There are four machines and it is known that eactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. The probability that only two tests are needed is (A) `1/6` (B) `1/3` (C) `1/2` (D) `1/4`

Options

A. `(1)/(3)`
B. `(1)/(6)`
C. `(1)/(2)`
D. ``(1)/(4)`

Answer

To solve the problem, we need to find the probability that only two tests are needed to identify both faulty machines out of four machines, where exactly two of them are faulty.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have four machines, and we know that exactly two of them are faulty. We need to test these machines one by one in a random order until both faulty machines are identified. We want to find the probability that we can identify both faulty machines in just two tests.

2. **Identifying Favorable Outcomes**: For us to identify both faulty machines in just two tests, both tests must result in selecting the faulty machines. This means that in the first test, we select one faulty machine, and in the second test, we select the second faulty machine.

3. **Calculating the Probability**:
- The total number of ways to choose 2 machines from 4 is given by the combination formula \( C(n, k) \), where \( n \) is the total number of machines and \( k \) is the number of machines to choose:
\[ C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]
- The favorable outcomes for our case (choosing both faulty machines in the first two tests) can happen in only one way: choosing the two faulty machines directly. Therefore, the number of favorable outcomes is 1.

4. **Calculating the Probability of Selecting Faulty Machines**:
- The probability of selecting the first faulty machine in the first test is \( \frac{2}{4} \) (since there are 2 faulty machines out of 4 total).
- After selecting one faulty machine, there is now 1 faulty machine left out of 3 remaining machines. Thus, the probability of selecting the second faulty machine in the second test is \( \frac{1}{3} \).
- Therefore, the combined probability of selecting both faulty machines in the first two tests is:
\[ P(\text{both faulty in first two tests}) = P(\text{first faulty}) \times P(\text{second faulty}) = \frac{2}{4} \times \frac{1}{3} = \frac{1}{6} \]

5. **Conclusion**: The probability that only two tests are needed to identify both faulty machines is \( \frac{1}{6} \).

### Final Answer:
The probability that only two tests are needed is \( \frac{1}{6} \). Thus, the correct option is (A) \( \frac{1}{6} \).