The probabilities of hitting a target by A, B and C are `3/5,3/4` and `1/3` respectively.
If all three hits the target simultaneously then find the probability of hitting the target by the least two of them.

Explanation

To solve the problem, we need to find the probability that at least two out of A, B, and C hit the target. We will first calculate the probabilities of each individual hitting the target and not hitting the target, and then use these to find the required probability.

### Step-by-Step Solution:

1. **Identify Individual Probabilities:**
- Probability of A hitting the target, \( P(A) = \frac{3}{5} \)
- Probability of B hitting the target, \( P(B) = \frac{3}{4} \)
- Probability of C hitting the target, \( P(C) = \frac{1}{3} \)

2. **Calculate Probabilities of Not Hitting the Target:**
- Probability of A not hitting the target, \( P(A') = 1 - P(A) = 1 - \frac{3}{5} = \frac{2}{5} \)
- Probability of B not hitting the target, \( P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \)
- Probability of C not hitting the target, \( P(C') = 1 - P(C) = 1 - \frac{1}{3} = \frac{2}{3} \)

3. **Calculate the Probability of Each Case:**
- **Case 1:** All three hit the target (A, B, C)
\[ P(E_1) = P(A) \cdot P(B) \cdot P(C) = \frac{3}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{3}{20} \]

- **Case 2:** A and B hit, C does not hit
\[ P(E_2) = P(A) \cdot P(B) \cdot P(C') = \frac{3}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{3}{10} \]

- **Case 3:** A and C hit, B does not hit
\[ P(E_3) = P(A) \cdot P(C) \cdot P(B') = \frac{3}{5} \cdot \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{20} \]

- **Case 4:** B and C hit, A does not hit
\[ P(E_4) = P(B) \cdot P(C) \cdot P(A') = \frac{3}{4} \cdot \frac{1}{3} \cdot \frac{2}{5} = \frac{1}{10} \]

4. **Calculate the Total Probability of Hitting the Target by At Least Two:**
\[ P(\text{at least 2 hit}) = P(E_1) + P(E_2) + P(E_3) + P(E_4) \]
\[ = \frac{3}{20} + \frac{3}{10} + \frac{1}{20} + \frac{1}{10} \]
To add these fractions, we convert them to a common denominator (20):
\[ = \frac{3}{20} + \frac{6}{20} + \frac{1}{20} + \frac{2}{20} = \frac{12}{20} = \frac{3}{5} \]

5. **Final Answer:**
The probability that at least two of them hit the target is \( \frac{3}{5} \).