A box contains 4 defective items and 16 non-defective items. If you randomly select 3 items from the box, what is the probability that at least one of them is defective?
To solve this problem, it's easier to first calculate the probability of the complementary event - that none of the selected items are defective - and then subtract that probability from 1.

First, let's find the probability of selecting a non-defective item on the first draw:
P(non-defective_1) = Number of non-defective items / Total number of items
                   = 16 / (4+16)
                   = 16 / 20
                   = 4 / 5

Since we're not replacing the items, there are now 15 items left in the box, with 15 non-defective items remaining. So, let's find the probability of selecting another non-defective item on the second draw:
P(non-defective_2 | non-defective_1) = Number of remaining non-defective items / Total number of remaining items
                                     = 15 / 19

Finally, there are now 14 items left in the box, with 14 non-defective items remaining. So, let's find the probability of selecting another non-defective item on the third draw:
P(non-defective_3 | non-defective_1 and non-defective_2) = Number of remaining non-defective items / Total number of remaining items
                                                        = 14 / 18

Now, we can multiply these probabilities to find the overall probability of all three selected items being non-defective:

P(all non-defective) = P(non-defective_1) * P(non-defective_2 | non-defective_1) * P(non-defective_3 | non-defective_1 and non-defective_2)
                     = (4/5) * (15/19) * (14/18)

To find the probability of at least one defective item, we subtract this value from 1:

P(at least one defective) = 1 - P(all non-defective)
                          ≈ 0.397

Therefore, the probability that at least one of the selected items is defective is approximately 39.7%.