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 $\text{Var}(X) = n \cdot p \cdot (1 - p)$.