Posted on
09
Mar 2021

## Independent vs. Dependent Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 8th March, 2021

## Independent vs. Dependent Probability

As promised last time, we’ll return to some strict GMAT probability today. Specifically, we’ll discuss the difference between independent and dependent probability. This simply refers to whether or not the events involved are dependent on one another. For example, let’s take a look at the following Official Guide problem:

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A) 11/8
B)
7/8
C) 9/64
D) 5/64
E) 3/64

In this case, we are dealing with independent events, because none of the probabilities affect the others. In other words, what Xavier does doesn’t affect Yvonne’s chances. We can treat each of the given probabilities as they are.

So mathematically, we would multiply, the probabilities involved. (Incidentally, the word “and” is often a good indication that multiplication is involved. We want Xavier AND Yvonne AND not Zelda to solve the problem.) And if Zelda has a chance of solving the problem, that means she has a chance of not solving it.

The answer would therefore be ¼ x ½ x ⅜  = 3/64 or answer choice E.

What if, however, we changed the problem to look like this:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

As you can see, the problem got much more complicated much more quickly, because now, the question stem is dependent upon a very specific series of events. Now, the events affect one another. Xavier will attempt the problem, but what happens at this stage affects what happens next. If he solves it, everything stops. But if he doesn’t, the problem moves to Yvonne. So in effect, there’s a ¼ chance that he’s the only person to attempt the problem, and there’s a ¾ chance the problem moves to Yvonne.

This is most likely how the GMAT will force you to think about probability: not in terms of formulas or complicated mathematical concepts, but rather in terms of narrative within a new problem with straightforward numbers.

That brings us to consideration of the question stem itself. What would have to happen for Zelda not to attempt the problem? Well, there are a couple of possibilities:

1. Xavier solves the problem

If Xavier solves the problem, the sequence ends, and Zelda does not see the problem. This is one case we’re interested in, and there’s a ¼ chance of that happening.

2. Xavier does not solve, but then Yvonne solves

There’s a ½ chance of Yvonne solving, but her seeing the problem is dependent upon the ¾ chance that Xavier does not solve. So in reality, we must multiply the two numbers together to acknowledge that the situation we want is “Xavier does not solve AND Yvonne does solve.” This results in ¾ x ½ = ⅜

The two above cases constitute two independent situations that we now must add together. For Zelda not to see the problem, either Xavier must solve it OR Yvonne must solve it. (The word “or” is often a good indication that addition will be used).

This leads us to our final probability of ¼ + ⅜ = that Zelda does not get to attempt the problem.

There is an alternative way to solve this problem, which we’ll talk about next time. It will segue nicely into the next topic, which we’ve already hinted at in our posts on GMAT combinatorics. Until then…

Posted on
14
Jan 2021

## Averages Problem No.1 : Test Averages

Hey guys, today we’re going to take a look at the test averages problem. This is a very straightforward mathematically oriented average problem or at least it can be. But there are very strong graphic solution paths here and there’s also a really strong sort of intuitive running tally counting solution path here. We’re going to start out with the math though, just because that’s how a lot of people are familiar with this problem. Before we jump into the heavier duty quicker sort of stuff.

## Doing the Math

So to solve this problem we want to take an average. But one of the components of our average is missing. So we have four things with an average of 78, and a fifth unknown. That means we can assume that each of the first four exams were 78. So we’ve got 4 times 78 plus X over 5. The total number of exams is going to give us our average of 80. Then through algebra, algebraic manipulation we multiply the 5 over, we get 400 equals 4 times 78 plus x. The 4 times 78 is 312. We subtract that off the 4 and that brings you to 88. Answer choice E.

## Graphic Solution Path: Poker Chips

Let’s take a look at this a little differently. One of the ways I like looking at averages is imagining stacks of poker chips and you can have stacks of anything. I like poker chips because they fit together and you can make two stacks equal very easily so what we’re being told here is we have four stacks of 78 a fifth unknown stack but if we equalize them all that is if we take chips off of the unknown stack and distribute them all the stacks will be 80. That means that the fifth stack needs to be 80 and then it needs two poker chips for each of the other four stacks to bring those 78’s up. We can also envision this as just a rectangle our goal is 80 but we have 78, and our goal is five tests but we have four so we have 78 by four here. And then 80 by 5 here what’s missing is the full 80 and then 2 on each of four stacks of 48.

## Running Tally Method: Intuitive Approach

The most powerful way to handle this problem though is probably by doing a running tally. Don’t even worry about the visualization but rather notice that, I’ve got 47 8s each of those are too short so I’m two, four, six. eight points short on the last test. I need to get the 80 that I want plus those eight points that I’m short bringing us to 88. And anybody who’s sweated like A+, B+, A- or a C+, B- has done this math. So if you characterize it like that a lot of times it becomes much more intuitive and once again allows you to cultivate confidence for a deeper treatment and more complex averages problems and mean problems check out the snack shop problem, check out the company production problem and there’s a great ds problem that we do the trade show problem you’ll find links to all of them just below and I hope this helped.

Enjoyed this Averages Problem ? Try another type of GMAT problem to get familiar with all question types on the exam: Remainder Number Theory Problem.