A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses rerquired.
The probability that `ge3` equals

Options

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

Explanation

To solve the problem, we need to find the probability that the number of tosses required to get a 6 is greater than or equal to 3. We denote this probability as \( P(X \geq 3) \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are tossing a fair die repeatedly until we get a 6. The random variable \( X \) represents the number of tosses required to get the first 6.

2. **Finding the Complement**:
To find \( P(X \geq 3) \), we can use the complement rule. We know that:
\[ P(X \geq 3) = 1 - P(X < 3) \]
Here, \( P(X < 3) \) includes the probabilities that \( X = 1 \) and \( X = 2 \).

3. **Calculating \( P(X = 1) \)**:
The probability that we get a 6 on the first toss is:
\[ P(X = 1) = \frac{1}{6} \]

4. **Calculating \( P(X = 2) \)**:
The probability that we do not get a 6 on the first toss and then get a 6 on the second toss is:
\[ P(X = 2) = P(\text{not 6 on 1st}) \times P(\text{6 on 2nd}) = \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) = \frac{5}{36} \]

5. **Calculating \( P(X < 3) \)**:
Now, we can find \( P(X < 3) \):
\[ P(X < 3) = P(X = 1) + P(X = 2) = \frac{1}{6} + \frac{5}{36} \]
To add these fractions, we need a common denominator. The common denominator of 6 and 36 is 36:
\[ P(X < 3) = \frac{6}{36} + \frac{5}{36} = \frac{11}{36} \]

6. **Calculating \( P(X \geq 3) \)**:
Finally, we can find \( P(X \geq 3) \):
\[ P(X \geq 3) = 1 - P(X < 3) = 1 - \frac{11}{36} = \frac{36}{36} - \frac{11}{36} = \frac{25}{36} \]

### Final Answer:
Thus, the probability that the number of tosses required is greater than or equal to 3 is:
\[ \boxed{\frac{25}{36}} \]