Three persons A , B and C shoot to hit a target. Their probabilities of hitting the target are `5/(6), 4/(5) and 3/(4)` respectively. Find the probability that :
At least one person hit the target.

Answer

To find the probability that at least one person hits the target, we can use the complement rule. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the probabilities of hitting the target:**
- Let \( P(A) \) be the probability that person A hits the target:
\[ P(A) = \frac{5}{6} \]
- Let \( P(B) \) be the probability that person B hits the target:
\[ P(B) = \frac{4}{5} \]
- Let \( P(C) \) be the probability that person C hits the target:
\[ P(C) = \frac{3}{4} \]

2. **Calculate the probabilities of missing the target:**
- The probability that person A misses the target:
\[ P(A') = 1 - P(A) = 1 - \frac{5}{6} = \frac{1}{6} \]
- The probability that person B misses the target:
\[ P(B') = 1 - P(B) = 1 - \frac{4}{5} = \frac{1}{5} \]
- The probability that person C misses the target:
\[ P(C') = 1 - P(C) = 1 - \frac{3}{4} = \frac{1}{4} \]

3. **Calculate the probability that all three miss the target:**
- Since the events are independent, the probability that all three miss the target is:
\[ P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \]
- Calculate this product:
\[ P(A' \cap B' \cap C') = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} = \frac{1}{120} \]

4. **Calculate the probability that at least one person hits the target:**
- Using the complement rule:
\[ P(\text{at least one hits}) = 1 - P(A' \cap B' \cap C') = 1 - \frac{1}{120} \]
- Simplifying this:
\[ P(\text{at least one hits}) = \frac{120}{120} - \frac{1}{120} = \frac{119}{120} \]

### Final Answer:
The probability that at least one person hits the target is:
\[ \frac{119}{120} \]