Find the variance of the number obtained on a throw of an unbiased die.

Answer

To find the variance of the number obtained on a throw of an unbiased die, we can follow these steps:

### Step 1: Identify the outcomes and their probabilities
The outcomes when throwing an unbiased die are 1, 2, 3, 4, 5, and 6. Each outcome has an equal probability of occurring, which is \( \frac{1}{6} \).

### Step 2: Calculate the expected value (E(X))
The expected value \( E(X) \) is calculated using the formula:
\[ E(X) = \sum_{i=1}^{n} x_i \cdot P(x_i) \]
Where \( x_i \) are the outcomes and \( P(x_i) \) are the probabilities.

Calculating \( E(X) \):
\[ E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \]
\[ E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5 \]

### Step 3: Calculate the expected value of \( X^2 \) (E(X²))
Next, we calculate \( E(X^2) \) using the formula:
\[ E(X^2) = \sum_{i=1}^{n} x_i^2 \cdot P(x_i) \]
Calculating \( E(X^2) \):
\[ E(X^2) = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{6} + 3^2 \cdot \frac{1}{6} + 4^2 \cdot \frac{1}{6} + 5^2 \cdot \frac{1}{6} + 6^2 \cdot \frac{1}{6} \]
\[ E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} \]

### Step 4: Calculate the variance
The variance \( Var(X) \) is calculated using the formula:
\[ Var(X) = E(X^2) - (E(X))^2 \]
Substituting the values we found:
\[ Var(X) = \frac{91}{6} - \left(\frac{21}{6}\right)^2 \]
Calculating \( (E(X))^2 \):
\[ (E(X))^2 = \left(\frac{21}{6}\right)^2 = \frac{441}{36} \]
Now, convert \( \frac{91}{6} \) to a fraction with a denominator of 36:
\[ \frac{91}{6} = \frac{91 \times 6}{6 \times 6} = \frac{546}{36} \]
Now substitute back into the variance formula:
\[ Var(X) = \frac{546}{36} - \frac{441}{36} = \frac{105}{36} \]
Simplifying:
\[ Var(X) = \frac{35}{12} \]

### Final Answer
Thus, the variance of the number obtained on a throw of an unbiased die is \( \frac{35}{12} \).

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