In Example 99, the conditional probability that `X ge 6 " given " X gt 3` equals

Options

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

Explanation

To find the conditional probability \( P(X \geq 6 \mid X > 3) \), we can use the formula for conditional probability:

\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]

In our case, let:
- \( A \) be the event \( X \geq 6 \)
- \( B \) be the event \( X > 3 \)

Thus, we need to compute:

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

Since \( X \geq 6 \) already implies \( X > 3 \), we can simplify \( P(X \geq 6 \cap X > 3) \) to \( P(X \geq 6) \). Therefore, we have:

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

### Step 1: Calculate \( P(X \geq 6) \)

Assuming \( X \) follows a geometric distribution with success probability \( p = \frac{1}{6} \), the probability mass function is given by:

\[ P(X = k) = (1-p)^{k-1} p = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6} \]

To find \( P(X \geq 6) \):

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

This can be expressed as:

\[ P(X \geq 6) = \sum_{k=6}^{\infty} P(X = k) = \sum_{k=6}^{\infty} \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6} \]

This is a geometric series where the first term \( a = \left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6} \) and the common ratio \( r = \frac{5}{6} \):

\[ P(X \geq 6) = \frac{a}{1 - r} = \frac{\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}}{1 - \frac{5}{6}} = \frac{\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}}{\frac{1}{6}} = \left(\frac{5}{6}\right)^{5} \]

### Step 2: Calculate \( P(X > 3) \)

Similarly, we compute \( P(X > 3) \):

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

This can be expressed as:

\[ P(X > 3) = \sum_{k=4}^{\infty} P(X = k) = \sum_{k=4}^{\infty} \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6} \]

This is also a geometric series where the first term \( a = \left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6} \):

\[ P(X > 3) = \frac{a}{1 - r} = \frac{\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}}{1 - \frac{5}{6}} = \frac{\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}}{\frac{1}{6}} = \left(\frac{5}{6}\right)^{3} \]

### Step 3: Combine Results

Now we can substitute these results back into our conditional probability formula:

\[ P(X \geq 6 \mid X > 3) = \frac{P(X \geq 6)}{P(X > 3)} = \frac{\left(\frac{5}{6}\right)^{5}}{\left(\frac{5}{6}\right)^{3}} = \left(\frac{5}{6}\right)^{2} = \frac{25}{36} \]

### Final Answer

Thus, the conditional probability \( P(X \geq 6 \mid X > 3) \) equals:

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