GMAT Tip of the Week: Taking the Least Amount of Time to Solve “At Least” Probability Problems

GMAT Tip of the WeekIn its efforts to keep everyone from getting perfect 800s, the GMAT has two powerful tools to stop you from perfection. For one, it can bait you into wrong answers (with challenging content, tempting trap answers, or a combination thereof). And secondly, it can waste your time, making it look like you need to do a lot of work when there’s a much simpler way.

Fortunately, and contrary to popular belief, the GMAT 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.

So, remember, when you see “at least one” probability on the GMAT, employ the “100% minus probability of none” strategy and you’ll save valuable time on at least one Quant problem on test day.

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By Brian Galvin.

GMAT Probability Practice: Questions and Answers

Roll the DiceThe Quantitative portion of the GMAT contains questions on a variety of math topics. One of those topics is probability. GMAT questions of this sort ask you to look for the likelihood that something will occur. Probability is not as familiar to many as Algebra, Geometry, and other topics on the test. This is why some test-takers hesitate when they see the word “probability” on a summary of the GMAT. However, this is just another topic that can be mastered with study and practice.

You may already know that there are certain formulas that can help solve GMAT probability questions, but there is more to these problems than teasing out the right answers. Take a look at some advice on how to tackle GMAT probability questions to calm your fears about the test:

Probability Formulas
As you work through GMAT probability practice questions, you will need to know a few formulas. One key formula to remember is that the probability equals the number of desired outcomes divided by the number of possible outcomes. Another formula deals with discrete events and probability – that formula is P(A and B) = P(A)*P(B). Figuring out the probability of an event not occurring is one minus the probability that the event will occur. Putting these formulas into practice is the most effective way to remember them.

Is it Enough to Know the Basic Formulas for Probability?
Some test-takers believe that once you know the formulas related to probability for GMAT questions, then you have the keys to success on this portion of the test. Unfortunately, that is not always the case. The creators of the GMAT are not just looking at your ability to plug numbers into formulas – you must understand what each question is asking and why you arrived at a particular answer. Successful business executives use reason and logic to arrive at the decisions they make. The creators of the GMAT want to see how good you are at using these same tools to solve problems.

The Value of Practice Exams
Taking a practice GMAT can help you determine your skill level when it comes to probability questions and problems on every other section of the test. Also, a practice exam gives you the chance to become accustomed to the amount of time you’ll have to finish the various sections of the test.

At Veritas Prep, we have one free GMAT practice test available to anyone who wants to get an idea of how prepared they are for the test. After you take the practice test, you will receive a score report and thorough performance analysis that lets you know how you fared on each section. Your performance analysis can prove to be one of the most valuable resources you have when starting to prepare for the GMAT. Follow-up practice tests can be just as valuable as the first one you take. These tests reveal your progress on probability problems and other skills on the GMAT. The results can guide you on how to adjust your study schedule to focus more time on the subjects that need it.

Getting the Right Kind of Instruction
When it comes to probability questions, GMAT creators have been known to set subtle traps for test-takers. In some cases, you may happen upon a question with an answer option that jumps out at you as the right choice. This could be a trap.

If you study for the GMAT with Veritas Prep, we can teach you how to spot and avoid those sorts of traps. Our talented instructors have not only taken the GMAT; they have mastered it. Each of our tutors received a score that placed them in the 99th percentile. Consequently, if you study with Veritas Prep, you’ll benefit from the experience and knowledge of tutors who have conquered the GMAT. When it comes to probability questions, GMAT tutors at Veritas Prep have you covered!

In addition to providing you with effective GMAT strategies, tips, and top-quality instruction, we also give you choices regarding the format of your courses. We have prep classes that are given online and in person – learn your lessons where you want, and when you want. You may want to go with our private tutoring option and get a GMAT study plan that is tailored to your needs. Contact Veritas Prep today and dive into your GMAT studies!

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Probability and Combinations: What You’ll Need to Know for the GMAT

Roll the DiceIf you’ve been paying attention to the exciting world of GMAT prep, you know that GMAC released two new practice tests fairly recently. I’d mentioned in a previous post that I was going to write about any conspicuous trends I noted, and one unmistakable pattern I’ve seen with my students is that probability questions seem to be cropping up with greater and greater frequency.

While these questions don’t seem fundamentally different from what we’ve seen in the past, there does seem to be a greater emphasis on probability questions for which a strong command of combinations and permutations will prove indispensable.

First, recall that the probability of x is the number ways x can occur/number of total possible outcomes (or p(x) = # desired/ # total). Another way to think about this equation is to see it as a ratio of two combinations or permutations. The number of ways x can occur is one combination (or permutation), and the total number of possible outcomes is another.

Keeping this in mind, let’s tackle this new official prompt:

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

(A) 3/7
(B) 5/12
(C) 27/70
(D) 2/7
(E) 9/35

Typically, I’ll start by calculating the total number of possible outcomes, as this calculation tends to be the more straightforward one. We’ve got 8 volunteers, and we want to know the number of total ways we can select 4 people from these 8 volunteers. Note, also, that the order does not matter – group of Tiffany, Mike, Louis, and Amy is the same as a group of Louis, Amy, Mike, and Tiffany. We’re not assigning titles or putting anyone in seats, so this is a combination.

If we use our combination formula N!/[(K!*(N-K)!] then N, our total pool of candidates, is 8, and K, the number we’re selecting, is 4. We get 8!/(4!*4!), which comes out to 70. At this point, we know that the denominator must be a factor of 70, so anything that doesn’t meet this criterion is out. In this case, this only allows us to eliminate B.

