Posted on
12
Aug 2021

GMAT Probability Problems – How to Tackle Them & What Mistakes to Avoid

By: Apex GMAT
Contributor: Ilia Dobrev
Date: August 12, 2021

The concept of probability questions is often pretty straightforward to understand, but when it comes to its application in the GMAT test it may trip even the strongest mathematicians.

Naturally, the place to find such types of problems is the Quantitative section of the exam, which is regarded as the best predictor of academic and career success by many of the most prestigious business schools out there – Stanford, Wharton, Harvard, Yale, INSEAD, Kellogg, and more. The simple concept of probability problems can be a rather challenging one because such questions appear more frequently as high-difficulty questions instead of low- or even medium-difficulty questions. This is why this article is designed to help test-takers who are pursuing a competitive GMAT score tackle the hazardous pitfalls that GMAT probability problems often create.

GMAT Probability – Fundamental Rules & Formulas

It is not a secret that the Quantitative section of the GMAT test requires you to know just the basic, high-school-level probability rules to carry out each operation of the practical solution path. The main prerequisite for success is mastering the Probability formula:
Probability = number of desired outcomes / total number of possible outcomes

 Probability P  = number of desired outcomes total number of possible outcomes

We can take one fair coin to demonstrate a simple example. Imagine you would like to find the probability of getting a tail. Flipping the coin can get you two possible comes – a tail or a head. However, you desire a specific result – getting only a tail – which can happen only one time. Therefore, the probability of getting a tail is the number of desired outcomes divided by the number of total possible outcomes, which is ½. Developing a good sense of the fundamental logic of how probability works is central to managing more events occurring in a more complex context.

Alternatively, as all probabilities add up to 1, the probability of an event not happening is 1 minus the probability of this event occurring. For example, 1 – ½ equals the chance of not flipping a tail.

Dependent  Events vs. Independent Events

On the GMAT exam, you will often be asked to find the probability of several events that happen either simultaneously or at different points in time. A distinction you must take under consideration is exactly what type of event you are exploring.

Dependent events or, in other words, disjoint events, are two or more events with a probability of simultaneous occurrence equalling zero. That is, it is absolutely impossible to have them both happen at the same time. The events of flipping either a tail or a head out of one single fair coin are disjoint.

If you are asked to find a common probability of two or more disjoint events, then you should consider the following formula:

 Probability P of events A and B   = (Probability of A) + (Probability of B)

Therefore, the probability of flipping one coin twice and getting two tails is ½ + ½.

If events A and B are not disjointed, meaning that the desired result can be in a combination between A and B, then we have to subtract the intersect part between the events in order to not count it twice:

 Probability P of events A and B   = P(A) + P(B) – Probability (A and B)

Independent events or discrete events are two or more events that do not have any effect on each other. In other words, knowing about the outcome of one event gives absolutely no information about how the other event will turn out. For example, if you roll not one but two coins, then the outcome of each event is independent of the other one. The formula, in this case, is the following:

 Probability P of events A and B   = (Probability of A) x (Probability of B)
How to approach GMAT probability problems

In the GMAT quantitative section, you will see probability incorporated into data sufficiency questions and even problems that do not have any numbers in their context. This can make it challenging for the test taker to determine what type of events he or she is presented with.
One trick you can use to approach such GMAT problems is to search for “buzzwords” that will signal out this valuable information.

• OR | If the question uses the word “or” to distinguish between the probabilities of two events, then they are dependent – meaning that they cannot happen independently of one another. In this scenario, you will need to find the sum of the two (or more) probabilities.
• AND | If the question uses the word “and” to distinguish between the probabilities of two events, then they are independent – meaning their occurrences have no influence on one another. In this case, you need to multiply the probabilities of the individual events to find the answer.

Additionally, you can draw visual representations of the events to help you determine if you should include or exclude the intersect. This is especially useful in GMAT questions asking about greatest probability and minimum probability.

If you experience difficulties while prepping, keep in mind that Apex’s GMAT instructors have not only mastered all probability and quantitative concepts, but also have vast experience tutoring clients from all over the world to 700+ scores on the exam. Private GMAT tutoring and tailored customized GMAT curriculum are ideal for gaining more test confidence and understanding the underlying purpose of each question, which might be the bridge between your future GMAT score and your desired business school admissions.

