# ROn Point: Probability on the GMAT The topics on the GMAT quantitative section are chosen because most test takers have some experience solving questions on these topics in high school. Subjects like algebra and geometry have given high school students white hairs and craned necks for generations (what? I was stretching, not copying off of her exam, honest!).

Areas such as calculus and linear algebra are usually only covered in college, putting students at varying levels of exposure depending on which programs they chose to study. Therefore, topics covered on the GMAT tend to come from high school level math, in order to give everyone a fair starting point. However, the problem with high school is that a lot of the learning is based on memorization and repetition. While these two procedures are important in the learning process, they are building blocks, not end points.

To that point, much of math in high school is based on memorizing and applying formulae (“formulas” is also an acceptable plural for formula, but I prefer the more apposite formulae). Mechanically applying the correct formula usually yields the correct answer, for example the probability of seeing a heads on two consecutive coin flips is P(Heads on first) + P(Heads on second) – P(Heads of both). Identifying the correct probabilities for each occurrence will yield you the correct answer, but understanding why the chosen formula works is more crucial to getting these questions right on the GMAT.  Automatically applying formulae without understanding what is being asked is a typical GMAT trap given that the exam is about logic and reasoning.

Consider the following probability question:

There is a 50% chance Jen will visit Chile this year, while there is a 25% chance that she will visit Madagascar this year. What is the probability that Jen will visit either Chile or Madagascar this year, but NOT both?

A. 25%
B. 50%
C. 62.5%
D. 63.5%
E. 75%

Putting aside her peculiar travel schedule for a moment, (visiting Western South America and South East Africa without stopping by the Falklands?) what is the probability that Jen travels to one of these locales but not the other? Dutifully applying our memorized probability formula: the probability of visiting Chile is 0.5, and the probability of visiting Madagascar is 0.25, we get 0.5 + 0.25 – (0.5*0.25) or 0.75 – 0.125 = 0.625. Or answer choice C. (As Admiral Ackbar tried to warn us in 1983: It’s a Trap!)

But why doesn’t the formula yield the correct choice? The answer lies in the question stem. This particular question asks us the probability of visiting Chile or Madagascar, but not both. The formula gives us the probability of visiting either, implicitly allowing the choice of visiting both. The probability of visiting either will indeed be 62.5% (or 5/8), but this is the probability of visiting Chile, Madagascar, or both. Verifying the converse, the probability of visiting neither is P(¬Chile) and P(¬Madagascar), or 0.5 * 0.75 = 0.375 (or 3/8), confirming that our 0.625 is merely the probability of not staying home this year.

The obvious question now is: Why doesn’t the formula work? Didn’t the formula already account for the possibility of both? How do I solve this question correctly? (Admittedly, these are three questions and not one). The key to answering all of them is the same, though. Let’s go through the logic of the formula P(C) + P(M) – P(C&M):

1. The first argument allows for all possibilities of visiting Chile, regardless of what happens with Madagascar.
2. The second argument allows for all possibilities of visiting Madagascar, regardless of what happens with Chile (or the Falklands)
3. The third argument is the possibility of both occurring.

The formula works because P(C) accounts for the both choice, and P(M) accounts for the both choice as well, indicating that this option has been double counted. In order to count it only once, we need to remove one instance of it. This is why the formula works and is popular; it addresses the inherent problem in probability, double counting certain situations. The same logic applies to Venn diagrams and other similar question types where counting the same argument twice (or thrice) can occur.

In practice, this question could then be solved using the default formula of P(C) + P(M) – P(C&M) and then subtracting  P(C&M) again… or simply P(C) + P(M) – 2*P(C&M). Simply put, the first two arguments in the formula account for the “both” possibility twice, so we must remove it twice to answer the question at hand.

A somewhat similar yet more straight forward alternative is to go bottom up instead of top down. The probability of going only to Chile is P(C) * P(¬M) = 0.5 * 0.75 = 0.375. The probability of going only to Madagascar is P(M) * P(¬C) = 0.25 * 0.5 = 0.125. Adding those two probabilities together yields the correct answer of 0.5 or 50%. The revised formula also yields this result (0.5 + 0.25 – 2 (0.125) = 0.5), so we can feel confident in our answer choice and not any of the other tempting answer choices.