Exactly one defective bulb in the sample

Question

A box holds 12 light bulbs, of which 4 are defective and 8 are good. An inspector draws 3 bulbs at random, all at once (so no bulb is drawn twice).

What is the probability that exactly 1 of the 3 drawn bulbs is defective?

Accepted Answer

When you grab a handful at once, order does not matter, so we count groups of bulbs using combinations. The notation \binom{n}{k} means "the number of ways to choose k items from n." Total ways to pick any 3 bulbs from 12: \binom{12}{3}=220. Favorable ways: exactly 1 defective and 2 good. Choose 1 of the 4 defective, and 2 of the 8 good, then multiply those two counts because each choice combines with each: \binom{4}{1}	imes\binom{8}{2}=4	imes 28=112. Divide favorable by total: \frac{112}{220}=\frac{28}{55}\approx 0.509. Takeaway. For "exactly k of a type" in an unordered handful, split the…

Exactly one defective bulb in the sample

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