To calculate the expected value $E(X)$ of the random variable $X$, where $X$ is the lower number when two dice are rolled, let's break down the problem step by step.

Step 1: Understanding the Random Variable $X$

When two dice are rolled, $X$ represents the lower of the two numbers. The possible values of $X$ are 1, 2, 3, 4, 5, or 6. Since both dice are identical and independent, we need to consider all the possible outcomes for each pair of dice.

Step 2: Determining the Probabilities

We will calculate the probabilities of each possible value of $X$, taking into account the symmetry and independence of the dice.

For $X = 1$:

• The lower number is 1 if either die shows 1. The possible combinations are:
  • (1,1), (1,2), (1,3), (1,4), (1,5), (1,6).
• So, there are 6 outcomes where $X = 1$.

For $X = 2$:

• The lower number is 2 if both dice show at least 2. The possible combinations are:
  • (2,2), (2,3), (2,4), (2,5), (2,6).
• So, there are 5 outcomes where $X = 2$.

For $X = 3$:

• The lower number is 3 if both dice show at least 3. The possible combinations are:
  • (3,3), (3,4), (3,5), (3,6).
• So, there are 4 outcomes where $X = 3$.

For $X = 4$:

• The lower number is 4 if both dice show at least 4. The possible combinations are:
  • (4,4), (4,5), (4,6).
• So, there are 3 outcomes where $X = 4$.

For $X = 5$:

• The lower number is 5 if both dice show at least 5. The possible combinations are:
  • (5,5), (5,6).
• So, there are 2 outcomes where $X = 5$.

For $X = 6$:

• The lower number is 6 only if both dice show 6. The only combination is:
  • (6,6).
• So, there is 1 outcome where $X = 6$.

Step 3: Computing the Probabilities

The total number of possible outcomes when rolling two dice is $6 \times 6 = 36$. So, the probabilities of each value of $X$ are as follows:

• $P(X = 1) = \frac{6}{36} = \frac{1}{6}$
• $P(X = 2) = \frac{5}{36}$
• $P(X = 3) = \frac{4}{36} = \frac{1}{9}$
• $P(X = 4) = \frac{3}{36} = \frac{1}{12}$
• $P(X = 5) = \frac{2}{36} = \frac{1}{18}$
• $P(X = 6) = \frac{1}{36}$

Step 4: Expected Value Formula

The expected value $E(X)$ is calculated as:

$E(X) = \sum_{x=1}^{6} x \cdot P(X = x)$

Substituting the probabilities:

$E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{5}{36} + 3 \cdot \frac{1}{9} + 4 \cdot \frac{1}{12} + 5 \cdot \frac{1}{18} + 6 \cdot \frac{1}{36}$

Step 5: Simplifying the Expression

Let me now compute the expected value.

$E(X) = \frac{1}{6} + \frac{10}{36} + \frac{12}{36} + \frac{12}{36} + \frac{10}{36} + \frac{6}{36}$

Simplifying:

$E(X) = \frac{1}{6} + \frac{50}{36} = \frac{6}{36} + \frac{50}{36} = \frac{56}{36} = \frac{14}{9}$

Thus, the expected value $E(X)$ is:

$E(X) = \frac{14}{9} \approx 1.56$

Final Answer:

The expected value $E(X)$ is $\frac{14}{9}$.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How would the expected value change if one die was weighted?
2. What is the probability distribution of the higher number when two dice are rolled?
3. How do you calculate the expected value of non-numeric outcomes?
4. What is the variance of the lower number when two dice are rolled?
5. How does this problem relate to conditional probability?

Tip: When dealing with dice problems, symmetry can simplify complex calculations.