Pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Explanation

To solve the problem, we need to find the probability distribution, mean, and variance of the random variable \(X\), which denotes the minimum of the two numbers that appear when a pair of fair dice is thrown.

### Step 1: Determine the Sample Space
When two dice are thrown, the total number of outcomes is \(6 \times 6 = 36\).

### Step 2: Identify the Possible Values of \(X\)
The random variable \(X\) can take values from 1 to 6, as the minimum of two dice can be any of these numbers.

### Step 3: Calculate the Probability Distribution of \(X\)

1. **For \(X = 1\)**:
- The outcomes where the minimum is 1 are:
\((1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1)\).
- Total outcomes = 11.
- Probability \(P(X = 1) = \frac{11}{36}\).

2. **For \(X = 2\)**:
- The outcomes where the minimum is 2 are:
\((2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2)\).
- Total outcomes = 9.
- Probability \(P(X = 2) = \frac{9}{36} = \frac{1}{4}\).

3. **For \(X = 3\)**:
- The outcomes where the minimum is 3 are:
\((3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)\).
- Total outcomes = 7.
- Probability \(P(X = 3) = \frac{7}{36}\).

4. **For \(X = 4\)**:
- The outcomes where the minimum is 4 are:
\((4,4), (4,5), (4,6), (5,4), (6,4)\).
- Total outcomes = 5.
- Probability \(P(X = 4) = \frac{5}{36}\).

5. **For \(X = 5\)**:
- The outcomes where the minimum is 5 are:
\((5,5), (5,6), (6,5)\).
- Total outcomes = 3.
- Probability \(P(X = 5) = \frac{3}{36} = \frac{1}{12}\).

6. **For \(X = 6\)**:
- The only outcome where the minimum is 6 is:
\((6,6)\).
- Total outcomes = 1.
- Probability \(P(X = 6) = \frac{1}{36}\).

### Step 4: Write the Probability Distribution
The probability distribution of \(X\) is as follows:

\[ \begin{align*} P(X = 1) & = \frac{11}{36} \\ P(X = 2) & = \frac{9}{36} = \frac{1}{4} \\ P(X = 3) & = \frac{7}{36} \\ P(X = 4) & = \frac{5}{36} \\ P(X = 5) & = \frac{3}{36} = \frac{1}{12} \\ P(X = 6) & = \frac{1}{36} \\ \end{align*} \]

### Step 5: Calculate the Mean of \(X\)
The mean (expected value) \(E(X)\) is calculated as follows:

\[ E(X) = \sum (x_i \cdot P(X = x_i)) \]
\[ E(X) = 1 \cdot \frac{11}{36} + 2 \cdot \frac{9}{36} + 3 \cdot \frac{7}{36} + 4 \cdot \frac{5}{36} + 5 \cdot \frac{3}{36} + 6 \cdot \frac{1}{36} \]
\[ E(X) = \frac{11 + 18 + 21 + 20 + 15 + 6}{36} = \frac{91}{36} \approx 2.53 \]

### Step 6: Calculate the Variance of \(X\)
The variance \(Var(X)\) is calculated using the formula:

\[ Var(X) = E(X^2) - (E(X))^2 \]

First, we calculate \(E(X^2)\):

\[ E(X^2) = \sum (x_i^2 \cdot P(X = x_i)) \]
\[ E(X^2) = 1^2 \cdot \frac{11}{36} + 2^2 \cdot \frac{9}{36} + 3^2 \cdot \frac{7}{36} + 4^2 \cdot \frac{5}{36} + 5^2 \cdot \frac{3}{36} + 6^2 \cdot \frac{1}{36} \]
\[ E(X^2) = \frac{11 + 36 + 63 + 80 + 75 + 36}{36} = \frac{301}{36} \approx 8.36 \]

Now, substituting back to find variance:

\[ Var(X) = E(X^2) - (E(X))^2 = \frac{301}{36} - \left(\frac{91}{36}\right)^2 \]
\[ Var(X) = \frac{301}{36} - \frac{8281}{1296} \]
\[ Var(X) = \frac{301 \cdot 36 - 8281}{1296} = \frac{10836 - 8281}{1296} = \frac{2555}{1296} \approx 1.96 \]

### Final Results
- **Probability Distribution**:
\[ P(X = 1) = \frac{11}{36}, P(X = 2) = \frac{9}{36}, P(X = 3) = \frac{7}{36}, P(X = 4) = \frac{5}{36}, P(X = 5) = \frac{3}{36}, P(X = 6) = \frac{1}{36} \]

- **Mean**: \(E(X) \approx 2.53\)

- **Variance**: \(Var(X) \approx 1.96\)