To solve the problem, we need to find the expected value of the random variable \( X \) defined as \( X = 2k \), where \( k \) is the sum of the numbers on two thrown dice.
### Step-by-Step Solution:
1. **Identify the Sample Space**:
When two dice are thrown, the total number of outcomes is \( 6 \times 6 = 36 \). Each die has 6 faces, so we can represent the outcomes as ordered pairs \((d_1, d_2)\) where \( d_1 \) and \( d_2 \) can take values from 1 to 6.
2. **Calculate the Possible Sums**:
The possible sums \( k \) when rolling two dice range from \( 2 \) (when both dice show 1) to \( 12 \) (when both dice show 6). The sums can be calculated as follows:
- \( k = 2 \): (1,1)
- \( k = 3 \): (1,2), (2,1)
- \( k = 4 \): (1,3), (2,2), (3,1)
- \( k = 5 \): (1,4), (2,3), (3,2), (4,1)
- \( k = 6 \): (1,5), (2,4), (3,3), (4,2), (5,1)
- \( k = 7 \): (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- \( k = 8 \): (2,6), (3,5), (4,4), (5,3), (6,2)
- \( k = 9 \): (3,6), (4,5), (5,4), (6,3)
- \( k = 10 \): (4,6), (5,5), (6,4)
- \( k = 11 \): (5,6), (6,5)
- \( k = 12 \): (6,6)
3. **Determine the Probability of Each Sum**:
We can summarize the probabilities for each sum \( k \):
- \( P(k=2) = \frac{1}{36} \)
- \( P(k=3) = \frac{2}{36} \)
- \( P(k=4) = \frac{3}{36} \)
- \( P(k=5) = \frac{4}{36} \)
- \( P(k=6) = \frac{5}{36} \)
- \( P(k=7) = \frac{6}{36} \)
- \( P(k=8) = \frac{5}{36} \)
- \( P(k=9) = \frac{4}{36} \)
- \( P(k=10) = \frac{3}{36} \)
- \( P(k=11) = \frac{2}{36} \)
- \( P(k=12) = \frac{1}{36} \)
4. **Assign Values to the Random Variable \( X \)**:
The random variable \( X \) is defined as \( X = 2k \). Therefore, the values of \( X \) corresponding to each \( k \) are:
- For \( k=2 \), \( X=4 \)
- For \( k=3 \), \( X=6 \)
- For \( k=4 \), \( X=8 \)
- For \( k=5 \), \( X=10 \)
- For \( k=6 \), \( X=12 \)
- For \( k=7 \), \( X=14 \)
- For \( k=8 \), \( X=16 \)
- For \( k=9 \), \( X=18 \)
- For \( k=10 \), \( X=20 \)
- For \( k=11 \), \( X=22 \)
- For \( k=12 \), \( X=24 \)
5. **Calculate the Expected Value \( E(X) \)**:
The expected value \( E(X) \) is calculated as follows:
\[
E(X) = \sum (X_i \cdot P(k_i))
\]
Plugging in the values:
\[
E(X) = 4 \cdot \frac{1}{36} + 6 \cdot \frac{2}{36} + 8 \cdot \frac{3}{36} + 10 \cdot \frac{4}{36} + 12 \cdot \frac{5}{36} + 14 \cdot \frac{6}{36} + 16 \cdot \frac{5}{36} + 18 \cdot \frac{4}{36} + 20 \cdot \frac{3}{36} + 22 \cdot \frac{2}{36} + 24 \cdot \frac{1}{36}
\]
Calculating each term:
\[
= \frac{4}{36} + \frac{12}{36} + \frac{24}{36} + \frac{40}{36} + \frac{60}{36} + \frac{84}{36} + \frac{80}{36} + \frac{72}{36} + \frac{60}{36} + \frac{44}{36} + \frac{24}{36}
\]
\[
= \frac{504}{36}
\]
Simplifying gives:
\[
E(X) = 14
\]
### Final Answer:
The expected value of \( X \) is \( 14 \).