Now we want our desired outcomes, in which Andrew is selected and Karen is not. Imagine that you’re responsible for assembling this group of four from a total pool of eight people. You plan on putting your group of four in a conference room. Your supervisor tells you that Andrew must be in and Karen must not be, so you take Andrew and put him in the conference room. Now you’ve got three more spots to fill and seven people remaining. But remember that Karen cannot be part of this group. That means you only have 6 people to choose from to fill those other 3 spots in the room.

Put another way, think of the combination as the number of choices you have. Andrew and Karen are not choices – you’ve been ordered to include one of them and not the other. Of the 4 spots in the conference room, you only get to choose 3. And you’re only selecting from the other 6 people for those spots. Now N = 6 and K = 3. Plugging these into our trusty combination formula, we get 6!/(3!*3!), which comes out to 20.

Summarizing, we know that there are 20 ways to create our desired group of 4, and 70 total ways to select 4 people from a pool of 8, giving us a probability of 20/70, or 2/7, so the correct answer is D.

Takeaway: Probability questions can be viewed as ratios of combinations or permutations, so when you brush up on combinatorics, you’re also bolstering your probability fundamentals. Anytime you’re stuck on a complex probability question, break your calculation down into its component parts – find the total number of possible outcomes first, then find the total number of desired outcomes. Like virtually every hard question on the GMAT, probability questions are never as hard as they first seem.

Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTube, Google+ and Twitter!

By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.

Quarter Wit, Quarter Wisdom: Using Visual Symmetry to Solve GMAT Probability Problems

Quarter Wit, Quarter WisdomToday, let’s take a look at an official GMAT question involving visual skills. It takes a moment to understand the given diagram, but at close inspection, we’ll find that this question is just a simple probability question – the trick is in understanding the symmetry of the figure:

The figure shown represents a board with 4 rows of pegs, and at the bottom of the board are 4 cells numbered 1 to 4. Whenever the ball shown passes through the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has the probability 1/2 of passing through the opening immediately to the left of that peg and probability 1/2 of passing through the opening immediately to the right. What is the probability that when the ball passes through the first two pegs at the top it will end in cell 2?

qwqw pegs pic





(A) 1/16
(B) 1/8
(C) 1/4
(D) 3/8
(E) 1/2

First, understand the diagram. There are small pegs arranged in rows and columns. The ball falls between two adjacent pegs and hits the peg directly below. When it does, there are two ways it can go – either to the opening on the left or to the opening on the right. The probability of each move is equal, i.e. 1/2.

The arrow show the first path the ball takes. It is dropped between the top two pegs, hits the peg directly below it, and then either drops to the left side or to the right. The same process will be repeated until the ball falls into one of the four cells – 1, 2, 3 or 4.

Method 1: Using Symmetry
Now that we understand this process, let’s examine the symmetry in this diagram.

Say we flip the image along the vertical axis – what do we get? The figure is still exactly the same, but now the order of cells is reversed to be 4, 3, 2, 1. The pathways in which you could reach Cell 1 are now the pathways in which you can use to reach Cell 4.

OR think about it like this:

To reach Cell 1, the ball needs to turn left-left-left.

To reach Cell 4, the ball needs to turn right-right-right.

Since the probability of turning left or right is the same, the situations are symmetrical. This will be the same case for Cells 2 and 3. Therefore, by symmetry, we see that:

The probability of reaching Cell 1 = the probability of reaching Cell 4.


The probability of reaching Cell 2 = the probability of reaching Cell 3. (There will be multiple ways to reach Cell 2, but the ways of reaching Cell 3 will be similar, too.)

The total probability = the probability of reaching Cell 1 + the probability of reaching Cell 2 + the probability of reaching Cell 3 + the probability of reaching Cell 4 = 1

Because we know the probability of reaching Cells 1 and 4 are the same, and the probabilities of reaching Cells 2 and 3 are the same, this equation can be written as:

2*(the probability of reaching Cell 1) + 2*(the probability of reaching Cell 2) = 1

Let’s find the probability of reaching Cell 1:

After the first opening (not the peg, but the opening between pegs 1 and 2 in the first row), the ball moves left (between pegs 1 and 2 in second row) or right (between pegs 2 and 3 in second row). It must move left to reach Cell 1, and the probability of this = 1/2.

After that, the ball must move left again – the probability of this occurring is also 1/2, since probability of moving left or right is equal. Finally, the ball must turn left again to reach Cell 1 – the probability of this occurring is, again, 1/2. This means that the total probability of the ball reaching Cell 1 = (1/2)*(1/2)*(1/2) = 1/8

Plugging this value into the equation above:

2*(1/8) + 2 * probability of reaching Cell 2 = 1

Therefore, the probability of reaching Cell 2 = 3/8

Method 2: Enumerating the Cases
You can also answer this question by simply enumerating the cases.

At every step after the first drop between pegs 1 and 2 in the first row, there are two different paths available to the ball – either it can go left or it can go right. This happens three times and, hence, the total number of ways in which the ball can travel is 2*2*2 = 8

The ways in which the ball can reach Cell 2 are:




So, the probability of the ball reaching Cell 2 is 3/8.

Note that here there is a chance that we might miss some case(s), especially in problems that involve many different probability options. Hence, enumerating should be the last option you use when tackling these types of questions on the GMAT.

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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!