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This problem involves flipping two coins 50 times, and it asks you to analyze the number of times both coins show "heads" ("Zahl"). The questions revolve around using the Bernoulli distribution and its derived binomial distribution.

Let’s go step by step:

### a) Why is this a Bernoulli process? Provide the parameters.
This scenario is a Bernoulli process because:
- **Bernoulli trials** are independent experiments with exactly two outcomes (success or failure). In this case, success means both coins show "heads."
- Each trial (coin flip) is independent of the others.
  
The parameter for each trial (denoted \( p \)) is the probability of success, which is the probability that both coins land on "heads." Since each coin flip has a probability of 0.5 to show "heads," the probability that both show "heads" simultaneously is:

\[
p = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25
\]

### b) Calculate the probabilities:
This is a binomial distribution problem with \( n = 50 \) trials and \( p = 0.25 \) as the probability of success.

#### i. Probability of exactly 12 successes (both coins showing "heads" 12 times):
The binomial formula is:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

For \( n = 50 \), \( k = 12 \), and \( p = 0.25 \), we can calculate \( P(X = 12) \).

#### ii. Probability of at least 12 successes:
\[
P(X \geq 12) = 1 - P(X < 12)
\]

We will sum the probabilities from 12 to 50.

#### iii. Probability of at most 12 successes:
This is simply \( P(X \leq 12) \), which can be calculated by summing up the probabilities from 0 to 12.

#### iv. Probability that the number of successes is between 15 and 30:
\[
P(15 \leq X \leq 30) = \sum_{k=15}^{30} P(X = k)
\]

### c) Expected value of \( X \):
The expected value \( E(X) \) of a binomial distribution is:

\[
E(X) = n \cdot p = 50 \cdot 0.25 = 12.5
\]

### d) Standard deviation of \( X \):
The standard deviation \( \sigma \) of a binomial distribution is:

\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{50 \cdot 0.25 \cdot 0.75} \approx 3.06
\]

### e) Interval for 95% probability:
For a binomial distribution, you can approximate this using a normal distribution. The interval will be:

\[
[\mu - 1.96 \cdot \sigma, \mu + 1.96 \cdot \sigma]
\]

This gives the range where approximately 95% of the data falls.

Would you like me to calculate any specific parts or clarify further?

---

#### 5 Follow-up Questions:
1. What other types of distributions are related to the Bernoulli process?
2. How does the binomial distribution relate to the normal distribution for large \( n \)?
3. Could the probabilities in part b be calculated using other methods (e.g., normal approximation)?
4. How does changing the number of coin flips affect the expected value and standard deviation?
5. Can you apply a similar binomial distribution model in real-world situations?

#### Tip:
For binomial distributions, as \( n \) becomes large, the distribution can be approximated by a normal distribution to simplify probability calculations.

Es werden fünfzigmal zwei Münzen gleichzeitig geworfen. Die Zufallsvariable X zählt, wie oft beide Münzen 'Zahl' zeigen. Berechnen Sie die Wahrscheinlichkeiten und statistischen Kennwerte für verschiedene Fälle.

Explore how to calculate probabilities using a binomial distribution based on 50 coin flips, where both coins must show 'heads'. This involves the Bernoulli process, expected value, standard deviation, and calculating probabilities for exact and range-based successes. Ideal for advanced high school or early college students.

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