A fair die is tossed repeatedly until a 6 is obtained. Let X denote the number of tosses rerquired.
The conditional probability that `Xge6` given `Xgt3` equals

Options

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

Explanation

To solve the problem, we need to find the conditional probability \( P(X \geq 6 \mid X > 3) \).

### Step 1: Understand the scenario
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. The probability of getting a 6 on one toss is \( \frac{1}{6} \), and the probability of not getting a 6 is \( \frac{5}{6} \).

### Step 2: Calculate \( P(X \geq 6) \)
To find \( P(X \geq 6) \), we need to consider that for \( X \) to be at least 6, the first five tosses must not result in a 6, followed by a 6 on the sixth toss or later.

\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + \ldots \]

This can be expressed as:

\[ P(X \geq 6) = P(\text{not getting a 6 in the first 5 tosses}) \cdot P(\text{getting a 6 on the 6th toss or later}) \]

The probability of not getting a 6 in the first 5 tosses is:

\[ \left(\frac{5}{6}\right)^5 \]

The probability of getting a 6 on the 6th toss or later can be modeled as a geometric series:

\[ P(X \geq 6) = \left(\frac{5}{6}\right)^5 \cdot \left(\frac{1}{6} + \frac{5}{6} \cdot \left(\frac{1}{6} + \frac{5}{6} \cdots \right)\right) \]

This series converges to:

\[ \frac{1}{6} \cdot \frac{1}{1 - \frac{5}{6}} = \frac{1}{6} \cdot 6 = 1 \]

Thus:

\[ P(X \geq 6) = \left(\frac{5}{6}\right)^5 \]

### Step 3: Calculate \( P(X > 3) \)
Next, we need to calculate \( P(X > 3) \):

\[ P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6) + \ldots \]

This can be expressed as:

\[ P(X > 3) = P(\text{not getting a 6 in the first 3 tosses}) \cdot P(\text{getting a 6 on the 4th toss or later}) \]

The probability of not getting a 6 in the first 3 tosses is:

\[ \left(\frac{5}{6}\right)^3 \]

Thus:

\[ P(X > 3) = \left(\frac{5}{6}\right)^3 \]

### Step 4: Calculate the conditional probability
Now, we can find the conditional probability:

\[ P(X \geq 6 \mid X > 3) = \frac{P(X \geq 6 \cap X > 3)}{P(X > 3)} = \frac{P(X \geq 6)}{P(X > 3)} \]

Substituting the values we calculated:

\[ P(X \geq 6 \mid X > 3) = \frac{\left(\frac{5}{6}\right)^5}{\left(\frac{5}{6}\right)^3} = \left(\frac{5}{6}\right)^{5-3} = \left(\frac{5}{6}\right)^2 \]

Calculating this gives:

\[ P(X \geq 6 \mid X > 3) = \frac{25}{36} \]

### Final Answer
Thus, the conditional probability that \( X \geq 6 \) given \( X > 3 \) is:

\[ \frac{25}{36} \]