A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses required.
The probability that `X=3` equals

Options

A. `25//216`
B. `25//36`
C. `5//36`
D. `125//216`

Explanation

To solve the problem, we need to find the probability that the number of tosses required to get a 6 (denoted by X) equals 3. This means that in the first two tosses, we do not get a 6, and in the third toss, we do get a 6.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are tossing a fair die repeatedly until we get a 6. We need to find the probability that it takes exactly 3 tosses to get the first 6.

2. **Identifying Probabilities**:
- The probability of rolling a 6 on a single toss of the die is \( P(6) = \frac{1}{6} \).
- The probability of not rolling a 6 (rolling a 1, 2, 3, 4, or 5) is \( P(\text{not 6}) = \frac{5}{6} \).

3. **Calculating the Probability for X = 3**:
- For \( X = 3 \), we need:
- The first toss to not be a 6: \( P(\text{not 6}) = \frac{5}{6} \)
- The second toss to not be a 6: \( P(\text{not 6}) = \frac{5}{6} \)
- The third toss to be a 6: \( P(6) = \frac{1}{6} \)

Therefore, the probability that \( X = 3 \) can be calculated as:
\[ P(X = 3) = P(\text{not 6 in 1st toss}) \times P(\text{not 6 in 2nd toss}) \times P(6 \text{ in 3rd toss}) \]
\[ P(X = 3) = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) \]

4. **Performing the Calculation**:
\[ P(X = 3) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{36} \times \frac{1}{6} = \frac{25}{216} \]

5. **Final Result**:
The probability that \( X = 3 \) is \( \frac{25}{216} \).

### Conclusion:
Thus, the probability that \( X = 3 \) is \( \frac{25}{216} \).