1 A drawer contains 50 bolts and 150 nuts Half of the bolts and half of the nuts are rusted If one item is chosen at random what is the probability that it is rusted or a bolt?

Answer 1

The best way to think about this question might be to think about a slightly different one: what is the probability that a randomly-chosen item is neither rusted nor a bolt? Well in this case it would have to be an un-rusted nut, and there are (1/2)*150=75 of these. So the probability of choosing an un-rusted nut is equal to the number of un-rusted nuts over the total number of items in the drawer, or 75/200=3/8=0.375. Now, everything in the drawer is either an un-rusted nut or it isn't, so the probability that the item we choose belongs to one or the other of those sets is 1 (it happens with certainty). Thus the probability that we choose something other than an un-rusted nut (something that is rusted or a bolt) is the difference in the two probabilities, or 1-3/8=5/8=0.625.

To get the same answer directly, you'd have to notice that there 25 rusty bolts and 75 rusty nuts, plus 25 bolts that aren't rusty, and so you're interested in the probability of choosing one of these 125 items out of the full drawer of 200 items. This is 125/200=5/8=0.625 as before. You can't just add the number of rusty items (100) to the number of bolts (50) and divide this by 200, because you then double-count the rusty bolts (each one belongs to the rusty group and to the bolt group). A backwards way of answering the question would be to add the number of rusty items to the number of bolts, and then subtract the number of rusty bolts, and finally divide by 200. Subtracting out the number of rusty bolts makes sure that these aren't double counted, and so the answer is again 125/200=5/8=0.625. This last approach is related to a more general idea called the Inclusion-Exclusion Principle.