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 find the probability that any one of the three children can win the race, we first need to determine the individual probabilities of each child winning and then calculate the total probability.

Let:
- The probability that child A wins the race, \( P(A) = \frac{1}{3} \)
- The probability that child B wins the race, \( P(B) = \frac{1}{4} \)
- The probability that child C wins the race, \( P(C) = \frac{1}{5} \)

### Step 1: Calculate the total probability of winning

The total probability that any one of the children can win the race is given by the sum of their individual probabilities. However, we need to ensure that the sum of the probabilities does not exceed 1. To do this, we will first find a common denominator.

The denominators are 3, 4, and 5. The least common multiple (LCM) of these numbers is 60.

### Step 2: Convert individual probabilities to have a common denominator

Now we will convert each probability to have a denominator of 60:

- For child A:
\[ P(A) = \frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} \]

- For child B:
\[ P(B) = \frac{1}{4} = \frac{1 \times 15}{4 \times 15} = \frac{15}{60} \]

- For child C:
\[ P(C) = \frac{1}{5} = \frac{1 \times 12}{5 \times 12} = \frac{12}{60} \]

### Step 3: Add the probabilities

Now, we can add the probabilities together:
\[ P(A) + P(B) + P(C) = \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{20 + 15 + 12}{60} = \frac{47}{60} \]

### Step 4: Conclusion

The total probability that any one of the three children can win the race is:
\[ \frac{47}{60} \]