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Model the draws sequentially with changing denominators (no replacement). Probability the first is defective: (\dfrac{3}{8}). After removing a defective, 2 defective remain out of 7 total: (P(\text{second defective} \mid \text{first defective}) = \dfrac{2}{7}). Multiply:

(\dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} = \dfrac{3}{28}).

Thus, A is correct. Option B ((\tfrac{2}{7})) ignores the first-draw probability and treats the second step as if it stood alone. C ((\tfrac{5}{28})) reflects arithmetic or counting errors. D ((\tfrac{3}{56})) divides by the wrong denominator twice (treating events as independent with a fixed 8 or erroneously halving). TEAS probability questions are simplified by a tree diagram or by combinations:

(\dfrac{\binom{3}{2}}{\binom{8}{2}} = \dfrac{3}{28}) gives the same result using unordered pairs. Always note with vs without replacement: if replaced, both factors would be (\frac{3}{8}), giving (\frac{9}{64}). Confirm your final fraction is simplified. A quick sense check: With only 3 defectives in 8, drawing two in a row should be unlikely (≈0.107), which matches (\frac{3}{28}).