The probability distribution of a random variable (X) is `P(X)={{:((x)/(12),":",X="1, 2, 3, 4, 5, 6"),(0,":","otherwise"):}`
Then, the conditional probability
`P(((3)/(2)ltXlt(7)/(2))/(X gt2))` is

Options

A. `(5)/(6)`
B. `(5)/(18)`
C. `(1)/(6)`
D. `(7)/(12)`

Answer

To solve the given problem, we need to calculate the conditional probability \( P\left( \frac{3}{2} < X < \frac{7}{2} \mid X > 2 \right) \).

### Step-by-Step Solution:

1. **Identify the Events**:
- Let \( A \) be the event \( \frac{3}{2} < X < \frac{7}{2} \).
- Let \( B \) be the event \( X > 2 \).

2. **Find the Intersection of Events \( A \) and \( B \)**:
- The condition \( \frac{3}{2} < X < \frac{7}{2} \) translates to \( 1.5 < X < 3.5 \).
- The values of \( X \) that satisfy \( X > 2 \) are \( 3, 4, 5, 6 \).
- The intersection \( A \cap B \) is the set of values of \( X \) that are both greater than \( 2 \) and between \( 1.5 \) and \( 3.5 \). Thus, \( A \cap B = \{3\} \).

3. **Calculate the Probability of \( A \cap B \)**:
- The probability \( P(A \cap B) \) is given by the probability of \( X = 3 \):
\[ P(X = 3) = \frac{3}{12} = \frac{1}{4} \]

4. **Calculate the Probability of \( B \)**:
- The event \( B \) includes the values \( 3, 4, 5, 6 \).
- Therefore, we calculate:
\[ P(B) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \]
- This gives:
\[ P(B) = \frac{3}{12} + \frac{4}{12} + \frac{5}{12} + \frac{6}{12} = \frac{3 + 4 + 5 + 6}{12} = \frac{18}{12} = \frac{3}{2} \]

5. **Calculate the Conditional Probability**:
- Using the formula for conditional probability:
\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
- Substituting the values we found:
\[ P(A \mid B) = \frac{\frac{1}{4}}{\frac{3}{2}} = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6} \]

### Final Answer:
Thus, the conditional probability \( P\left( \frac{3}{2} < X < \frac{7}{2} \mid X > 2 \right) = \frac{1}{6} \).