A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is:

Options

A. `(1)/(3)`
B. `(1)/(4)`
C. `(1)/(8)`
D. `(1)/(9)`

Answer

To solve the problem, we need to find the conditional probability that the score of 4 has appeared at least once given that the sum of the scores from two throws of a die is a multiple of 4.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( A \) be the event that the score 4 appears at least once.
- Let \( B \) be the event that the sum of the scores is a multiple of 4.

2. **Determine the Sample Space**:
- When a die is thrown twice, the total number of outcomes is \( 6 \times 6 = 36 \).

3. **Identify Multiples of 4**:
- The possible sums from two dice that are multiples of 4 are: 4, 8, and 12.

4. **Count Outcomes for Event B**:
- **Sum = 4**: The combinations are (1,3), (2,2), (3,1). Total = 3 outcomes.
- **Sum = 8**: The combinations are (2,6), (3,5), (4,4), (5,3), (6,2). Total = 5 outcomes.
- **Sum = 12**: The combination is (6,6). Total = 1 outcome.
- **Total outcomes for B**: \( 3 + 5 + 1 = 9 \).

5. **Calculate Probability of B**:
\[ P(B) = \frac{\text{Number of outcomes in B}}{\text{Total outcomes}} = \frac{9}{36} = \frac{1}{4}. \]

6. **Count Outcomes for Event \( A \cap B \)**:
- We need to find outcomes where the sum is a multiple of 4 and at least one die shows a 4.
- **Sum = 8**: The combinations that include 4 are (4,4) and (4, 4). Total = 1 outcome.
- **Sum = 12**: The combination (6,6) does not include 4.
- **Sum = 4**: The combinations (1,3), (2,2), (3,1) do not include 4.
- **Total outcomes for \( A \cap B \)**: Only (4,4) gives us 1 outcome.

7. **Calculate Probability of \( A \cap B \)**:
\[ P(A \cap B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Total outcomes}} = \frac{1}{36}. \]

8. **Calculate Conditional Probability \( P(A|B) \)**:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{36}}{\frac{1}{4}} = \frac{1}{36} \times \frac{4}{1} = \frac{4}{36} = \frac{1}{9}. \]

### Final Answer:
The conditional probability that the score 4 has appeared at least once given that the sum is a multiple of 4 is \( \frac{1}{9} \).