Find the mean and variance of the number of heads when two coins are tossed simultaneously.

Explanation

To solve the problem of finding the mean and variance of the number of heads when two coins are tossed simultaneously, we can follow these steps:

### Step 1: Define the Sample Space
When two coins are tossed, the possible outcomes (sample space) are:
- HH (2 heads)
- HT (1 head)
- TH (1 head)
- TT (0 heads)

Thus, the sample space \( S \) is:
\[ S = \{ HH, HT, TH, TT \} \]

### Step 2: Define the Random Variable
Let \( X \) be the random variable representing the number of heads obtained. The possible values of \( X \) are:
- \( X = 0 \) (for TT)
- \( X = 1 \) (for HT and TH)
- \( X = 2 \) (for HH)

### Step 3: Determine the Probability Distribution
Next, we calculate the probabilities for each value of \( X \):
- \( P(X = 0) = \frac{1}{4} \) (1 outcome: TT)
- \( P(X = 1) = \frac{2}{4} = \frac{1}{2} \) (2 outcomes: HT, TH)
- \( P(X = 2) = \frac{1}{4} \) (1 outcome: HH)

Thus, the probability distribution is:
- \( P(X = 0) = \frac{1}{4} \)
- \( P(X = 1) = \frac{1}{2} \)
- \( P(X = 2) = \frac{1}{4} \)

### Step 4: Calculate the Mean (Expected Value)
The mean (expected value) \( \mu \) is calculated using the formula:
\[ \mu = E(X) = \sum (x \cdot P(X = x)) \]
Calculating this:
\[ \mu = (0 \cdot \frac{1}{4}) + (1 \cdot \frac{1}{2}) + (2 \cdot \frac{1}{4}) \]
\[ = 0 + \frac{1}{2} + \frac{2}{4} \]
\[ = 0 + \frac{1}{2} + \frac{1}{2} = 1 \]

### Step 5: Calculate the Variance
The variance \( \sigma^2 \) is calculated using the formula:
\[ \sigma^2 = E(X^2) - \mu^2 \]
First, we need to calculate \( E(X^2) \):
\[ E(X^2) = \sum (x^2 \cdot P(X = x)) \]
Calculating this:
\[ E(X^2) = (0^2 \cdot \frac{1}{4}) + (1^2 \cdot \frac{1}{2}) + (2^2 \cdot \frac{1}{4}) \]
\[ = 0 + \frac{1}{2} + \frac{4}{4} \]
\[ = 0 + \frac{1}{2} + 1 = \frac{3}{2} \]

Now, substituting into the variance formula:
\[ \sigma^2 = E(X^2) - \mu^2 = \frac{3}{2} - 1^2 \]
\[ = \frac{3}{2} - 1 = \frac{3}{2} - \frac{2}{2} = \frac{1}{2} \]

### Final Results
- Mean \( \mu = 1 \)
- Variance \( \sigma^2 = \frac{1}{2} \)