Linearity of expectation saves you here, which I didn't immediately reach for.
Use linearity of expectation by defining indicator variables for each rank, where the indicator is 1 if both cards of that rank appear in the 5-card hand. Compute the probability for one rank and multiply by 5, since expectation is linear regardless of dependence.
Pro tip: Mention that linearity of expectation works even when events are dependent, and note that the maximum number of pairs is 2 (since 5 cards can contain at most two complete pairs). This shows you understand the problem's constraints and the power of the technique.
For each rank i (1 to 5), let X_i be 1 if both cards of rank i are in the hand, and 0 otherwise. Then the total number of pairs is X = X_1 + X_2 + ... + X_5.
Calculate P(X_i = 1) = C(8,3)/C(10,5) = 56/252 = 2/9. This is the probability that both cards of a specific rank are included in the 5-card hand.
Since E[X] = E[X_1] + ... + E[X_5] = 5 * (2/9) = 10/9, the expected number of pairs is 10/9.
Verify that the answer is reasonable: the maximum possible pairs is 2, and 10/9 ≈ 1.11, which is plausible given the probabilities.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.