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

### Step 1: Identify the events
Let:
- Event A: \( \frac{3}{2} < X < \frac{7}{2} \)
- Event B: \( X > 2 \)

### Step 2: Find the intersection of events A and B
We need to find \( P(A \cap B) \), which is the probability that \( X \) is greater than 2 and less than \( \frac{7}{2} \).

Since \( \frac{7}{2} = 3.5 \), we can see that the values of \( X \) that satisfy both conditions are \( X = 3 \).

### Step 3: Calculate \( P(A \cap B) \)
The probability distribution is given as:
\[ P(X) = \frac{x}{12} \quad \text{for } X = 1, 2, 3, 4, 5, 6 \]
Thus, we calculate:
- \( P(3) = \frac{3}{12} = \frac{1}{4} \)

### Step 4: Find \( P(B) \)
Next, we calculate \( P(B) \), which is the probability that \( X > 2 \). The values of \( X \) that satisfy this condition are \( X = 3, 4, 5, 6 \).

Calculating \( P(B) \):
\[ P(B) = P(3) + P(4) + P(5) + P(6) \]
Calculating each:
- \( P(4) = \frac{4}{12} = \frac{1}{3} \)
- \( P(5) = \frac{5}{12} \)
- \( P(6) = \frac{6}{12} = \frac{1}{2} \)

Now summing these probabilities:
\[ P(B) = \frac{1}{4} + \frac{1}{3} + \frac{5}{12} + \frac{1}{2} \]
Finding a common denominator (which is 12):
\[ P(B) = \frac{3}{12} + \frac{4}{12} + \frac{5}{12} + \frac{6}{12} = \frac{18}{12} = \frac{3}{2} \]

### Step 5: Calculate the conditional probability
Now we can use the formula for conditional probability:
\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
Substituting the values we found:
\[ P(A \mid B) = \frac{\frac{1}{4}}{\frac{3}{2}} = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6} \]

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

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