To calculate the expected value $E(X)$ of the given random variable $X$, we need to consider the outcomes when two dice are rolled. The value of $X$ is defined as:

• The higher of the two numbers if the dice show different values.
• The common number (the number shown on both dice) if doubles are rolled.

Step 1: Define the possible outcomes

Each die has 6 faces, so there are a total of $6 \times 6 = 36$ possible outcomes when two dice are rolled.

The possible values of $X$ are from 1 to 6, with different probabilities depending on whether the dice show doubles or not.

Step 2: Compute the probabilities of each value of $X$

• When $X = 1$: This happens only when both dice show a 1, i.e., (1,1). There is 1 outcome: $P(X = 1) = \frac{1}{36}$.
• When $X = 2$: This occurs in two cases: (2,2) and (2,1). The probability is $P(X = 2) = \frac{3}{36} = \frac{1}{12}$.
• When $X = 3$: This occurs in three cases: (3,3), (3,2), (3,1). The probability is $P(X = 3) = \frac{5}{36}$.
• When $X = 4$: This occurs in four cases: (4,4), (4,3), (4,2), (4,1). The probability is $P(X = 4) = \frac{7}{36}$.
• When $X = 5$: This occurs in five cases: (5,5), (5,4), (5,3), (5,2), (5,1). The probability is $P(X = 5) = \frac{9}{36} = \frac{1}{4}$.
• When $X = 6$: This occurs in six cases: (6,6), (6,5), (6,4), (6,3), (6,2), (6,1). The probability is $P(X = 6) = \frac{11}{36}$.

Step 3: Calculate the expected value

The expected value $E(X)$ is calculated as the weighted sum of all possible values of $X$, using their respective probabilities:

$E(X) = 1 \cdot \frac{1}{36} + 2 \cdot \frac{3}{36} + 3 \cdot \frac{5}{36} + 4 \cdot \frac{7}{36} + 5 \cdot \frac{9}{36} + 6 \cdot \frac{11}{36}$

$E(X) = \frac{1}{36} + \frac{6}{36} + \frac{15}{36} + \frac{28}{36} + \frac{45}{36} + \frac{66}{36}$

$E(X) = \frac{161}{36} \approx 4.472$

Final Answer:

The expected value $E(X)$ is approximately 4.472.

Would you like more details or further clarification? Here are some related questions:

1. How does the calculation of the expected value change if the dice are not fair?
2. What is the variance of this random variable $X$?
3. How would the expected value change if the definition of $X$ was the sum of the two dice instead?
4. What is the expected value if the random variable is defined as the minimum of the two dice rolls?
5. How would the probabilities and expected value change if we used more than two dice?

Tip: When calculating expected value, always ensure that the sum of probabilities adds up to 1, as a check of your work.