of 3, 4, and 5 is 60.

### Step 5: Convert each fraction
Now, we convert each fraction to have the common denominator of 60:
- \( P(A) = \frac{1}{3} = \frac{20}{60} \)
- \( P(B) = \frac{1}{4} = \frac{15}{60} \)
- \( P(C) = \frac{1}{5} = \frac{12}{60} \)

### Step 6: Add the fractions
Now, we can add these fractions:
\[ P(A \text{ or } B \text{ or } C) = \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{20 + 15 + 12}{60} = \frac{47}{60} \]

### Step 7: Conclusion
Thus, the probability that any one of the three children can win the race is:
\[ \frac{47}{60} \]