Probability GMAT Problems can be super complex if you don’t frame it correctly. One of the keys to looking at probability problems, particularly conditional probability and independent probability problems, is breaking each part up into its own entity, and a lot of times this clarifies the problem.
Introduction To The XYZ Probability Problem
Let’s take a look at this ‘XYZ’ probability problem. Xavier, Yvonne, and Zelda are solving problems. We’re given the 3 probabilities for correct answers and we’re being asked what’s the probability of X being right and solving it, Y solving it, and Z not solving it.
The first thing we can look at is, say: “Well what’s the probability of Zelda not solving it?” And it’s just going to be the flip, the other side of 5/8 to bring us up to 1. If she solves it 5 out of 8 times, she’s not going to solve it the other 3 out of 8 times. So, we’re dealing with 1/4, 1/2, and 3/8.
Doing The Math May Seem Simple
The math here is straightforward, multiply them together. But that might not be readily apparent, or at the very least, just plugging it into that formula can get you into trouble. So, here’s where owning it conceptually and mapping it out with a visualization helps you take command of this problem.
Xavier Getting It Correct
Since each probability is independent of the others we can look at them independently. What’s the probability of Xavier getting this correct? 1 out of 4 times. So, we can say in general, for every four attempts, he gets it correct once or 25%. If, and only if Xavier gets it correct can we move on to the next part – Yvonne.
Yvonne Getting It Correct
Xavier gets a correct 1 out 4 times then what are the chances that Yvonne gets a correct? 1 out of 2. So to have Xavier get it correct and then Yvonne get it correct it’s going to be 1 out of 8 times – 1/4 times 1/2.
It’s not that we can’t look at a Yvonne when Xavier gets it incorrect, it’s that it doesn’t matter. From a framing perspective, this is all about only looking at the probability for the outcome that we want and ignoring the rest.
Zelda Getting It Incorrect
Xavier: 1 out of 4, Yvonne: 1 out of 2, gets us to 1 out of 8. Then and only then, what are the chances that Zelda gets it incorrect? 1 out of 8 trials brings us to X and Y are correct, then we multiply it by the 3/8 that Zelda gets it incorrect. That gets us to 3/64. 3 out of every 64 attempts will end in ‘correct’, ‘correct’, ‘incorrect’.
This is one of those problems that may have to go through a few times but once you attach the explanation to it, you can’t mess up the math.