Problem 4 - Arbeitsbuch Mathematik für Ingenieure

Question

Zwei unterscheidbare Würfel werden gleichzeitig geworfen. Prüfen Sie, ob die folgenden Ereignisse \(A, B\) und \(C\) jeweils paarweise unabhängig sind: a) Ereignis A: „Die Würfel zeigen gleiche Augenzahl". b) Ereignis \(B\) :, Der zweite Würfel zeigt eine 4 ". c) Ereignis \(C\) : „Die Augensumme der beiden Würfel ist durch 4 teilbar".

Step-by-Step Solution

Answer

A and B are independent, but neither of them is independent from C.

Step 1 - Define Events and Sample Space

Determine the sample space for two distinguishable dice. Each die has 6 faces, so the sample space has 36 outcomes: \[ S = \{ (1,1), (1,2), (1,3), ..., (6,6) \} \]. Define the events: \( A \): 'Both dice show the same number' \( B \): 'The second die shows a 4' \( C \): 'The sum of both dice is divisible by 4'.

Step 2 - Determine Probabilities of Events

Calculate the probabilities of each event:\[ P(A) = \frac{6}{36} = \frac{1}{6} \]There are 6 outcomes where both dice show the same number. \[ P(B) = \frac{6}{36} = \frac{1}{6} \]There are 6 outcomes where the second die shows 4. Identify the sums divisible by 4: 4, 8, 12. The corresponding favorable outcomes are: (1,3), (2,2), (3,1), (3,5), (4,4), (5,3), (6,2). \[ P(C) = \frac{9}{36} = \frac{1}{4} \]There are 9 such outcomes.

Step 3 - Check Pairwise Independence

Check independence for each pair of events by verifying \( P(A \cap B) = P(A) \cdot P(B) \): \[ P(A \cap B) = P((4,4)) = \frac{1}{36} \]\[ P(A) \cdot P(B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \]Since \( P(A \cap B) = P(A) \cdot P(B) \), A and B are independent. Check for A and C:\[ P(A \cap C) = \frac{2}{36} = \frac{1}{18} \]There are 2 outcomes: (2,2), (4,4).\[ P(A) \cdot P(C) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} \]Since \( P(A \cap C) e P(A) \cdot P(C) \), A and C are not independent. Check for B and C:\[ P(B \cap C) = \frac{3}{36} = \frac{1}{12} \]There are 3 outcomes: (1,3), (4,4), (6,2).\[ P(B) \cdot P(C) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} \]Since \( P(B \cap C) e P(B) \cdot P(C) \), B and C are not independent.

Key Concepts

independent events

In probability theory, two events are considered 'independent' if the occurrence of one event does not affect the probability of the other event occurring. This concept is crucial because it helps to determine how events interplay in a given scenario. To test for independence, we check whether the probability of the intersection of two events equals the product of their individual probabilities. Mathematically, two events A and B are independent if and only if:
\( P(A \cap B) = P(A) \cdot P(B) \)
In the provided exercise, the independence of events A (both dice show the same number) and B (the second die shows a 4) is tested by checking if \( P(A \cap B) = P(A) \cdot P(B) \) holds true.

sample space

The 'sample space' in probability theory refers to the set of all possible outcomes of a random experiment. For example, when rolling two distinguishable dice, each face of the first die can pair with each face of the second die, resulting in 36 possible outcomes: \[ S = \{ (1,1), (1,2), (1,3), ..., (6,6) \} \]
Understanding the sample space helps to visualize and calculate the likelihood of different events. In the given exercise, specific events like the dice showing the same number, the second die showing 4, and the sum of dice being divisible by 4 are subsets of this sample space. Each event has its designated outcomes within the sample space which are used to calculate their probabilities.

probability calculation

Calculating probability involves determining how likely an event is to occur within the sample space. This probability is the ratio of the number of favorable outcomes to the total number of outcomes in the sample space. For example, if there are 6 favorable outcomes for event A and 36 possible outcomes in total, the probability of A is:
\[ P(A) = \frac{6}{36} = \frac{1}{6} \]
In the exercise, the probabilities for each event are calculated in this way:
\$ P(A) = \frac{6}{36} = \frac{1}{6} \$
\$ P(B) = \frac{6}{36} = \frac{1}{6} \$
\$ P(C) = \frac{9}{36} = \frac{1}{4} \$
These values are used to verify pairwise independence among the events by checking the intersection probabilities.

pairwise independence

Pairwise independence occurs when any two events in a set are independent of each other. To check if three events A, B, and C are pairwise independent, you need to verify the independence of each pair:

• A and B
• A and C
• B and C

For each pair, you confirm:
\ P(A \cap B) = P(A) \cdot P(B) \
\ P(A \cap C) = P(A) \cdot P(C) \
\ P(B \cap C) = P(B) \cdot P(C) \
In the exercise, event pairs were checked as follows:
For A and B: \[ P(A \cap B) = P((4,4)) = \frac{1}{36} \cdot P(A) = \cdot P(B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \]
For A and C: \[ P(A \cap C) = \frac{2}{36} = \frac{1}{18} \cdot P(A) = \cdot P(C) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} \]
For B and C: \[ P(B \cap C) = \frac{3}{36} = \frac{1}{12} \cdot P(B) = \cdot P(C) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} \]
Thus, only A and B were found to be independent.