GATE 2023 ST – Question 31
Two of five bulbs are defective. Bulbs are chosen uniformly without replacement and tested one at a time. Let X be the minimum number of tests needed to identify both defective bulbs. Find P(X=3) to two decimals.
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Correct answer: 0.28 to 0.32
Explanation
All 10 pairs of defective positions are equally likely. Identification after exactly three tests occurs if the second defective is at position 3 (pairs {1,3},{2,3}),or if the first three bulbs are all good (pair {4,5},so the remaining two must be defective). These are 3 pairs out of 10,giving **0.30**. Identification can use the known total of two defects; actually testing both defects would be a different stopping rule.