How to solve “at least” probability problems in the least amount of time

Fortunately, and contrary to popular belief, the CAT isn’t “pure evil.” Wherever it provides opportunities for less-savvy examinees to waste their time, it also provides a shortcut for those who have put in the study time to learn it or who have the patience to look for the elevator, so to speak, before slogging up the stairs. And one classic example of that comes with the “at least one” type of probability question. To illustrate, let’s consider an example: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow? (A) 5/17 (B) 12/17 (C) 25/81 (D) 56/81 (E) 4/9 Here, you can first streamline the process along the lines of one of those “There are two types of people in the world: those who _______ and those who don’t _______” memes. Your goal is to determine whether you get a yellow marble, so you don’t care as much about “blue” and “black”…those can be grouped into “not yellow,” thereby giving you only two groups: 8 yellow marbles and 10 not-yellow marbles. Fewer groups means less ugly math! But even so, trying to calculate the probability of every sequence that gives you one or two yellow marbles is labor intensive. You could accomplish that “not yellow” goal several ways: First marble: Yellow; Second: Not Yellow First: Not Yellow; Second: Yellow First: Yellow; Second: Yellow That’s three different math problems each involving fractions and requiring attention to detail. There ought to be an easier way…and there is. When a probability problem asks you for the probability of “at least one,” consider the only situation in which you WOULDN’T get at least one: if you got none. That’s a single calculation, and helpful because if the probability of drawing two marbles is 100% (that’s what the problem says you’re doing), then 100% minus the probability of the unfavorable outcome (no yellow) has to equal the probability of the favorable outcome. So if you determine “the probability of no yellow” and subtract from 1, you’re finished. That means that your problem should actually look like: PROBABILITY OF NO YELLOW, FIRST DRAW: 10 non-yellow / 18 total PROBABILITY OF NO YELLOW, SECOND DRAW: 9 remaining non-yellow / 17 remaining total 10/18 * 9/17 reduces to 10/2 * 1/17 = 5/17. Now here’s the only tricky part of using this technique: 5/17 is the probability of what you DON’T want, so you need to subtract that from 1 to get the probability you do want. So the answer then is 12/17, or B. More important than this problem is the lesson: when you see an “at least one” probability problem, recognize that the probability of “at least one” equals 100% minus the probability of “none.” Since “none” is always a single calculation, you’ll always be able to save time with this technique. Had the question asked about three marbles, the number of favorable sequences for “at least one yellow” would be: Yellow Yellow Yellow Yellow Not-Yellow Not-Yellow Yellow Not-Yellow Yellow Yellow Yellow Not-Yellow Not-Yellow Yellow Yellow … (And note here – this list is not yet exhaustive, so under time pressure you may very well forget one sequence entirely and then still get the problem wrong even if you’ve done the math right.) Whereas the probability of No Yellow is much more straightforward: Not-Yellow, Not-Yellow, Not-Yellow would be 10/18 * 9/17 * 8/16 (and look how nicely that last fraction slots in, reducing quickly to 1/2). What would otherwise be a terrifying slog, the “long way” becomes quite quick the shorter way.