2 dice are thrown. Suppose a random variable X is assigned a value 2k, if the sum on the dice is equal to k, then the expected value of X is

Options

A. 10
B. 12
C. 14
D. `(50)/(9)`

Answer

To find the expected value of the random variable \( X \) assigned a value \( 2k \) when the sum of two dice is equal to \( k \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Possible Sums**: When two dice are thrown, the possible sums range from 2 to 12. The minimum sum (2) occurs when both dice show 1, and the maximum sum (12) occurs when both dice show 6.

2. **Calculate Probabilities for Each Sum**:
- The total number of outcomes when throwing two dice is \( 6 \times 6 = 36 \).
- We need to find the number of ways to achieve each sum from 2 to 12:
- Sum = 2: (1,1) → 1 way
- Sum = 3: (1,2), (2,1) → 2 ways
- Sum = 4: (1,3), (2,2), (3,1) → 3 ways
- Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4 ways
- Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 ways
- Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
- Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 ways
- Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 ways
- Sum = 10: (4,6), (5,5), (6,4) → 3 ways
- Sum = 11: (5,6), (6,5) → 2 ways
- Sum = 12: (6,6) → 1 way

3. **Probability Calculation**:
- Now, we can calculate the probability \( P(k) \) for each sum \( k \):
- \( P(2) = \frac{1}{36} \)
- \( P(3) = \frac{2}{36} = \frac{1}{18} \)
- \( P(4) = \frac{3}{36} = \frac{1}{12} \)
- \( P(5) = \frac{4}{36} = \frac{1}{9} \)
- \( P(6) = \frac{5}{36} \)
- \( P(7) = \frac{6}{36} = \frac{1}{6} \)
- \( P(8) = \frac{5}{36} \)
- \( P(9) = \frac{4}{36} = \frac{1}{9} \)
- \( P(10) = \frac{3}{36} = \frac{1}{12} \)
- \( P(11) = \frac{2}{36} = \frac{1}{18} \)
- \( P(12) = \frac{1}{36} \)

4. **Calculate Expected Value**:
- The expected value \( E(X) \) is calculated as follows:
\[ E(X) = \sum (2k \cdot P(k)) \]
- Substitute the values:
\[ E(X) = 2 \cdot 2 \cdot \frac{1}{36} + 2 \cdot 3 \cdot \frac{2}{36} + 2 \cdot 4 \cdot \frac{3}{36} + 2 \cdot 5 \cdot \frac{4}{36} + 2 \cdot 6 \cdot \frac{5}{36} + 2 \cdot 7 \cdot \frac{6}{36} + 2 \cdot 8 \cdot \frac{5}{36} + 2 \cdot 9 \cdot \frac{4}{36} + 2 \cdot 10 \cdot \frac{3}{36} + 2 \cdot 11 \cdot \frac{2}{36} + 2 \cdot 12 \cdot \frac{1}{36} \]
- Simplifying this gives:
\[ = \frac{4}{36} + \frac{12}{36} + \frac{24}{36} + \frac{40}{36} + \frac{60}{36} + \frac{84}{36} + \frac{80}{36} + \frac{72}{36} + \frac{60}{36} + \frac{44}{36} + \frac{24}{36} \]
- Adding these fractions:
\[ = \frac{504}{36} = 14 \]

5. **Conclusion**:
- The expected value of \( X \) is \( 14 \).