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A bin contains 8 identical-looking widgets, of which exactly 3 are defective and 5 are good. An inspector draws the widgets one at a time at random without replacement. What is the probability that the second defective widget is drawn on exactly the 4th draw?
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Correct answer
B
The second defective falls on the 4th draw exactly when the first three draws hold exactly one defective and the 4th draw is defective. The chance of exactly one defective among the first three is C(3,1) × C(5,2) ÷ C(8,3) = (3 × 10) ÷ 56 = 30/56 = 15/28. After those three draws, 1 defective and 2 good are gone, leaving 2 defective and 3 good among 5, so the 4th draw is defective with probability 2/5. Multiply: (15/28)(2/5) = 30/140 = 3/14.
So the answer is B. Choice A (3/28) computes only two defectives landing back to back, (3/8)(2/7), ignoring the required structure. Choice C (15/56) gets stage one right but uses 1/2 for the 4th draw instead of the conditional 2/5. Choice D (3/8) gives the unconditional chance that the 4th draw alone is defective. Choice E (15/28) is the stage-one value alone, forgetting that the 4th draw must also be defective.
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