Posted on
16
Mar 2021

When Probability Meets Combinatorics: One Problem, Two Approaches

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th March, 2021

Now, we’d like to take a look at an Official GMAT Probability problem to pull everything together. The following is a good example for two reasons:

1. It illustrates a quirky case that is difficult more conceptually than mathematically, and thus is better for the GMAT.

2. It can be tackled either through straight probability or through a combination of probability and combinatorics.

Here’s the question:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A) 1/24
B)
1/8
C) 1/4
D) 1/3
E) 3/8

First, as always, give the problem a shot before reading on for the explanation. If possible, see if you can tackle it with both methods (pure probability and probability w/ combinatorics).

Explanation #1:

First, we’ll tackle pure probability. Let’s label the letters A, B, C, and D, and let’s say that A is the letter we wish to match with its correct envelope. The other three will be matched with incorrect envelopes. We now must examine the individual probabilities of the following events happening (green for correct, red for incorrect):

_A_   _B_   _C_   _D_

For the above, each slot represents a letter matched with an envelope. There are four envelopes and only one is correct for letter A. That means Tanya has a 1/4 chance of placing letter A in its correct envelope:

_1/4__   _B_   _C_   _D_

We now desire letter B to be placed in an incorrect envelope. Two of the remaining three envelopes display incorrect addresses, so there is a 2/3 chance of that happening:

_1/4__   _2/3_   _C_   _D_

We then desire letter C to also be placed in an incorrect envelope. Only one of the remaining two envelopes displays an incorrect address, so there is a 1/2 chance of that happening:

_1/4__   _2/3_   _1/2_   _D_

At that point, the only remaining option is to place the last remaining letter in the last remaining envelope (i.e. a 100% chance, so we place a 1 in the final slot):

_1/4__   _2/3_   _1/2_   _1_

Multiplying the fractions, we can hopefully see that some cancelling will occur:

¼ x ⅔ x ½ x 1

= 1 x 2 x 1
———–
4 x 3 x 2

= 1/12

But lo and behold, 1/12 is not in our answer choices. Did you figure out why?

We can’t treat letter A as the only possible correct letter. Any of the four letters could possibly be the correct one. However, the good news is that in any of the four cases, the math will be exactly the same. So all we have to do is take the original 1/12 we just calculated and multiply it by 4 to get the final answer: 4 x 1/12 = 4/12 = 1/3. The correct answer is D.

Explanation #2:

So what about a combinatorics approach?

As we’ve discussed in our previous GMAT probability posts, all probability can be boiled down to Desired Outcomes / Total Possible Outcomes. And as we discussed in our posts on GMAT combinatorics, we can use factorials to figure out the total possible outcomes in a situation such as this, which is actually a simple PERMUTATION. There are four envelopes, so for the denominator of our fraction (total possible outcomes), we can create a slot for each envelope and place a number representing the letters in each slot to get:

_4_  _3_  _2_  _1_  =  4! = 24  possible outcomes

This lets us know that if we were to put the four letters into the four envelopes at random, as the problem says, there would be 24 permutations, giving us the denominator of our fraction (total possible outcomes).

So what about the desired outcomes? How many of those 24 involve exactly one correctly placed letter? Well, let’s again treat letter A as the correctly placed letter. Once it’s placed, there are three slots (envelopes) left:

___  ___  ___

But the catch is: the next envelope has only two letters that could go into it. Remember, one of the letters correctly matches the envelope in address, and we want a mismatch:

_2_  ___  ___

Likewise, that would leave two letters available for the next envelope, but only one of them would have the wrong address:

_2_  _1_  ___

And finally, there would be only one choice left for the final envelope:

_2_  _1_  _1_

That would mean for the correctly-placed A letter, there are only two permutations in which each of the other letters is placed incorrectly:

_2_ x  _1_ x  _1_ = 2 possible outcomes.

But as before, we must consider that any of the four letters could be the correct letter, not just letter A. So we must multiply the 2 possible outcomes by four to get 8 desired outcomes involving exactly one letter being placed in its correct envelope. That gives us our numerator of Desired Outcomes. Our denominator, remember, was 24 total possible outcomes. So our final answer, once again, is 8/24 = 1/3.

This is a great example of how GMAT combinatorics can intersect with probability.

