There are four calculators and it is known it is known that exactly two of them are defective. They are tested one by one , in a random order till both the defective calculators are identified. Then the probability that only two tests are required is :

Options

A. `5/6`
B. `1/2`
C. `1/6`
D. `1/3`

Answer

To solve the problem, we need to find the probability that exactly two tests are required to identify both defective calculators out of the four available.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 4 calculators, out of which 2 are defective (D) and 2 are working (W). We need to find the probability that both defective calculators are identified in the first two tests.

2. **Identifying Favorable Outcomes**:
The only way to identify both defective calculators in exactly two tests is to select both defective calculators in the first two tests. The successful outcomes can be represented as:
- (D1, D2)
where D1 and D2 are the defective calculators.

3. **Total Possible Outcomes**:
The total number of ways to select any 2 calculators from the 4 is given by the combination formula:
\[ \text{Total ways} = \binom{4}{2} = 6 \]
The possible pairs of calculators are:
- (D1, D2)
- (D1, W1)
- (D1, W2)
- (D2, W1)
- (D2, W2)
- (W1, W2)

4. **Favorable Outcomes for Selecting Defective Calculators**:
There is only 1 favorable outcome where both selected calculators are defective:
- (D1, D2)

5. **Calculating the Probability**:
The probability of selecting both defective calculators in the first two tests is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{both defective in 2 tests}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{6} \]

6. **Conclusion**:
Thus, the probability that only two tests are required to identify both defective calculators is:
\[ \boxed{\frac{1}{6}} \]