Compute the mean and variance of the number of heads that appear in
 3 flips of a fair coin.

To compute the **mean** and **variance** of the number of heads in 3 flips of a fair coin, we approach the problem as follows:

### Step 1: Define the random variable
Let \( X \) be the random variable representing the number of heads that appear in 3 flips of a fair coin. Each coin flip is a Bernoulli trial with two outcomes: heads (success) or tails (failure), where the probability of heads is \( p = \frac{1}{2} \).

Thus, \( X \) is the sum of 3 independent Bernoulli trials. We can write:
\[
X = X_1 + X_2 + X_3
\]
where each \( X_i \) is a Bernoulli random variable that takes the value 1 with probability \( \frac{1}{2} \) (heads) and 0 with probability \( \frac{1}{2} \) (tails).

### Step 2: Mean (Expected value)
The expected value of a Bernoulli random variable \( X_i \) is:
\[
E(X_i) = 1 \cdot p + 0 \cdot (1 - p) = p = \frac{1}{2}
\]
Since \( X \) is the sum of 3 independent Bernoulli trials:
\[
E(X) = E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3) = 3 \cdot \frac{1}{2} = \frac{3}{2}
\]
Thus, the mean (expected number of heads) is:
\[
E(X) = \frac{3}{2} = 1.5
\]

### Step 3: Variance
The variance of a Bernoulli random variable \( X_i \) is:
\[
\text{Var}(X_i) = p(1 - p) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
\]
Since \( X_1, X_2, X_3 \) are independent, the variance of their sum is the sum of their variances:
\[
\text{Var}(X) = \text{Var}(X_1 + X_2 + X_3) = \text{Var}(X_1) + \text{Var}(X_2) + \text{Var}(X_3) = 3 \cdot \frac{1}{4} = \frac{3}{4}
\]

### Final Answer:
- **Mean (Expected value)** of the number of heads: \( \frac{3}{2} = 1.5 \)
- **Variance** of the number of heads: \( \frac{3}{4} = 0.75 \)

Do you want further details on any specific part or another question?

Here are some related questions to expand on this concept:
1. What is the probability of getting exactly 2 heads in 3 flips of a fair coin?
2. How does the variance change if the number of flips is increased to 5?
3. What is the standard deviation of the number of heads in 3 flips of a fair coin?
4. Can you find the probability distribution of the number of heads in 3 flips?
5. What is the expected value if the coin is biased with \( p = 0.6 \) for heads?

**Tip**: The mean of a binomial distribution can be calculated as \( E(X) = n \cdot p \) and the variance as \( \text{Var}(X) = n \cdot p \cdot (1 - p) \).

Compute the mean and variance of the number of heads that appear in 3 flips of a fair coin.

Explore how to calculate the mean (expected value) and variance of the number of heads that appear in 3 flips of a fair coin. This solution includes key concepts like Bernoulli trials and the binomial distribution. Suitable for students in Grades 9-12.

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