A box contains 12 bulbs with 3 defective ones. If two bulbs are drawn from the box, what is the probability that both bulbs are defective. both are non defective and one bulb is defective?

Answer 1

prob(both defective) = (3/12)(2/11) = 6/132
prob(both non-defective) = (9/12)(8/11) = 72/132
prob(one defective) = (3/12)(9/11) + (9/12)(3/11) = 54/132

(did you notice that 6/132 + 72/132 + 54/132 = 1 ? )