A die is tossed once. If the random variable X is defined as X = cases 1, & if the die result in an odd number \\ -1, & if the die result in an even number cases, then the variance of X is Option 1: 0 Option 2: 1 Option 3: 1/2 Option 4: 1/4

Answer: 1. Solution: When a die is rolled, the outcomes are: 1, 2, 3, 4, 5, or 6 Odd numbers (1, 3, 5) give X = 1 Even numbers (2, 4, 6) give X = -1 Out of 6 possible outcomes: 3 are odd, so P(X = 1) = 3/6 = 1/2 3 are even, so P(X = -1) = 3/6 = 1/2 The expected value is: E(X) = (1) x 1/2 + (-1) x 1/2 E(X) = 1/2 - 1/2 E(X) = 0 The expected value of X^2 is: E(X^2) = (1)^2 x 1/2 + (-1)^2 x 1/2 E(X^2) = 1 x 1/2 + 1 x 1/2 E(X^2) = 1/2 + 1/2 E(X^2) = 1 Using the variance formula: Var(X) = E(X^2) - [E(X)]^2 Var(X) = 1 - (0)^2 Var(X) = 1 - 0 Var(X) = 1 Therefore, the variance of X is 1.

Algebra

Probability

Medium

core

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

$\frac{1}{2}$

$\frac{1}{4}$

✅ Correct Option: 2

When a die is rolled, the outcomes are: 1, 2, 3, 4, 5, or 6

Odd numbers (1, 3, 5) give $X = 1$

Even numbers (2, 4, 6) give $X = -1$

Out of 6 possible outcomes:

3 are odd, so $P(X = 1) = \dfrac{3}{6} = \dfrac{1}{2}$

3 are even, so $P(X = -1) = \dfrac{3}{6} = \dfrac{1}{2}$

The expected value is:

$E(X) = (1) \times \dfrac{1}{2} + (-1) \times \dfrac{1}{2}$

$E(X) = \dfrac{1}{2} - \dfrac{1}{2}$

$E(X) = 0$

The expected value of $X^2$ is:

$E(X^2) = (1)^2 \times \dfrac{1}{2} + (-1)^2 \times \dfrac{1}{2}$

$E(X^2) = 1 \times \dfrac{1}{2} + 1 \times \dfrac{1}{2}$

$E(X^2) = \dfrac{1}{2} + \dfrac{1}{2}$

$E(X^2) = 1$

Using the variance formula:

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

$Var(X) = 1 - (0)^2$

$Var(X) = 1 - 0$

$Var(X) = 1$

Therefore, the variance of $X$ is $1$ .

2026: 30 May Shift 1

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If answer is correct, the probability that he guesses, is :

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A farmer has a tubewell which works by connecting two switches $S_1$ and $S_2$ . The probability that $S_1$ works is 9/10 and the probability that $S_2$ works is 8/10. The farmer connects them in series in 30% cases and in parallel in 70% cases. If it is known that the tubewell was working, what is the probability that it was a series connection.

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The value of $P(E)$ is

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