Question

Zwei Würfel werden geworfen. Berechnen Sie den Mittelwert der Zufallsvariablen, „Summe der Augenzahlen".

Short answer

The mean of the sum of numbers on two dice is 7.

Step-by-step solution

Understand the Problem

Two dice are thrown, and we need to find the mean (average) of the sum of the numbers that come up.

List Possible Outcomes

Consider the possible outcomes when two dice are thrown. Each die has 6 faces, so there are a total of 6 × 6 = 36 possible outcomes.

Compute the Possible Sums

List all possible sums of the numbers on the two dice. These sums range from 2 (1+1) to 12 (6+6).

Determine the Frequency of Each Sum

Calculate how many times each sum occurs: - 2: (1 outcome) - 3: (2 outcomes) - 4: (3 outcomes) - 5: (4 outcomes) - 6: (5 outcomes) - 7: (6 outcomes) - 8: (5 outcomes) - 9: (4 outcomes) - 10: (3 outcomes) - 11: (2 outcomes) - 12: (1 outcome)

Calculate the Total Sum and Frequency

Multiply each sum by its frequency and add them together to get the total sum: \[ 2 \times 1 + 3 \times 2 + 4 \times 3 + 5 \times 4 + 6 \times 5 + 7 \times 6 + 8 \times 5 + 9 \times 4 + 10 \times 3 + 11 \times 2 + 12 \times 1 = 252 \]Total frequency is the number of possible outcomes, which is 36.

Calculate the Mean

Divide the total sum by the total frequency to find the mean: \[ \frac{252}{36} = 7 \]

Key concepts

Probability

Probability is a measure of how likely an event is to occur. When throwing two dice, each face has an equal chance of showing up. This means each of the 36 possible outcomes (6 faces on die 1 x 6 faces on die 2) has an equal probability of occurring. The probability of any specific outcome is calculated by taking the number of ways that outcome can occur and dividing it by the total number of possible outcomes.
For instance, the probability of rolling a sum of 7, which can occur in 6 different ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), is \( \frac{6}{36} \) or \( \frac{1}{6} \).

Random Variables

In probability, a random variable is a numerical description of the outcome of a random phenomenon. For the exercise, the random variable is the sum of the numbers on two dice. Random variables can be discrete or continuous. Here, we deal with a discrete random variable since the sum can only take certain distinct values (from 2 to 12). Each possible sum is assigned a probability based on its frequency of occurrence.

Expected Value

The expected value (or mean) of a random variable provides a measure of the 'center' of its distribution. It is calculated as the sum of all possible values, each weighted by its probability of occurrence.
In our example, the expected value of the sum of two dice is determined by summing all possible sums (2 through 12), each multiplied by the corresponding probability.
Mathematically, it is written as:
\[ E(X) = \sum{(x_i \cdot p(x_i))} \]
where \(x_i\) are the possible sums and \(p(x_i)\) are their corresponding probabilities. The expected value in this case is 7, as calculated step-by-step.

Dice Outcomes

Dice outcomes are straightforward since each die has 6 faces numbered from 1 to 6. When two dice are thrown, each die operates independently. Hence, each possible outcome (like rolling a 3 on the first die and a 4 on the second) is equally likely.
The sums of these outcomes range from 2 (1+1) to 12 (6+6). Some sums are more frequent because there are more combinations to achieve them. For example, a sum of 7 occurs in 6 different ways, while a sum of 2 only has 1 possible combination (1+1).

Frequency Distribution

A frequency distribution shows how often each possible outcome of a random variable occurs. For rolling two dice, the frequency distribution of sums wouldlist the sums and how often each one occurs in a total of 36 rolls.
Here’s the frequency of each sum:

• Sum 2: 1 time
• Sum 3: 2 times
• Sum 4: 3 times
• Sum 5: 4 times
• Sum 6: 5 times

Continue this for sums up to 12. A frequency distribution is fundamental for determining probabilities and expected values as it provides the data needed to perform these calculations.