A box contains 10 bulbs of which 3 are defective. If 2 bulbs are chosen at random, what is the probability that both are defective?

Instruction: Calculate the probability without replacement.

Context: This question evaluates understanding of combinatorial probability in a practical context, focusing on quality control.

Updated

Official answer

There are C(10,2) = 45 equally likely unordered pairs of bulbs, and C(3,2) = 3 pairs contain two defective bulbs. Therefore P(both defective) = 3/45 = 1/15, about 6.67%.

Sequentially, the first bulb is defective with probability 3/10. Given that result, two defective bulbs remain among nine, so (3/10)(2/9) = 1/15. Squaring 3/10 would assume replacement and independent draws, which is a different experiment.

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