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A bakery's morning tray holds 12 croissants, of which 3 were underbaked and 9 were baked properly. A quality inspector picks 2 of the croissants from the tray at random, one after the other and without returning the first before picking the second. What is the probability that at least one of the 2 croissants the inspector picks is underbaked?
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Correct answer
C
Complement counting on two draws made without replacement. The event "at least one underbaked" is the opposite of "both baked properly," and the complement is far cleaner to compute than the direct casework.
Step 1, find the probability both are properly baked. The first pick is properly baked with probability 9/12. After removing that croissant the tray holds 11, of which 8 are properly baked, so the second pick is properly baked with probability 8/11. Multiply these dependent stages: (9/12) × (8/11) = (3/4) × (8/11) = 24/44 = 6/11.
Step 2, take the complement: P(at least one underbaked) = 1 - 6/11 = 5/11. That is choice C.
Check by direct counting: there are C(12,2) = 66 equally likely pairs. Pairs with at least one underbaked = C(3,1) × C(9,1) + C(3,2) = 27 + 3 = 30, and 30/66 = 5/11. The two methods agree.
Why the distractors are tempting. Choice A, 1/4, is the chance a single croissant is underbaked and ignores that two are drawn. Choice B, 7/16, comes from treating the draws as with replacement, 1 minus (9/12)(9/12), which wrongly keeps 12 croissants for the second pick. Choice D, 1/2, adds the per-draw chances 1/4 + 1/4 instead of subtracting the both-baked-properly case, which double counts the both-underbaked outcome. Choice E, 6/11, is the probability that both are baked properly and skips the final step of subtracting from 1.
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