If we take 8 identical slips of paper and write the number 0 on one of them, the number 1 on three of the slips, the number 2 on three of the slips and the number 3 on one of the slips. These slips are folded, put in a box and roughly mixed. One slip is drawn at random from the box. If X is the random variable denoting the number written on the drawn slip, the variance of X is:

We have 8 slips of paper with numbers:

• 1 slip has the number 0
• 3 slips have the number 1
• 3 slips have the number 2
• 1 slip has the number 3

When randomly picking one slip, X represents the number on that slip.

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The probability distribution for X:

$P(X = 0) = \frac{1}{8}$

$P(X = 1) = \frac{3}{8}$

$P(X = 2) = \frac{3}{8}$

$P(X = 3) = \frac{1}{8}$

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The expected value $E(X)$:

$E(X) = \sum x \cdot P(x)$

$E(X) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}$

$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}$

$E(X) = \frac{12}{8}$

$E(X) = \frac{3}{2}$

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The value of $E(X^2)$:

$E(X^2) = \sum x^2 \cdot P(x)$

$E(X^2) = 0^2 \times \frac{1}{8} + 1^2 \times \frac{3}{8} + 2^2 \times \frac{3}{8} + 3^2 \times \frac{1}{8}$

$E(X^2) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 4 \times \frac{3}{8} + 9 \times \frac{1}{8}$

$E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8}$

$E(X^2) = \frac{24}{8}$

$E(X^2) = 3$

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The variance of X:

$\text{Var}(X) = E(X^2) - [E(X)]^2$

$\text{Var}(X) = 3 - \left(\frac{3}{2}\right)^2$

$\text{Var}(X) = 3 - \frac{9}{4}$

$\text{Var}(X) = \frac{12}{4} - \frac{9}{4}$

$\text{Var}(X) = \frac{3}{4}$

Therefore, the variance of X is $\frac{3}{4}$.