A box holds 10 pens, of which 3 are defective. Three pens are taken out together at random. What is the probability that exactly one of them is defective?

Three pens can be chosen in $C(10,3)=120$ ways. Exactly one defective means 1 of the 3 defective pens together with 2 of the 7 good ones: $C(3,1)\times C(7,2)=3\times 21=63$ ways, a probability of 63/120 = 21/40. The value $7/24=C(7,3)/120=35/120$ is the probability of NO defective pen and 17/24 is its complement, at least one defective; 3/10 is the chance that a single pen picked on its own is defective.