To tide you over until next time, give this Official GMAT problem a try. It will also give a nice segue into Number Theory, which we’ll begin to talk more about going forward. Explanation next time…

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Posted on
11
Mar 2021

An “Undesired” Approach to GMAT Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 11th March, 2021

In our last post, we discussed a solution for the following question, which is a twist on an Official Guide GMAT probability problem:

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

We also mentioned that there was an alternate way to solve it. Did you find it? In truth, it relates to something we discussed in a previous post we did on GMAT Combinatorics, specifically Combinations with Restrictions. In that post, we discussed the idea of considering combinations in which you’re not interested. It might seem counterintuitive, but if you subtract those out from the total number of combinations possible, you’re left with the number of combinations in which you are interested:

You can actually do something similar with probability. Take the following basic example:

Suppose I told you to flip a fair coin five times, “fair” meaning that it has an equal chance of landing heads-up or tails-up. I then wanted to know the probability that I flip at least one head. Now, when you think about it, the language “at least one” involves so many desired possibilities here. It could be 1 head, 2 heads, …, all the way up to 5 heads. I’d have to calculate each of those probabilities individually and add them up.

Or…

I could consider what is not desired, since the possibilities are so much fewer:

All of the above must add to 100% or 1, meaning all possible outcomes. So why not figure out the probability that I get 0 heads (or all tails), and then subtract it from 100% or 1 (depending on whether I’m using a percentage or decimal/fraction)? I’ll then be left with all the possibilities in which I’m actually interested, without the need to do any more calculations.

Each time I flip the coin, there is a ½ chance that I flip a tail. This is the same each of the five times I flip the coin. I then multiply all of the probabilities together:

½ x ½ x ½ x ½ x ½ = 1 / 25  = 1 / 32

Another way to view this is through combinatorics. Remember, probability is always Desired outcomes / Total possible outcomes. If we start with the denominator, there are two outcomes each time we flip the coin. That means for five flips, we have 25 or 32 possible outcomes, as illustrated here with our slot method:

_2_  _2_  _2_  _2_  _2_ = 32

Out of those 32 outcomes, how many involve our (not) desired outcome of all tails? Well, there’s only one possible way to do that:

_T_  _T_  _T_  _T_  _T_    ← Only 1 outcome possible

It really is that straightforward: with one outcome possible out of 32 total, the probability is 1/32 that we flip all tails.

Now remember, that is our, not desired. Our desired is the probability of getting at least one head

So, since the probability of getting 0 heads (all tails) is 1/32, we simply need to subtract that from 1 (or 32/32) to get our final result. The probability that we flip at least one head if we flip a fair coin five times is 31/32.

Application to problem from previous post

So now, how do we work that into the problem we did last time? Well, in the previous post, we took a more straightforward approach in which we considered the outcomes we desired. But can we use the above example and consider not desired instead? Think about it and give it a shot before reading the explanation:

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

Explanation

In this question, our desired case is that Zelda does not attempt the problem. That means, quite simply, that our not desired case is that Zelda does get to attempt it. This requires us analytically to consider how such a case would arise. Let’s map out the possibilities with probabilities:

Notice that two complementary probabilities are presented for each box. For example, since there is a 1/4 chance Xavier solves the problem (left arrow), we include the 3/4 probability that he does not solve the problem (right arrow).

If Zelda does get to attempt it, it’s clear from the above that first Xavier and Yvonne must each not solve it. There is a 3/4 and a 1/2 chance, respectively, of that happening. This is also a dependent situation. Xavier must not solve AND Yvonne must not solve. Therefore, we will multiply the two probabilities together to get ¾ x ½ = ⅜. So there is a 3/8 chance of getting our not desired outcome of Zelda attempting the problem.

So, we can finally subtract this number from 1 (or 8/8) and see that there is a 5/8 chance of Zelda not getting to attempt the problem. The correct answer is B.

Next time, we’ll discuss how GMAT Probability and Combinatorics can combine to form some interesting problems…

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
25
Feb 2021

What Happens When Permutations Have Repeat Elements?

As promised in the last post, today we’ll discuss what happens when we have a PERMUTATIONS situation with repeat elements. What does this mean exactly? Well, let’s return to the basic example in our intro post on GMAT combinatorics:

If we have five distinct paintings, and we want to know how many arrangements can be created from those five, we simply use the factorial to find the answer (i.e. 5! = 5*4*3*2*1 = 120). Let’s say those paintings were labeled A, B, C, D, and E.

