A die is tossed once. If the random variable X is defined as $X =\left\{\begin{matrix}1,&\text{if the die result in an odd number}\\-1,&\text{if the die result in an even number}\end{matrix}\right.$, then the variance of X is

1

The correct answer is Option (2) → 1

Given: A die is tossed once and random variable $X$ is defined as:

X = \[ \begin{cases} 1, & \text{if die result is odd (1, 3, 5)} \\ -1, & \text{if die result is even (2, 4, 6)} \end{cases} \]

Each die outcome has probability $\frac{1}{6}$. So,

$P(X = 1) = \frac{3}{6} = \frac{1}{2}$ and $P(X = -1) = \frac{3}{6} = \frac{1}{2}$

Mean: $\mathbb{E}(X) = 1 \cdot \frac{1}{2} + (-1) \cdot \frac{1}{2} = 0$

Variance: $\text{Var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$

Now, $X^2 = 1$ always. So $\mathbb{E}(X^2) = 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = 1$