Two cards are dealt from a well-shuffled pack of 52 cards, one after the other, without replacement…

Step 1. The distribution. $P(X = 0) = \dfrac{48}{52} \times \dfrac{47}{51} = \dfrac{188}{221}$. $P(X = 2) = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{1}{221}$. $P(X = 1) = 2 \times \dfrac{4}{52} \times \dfrac{48}{51} = \dfrac{32}{221} \approx 0.1448$. The three add up to $1$.

Step 2. The mean. $E(X) = 1 \times \dfrac{32}{221} + 2 \times \dfrac{1}{221} = \dfrac{34}{221} = \dfrac{2}{13} \approx 0.1538$.

Step 3. The mean of the squares. $E(X^2) = \dfrac{32}{221} + 4 \times \dfrac{1}{221} = \dfrac{36}{221}$.

Step 4. The variance. $\dfrac{36}{221} - \left(\dfrac{2}{13}\right)^2 = \dfrac{400}{2873} \approx 0.1392$.