A fair coin is tossed four times. Let X denote the number of heads occurring. Find the probability distribution, mean and variance of X.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Random Variable and Possible Outcomes
Let \( X \) denote the number of heads occurring when a fair coin is tossed four times. The possible values of \( X \) are \( 0, 1, 2, 3, \) and \( 4 \).

### Step 2: Calculate the Total Number of Outcomes
When a fair coin is tossed four times, the total number of outcomes is:
\[ 2^4 = 16 \]

### Step 3: Determine the Probability Distribution
We need to find the probability of each value of \( X \).

1. **Probability \( P(X = 0) \)**:
- The only outcome is TTTT (4 tails).
- Number of outcomes = 1.
- Probability:
\[ P(X = 0) = \frac{1}{16} \]

2. **Probability \( P(X = 1) \)**:
- The outcomes are: HTTT, THTT, TTHT, TTTT (1 head).
- Number of outcomes = 4.
- Probability:
\[ P(X = 1) = \frac{4}{16} = \frac{1}{4} \]

3. **Probability \( P(X = 2) \)**:
- The outcomes are: HHTT, HTHT, HTTH, THHT, THTH, TTHH (2 heads).
- Number of outcomes = 6.
- Probability:
\[ P(X = 2) = \frac{6}{16} = \frac{3}{8} \]

4. **Probability \( P(X = 3) \)**:
- The outcomes are: HHHT, HHTH, HTHH, THHH (3 heads).
- Number of outcomes = 4.
- Probability:
\[ P(X = 3) = \frac{4}{16} = \frac{1}{4} \]

5. **Probability \( P(X = 4) \)**:
- The only outcome is HHHH (4 heads).
- Number of outcomes = 1.
- Probability:
\[ P(X = 4) = \frac{1}{16} \]

### Step 4: Summarize the Probability Distribution
The probability distribution of \( X \) is:
\[ \begin{align*} P(X = 0) & = \frac{1}{16} \\ P(X = 1) & = \frac{1}{4} \\ P(X = 2) & = \frac{3}{8} \\ P(X = 3) & = \frac{1}{4} \\ P(X = 4) & = \frac{1}{16} \end{align*} \]

### Step 5: Calculate the Mean (Expected Value)
The mean \( E(X) \) is calculated as follows:
\[ E(X) = \sum (x_i \cdot P(X = x_i)) \]
Calculating each term:
\[ E(X) = 0 \cdot \frac{1}{16} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{16} \]
Calculating each term:
\[ = 0 + \frac{1}{4} + \frac{6}{8} + \frac{3}{4} + \frac{4}{16} \]
Converting to a common denominator (16):
\[ = 0 + \frac{4}{16} + \frac{12}{16} + \frac{12}{16} + \frac{4}{16} = \frac{32}{16} = 2 \]

### Step 6: Calculate the Variance
The variance \( Var(X) \) is calculated using:
\[ Var(X) = E(X^2) - (E(X))^2 \]
First, we need to find \( E(X^2) \):
\[ E(X^2) = \sum (x_i^2 \cdot P(X = x_i)) \]
Calculating each term:
\[ E(X^2) = 0^2 \cdot \frac{1}{16} + 1^2 \cdot \frac{1}{4} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{4} + 4^2 \cdot \frac{1}{16} \]
Calculating each term:
\[ = 0 + \frac{1}{4} + \frac{12}{8} + \frac{9}{4} + \frac{16}{16} \]
Converting to a common denominator (16):
\[ = 0 + \frac{4}{16} + \frac{24}{16} + \frac{36}{16} + \frac{16}{16} = \frac{80}{16} = 5 \]
Now, substituting into the variance formula:
\[ Var(X) = E(X^2) - (E(X))^2 = 5 - 2^2 = 5 - 4 = 1 \]

### Final Results
- Probability Distribution:
\[ \begin{align*} P(X = 0) & = \frac{1}{16} \\ P(X = 1) & = \frac{1}{4} \\ P(X = 2) & = \frac{3}{8} \\ P(X = 3) & = \frac{1}{4} \\ P(X = 4) & = \frac{1}{16} \end{align*} \]
- Mean \( E(X) = 2 \)
- Variance \( Var(X) = 1 \)