A box contains 12 light bulbs 2 of which are defective If a

A box contains 12 light bulbs 2 of which are defective. If a quality controller picks 3 bulbs at random, what is the probabilty that one of them is defective?

please show work.

Solution

There are 12C3 ways to choose any three bulbs.

There are 10C2 ways to choose two nondefectives, and 2 ways to choose a defective bulb.

Thus,

P(one is defective) = [(10C2)*2]/(12C3)

= 0.409090909 [ANSWER]