Competency-Based
Assessment Made Simple.

Let $X$ be the number of defective pens in the sample of 2.

Possible values of $X$ are 0, 1, and 2.

Total number of ways to choose 2 pens from 10 is $\binom{10}{2} = \frac{10 \times 9}{2} = 45$.

Case 1: $X = 0$ (no defective pens)

Number of ways to choose 2 non-defective pens from 7 is $\binom{7}{2} = \frac{7 \times 6}{2} = 21$.

$P(X=0) = \frac{21}{45} = \frac{7}{15}$.

Case 2: $X = 1$ (one defective pen)

Number of ways to choose 1 defective pen from 3 and 1 non-defective pen from 7 is $\binom{3}{1} \binom{7}{1} = 3 \times 7 = 21$.

$P(X=1) = \frac{21}{45} = \frac{7}{15}$.

Case 3: $X = 2$ (two defective pens)

Number of ways to choose 2 defective pens from 3 is $\binom{3}{2} = \frac{3 \times 2}{2} = 3$.

$P(X=2) = \frac{3}{45} = \frac{1}{15}$.

Now, we calculate the expected value $E(X)$:

$E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}$.

Next, we calculate $E(X^2)$:

$E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) + 2^2 \cdot P(X=2) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 4 \cdot \frac{1}{15} = \frac{7}{15} + \frac{4}{15} = \frac{11}{15}$.

Finally, we calculate the variance $Var(X)$:

$Var(X) = E(X^2) - [E(X)]^2 = \frac{11}{15} - \left(\frac{3}{5}\right)^2 = \frac{11}{15} - \frac{9}{25} = \frac{55 - 27}{75} = \frac{28}{75}$.