Now, each re-arrangement of those five is a different PERMUTATION, because the order is different:

ABCDE

etc

Remember, there are 120 permutations because if we use the blank (or slot) method, we would have five choices for the first blank, and once that painting is in place, there would be four left for the second blank, etc…

_5_  _4_  _3_  _2_  _1_

…and we would multiply these results to get 5! or 120.

However, what if, say we suddenly changed the situation such that some of the paintings were identical? Let’s replace painting C with another B and E with another D:

ABBDD

Suddenly, the number of permutations decreases, because some paintings are no longer distinct. And believe it or not, there’s a formulaic way to handle the exact number of permutations. All you have to do is take the original factorial, and divide it by the factorials of each repeat. In this case, we have 5! for our original five elements, and we now must divide by 2! for the two B’s and another 2! for the two D’s:

5!
——
2! 2!

= 5*4*3*2*1
————-
(2*1)(2*1)

= 5*2*3
= 30 permutations

As another example, try to figure out how many permutations you can make out of the letters in the word BOOKKEEPER? Give it a shot before reading the next paragraph.

In the case of BOOKKEEPER, there are 10 letters total, so we start with a base of 10!

We then have two O’s, two K’s and three E’s for repeats, so our math will look like this:

10!
———
2! 2! 3!

Definitely don’t calculate this, though, as GMAT math stays simple and likes to come clean. Remember, we’ll have to divide out the repeats. You are extremely unlikely to have to do this calculation for a GMAT problem, however, since it relies heavily on busy-work mechanics. The correct answer choice would thus look like the term above.

Let’s now take a look at an Official Guide question in which this principle has practical use. I’ll leave it to you to discover how. As usual, give the problem a shot before reading on:

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

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

Quick Probability Review

Remember from our post of GMAT Probability that, no matter how complicated the problem, probability always boils down to the basic concept of:

Desired Outcomes
———————————–
Total Possible Outcomes

In this case, each child has two equally likely outcomes: boy and girl. And since there are four children, we can use are blank method to realize that we’ll be multiplying two 4 times:

_2_  _2_  _2_  _2_   =  16 total possible outcomes (denominator)

This may give you the premature notion that C or E must be correct, simply because you see a 16 in the denominator, but remember, fractions can reduce! We could have 4 in the numerator, giving us a fraction of 4/16, which would reduce to 1/4. And every denominator in the answer choices contains a factor of 16, so we can’t eliminate any answers based on this.

Now, for the Desired Outcomes component, we must figure out how many outcomes consist of exactly two boys and two girls. The trick here is to recognize that it could be in any order. You could have the two girls followed by the two boys, vice versa, or have them interspersed. Now, you could brute-force this and simply try writing out every possibility. However, you must be accurate, and there’s a chance you’ll forget some examples.

What if we instead write out an example as GGBB for two girls and two boys? Does this look familiar? Well, this should recall PERMUATIONS, as we are looking for every possible ordering in which the couple could have two girls and two boys. And yes, we have two G’s and two B’s as repeats. Here’s the perfect opportunity to put our principle into play:

We have four children, so we use 4! for our numerator, then we divide by 2! twice for each repeat:

4!
——
2! 2!

This math is much simpler, as the numerator is 24, while the denominator is 4. (Remember, memorize those factorials up to 6!)

This yields 6 desired outcomes of two boys and two girls.

With 6 desired outcomes of 16 total possible outcomes, our final probability fraction is 6/16, which reduces to 3/8. The correct answer is A.

Next time, we’ll look into combinatorics problems that involve restrictions, which can present interesting conceptual challenges.

By: Rich Zwelling, Apex GMAT Instructor
Date: 25th February, 2021

Posted on
20
Nov 2020

GMAT How-to: Probability Problems

GMAT probability questions, which test logical reasoning skills, tend to be quite daunting. The good news is that they don’t appear very frequently; the Quant section contains no more than three or four probability questions. However, since so many test-takers struggle with these questions, mastering probability can be an excellent way to boost your overall score.

GMAT probability questions aren’t so hard once you’ve grasped the basic concepts. Like the majority of the Quant section, probability questions only cover high school level material. The principal challenge is the tricky wording.

What Is Probability?

The first step to mastering probability is to break down the basic idea:

Probability = the number of desired outcomes / the total number of outcomes

Or in other words, the chance of something happening is the quotient of the number of desired outcomes and the total number of possible outcomes.

A coin flip is one generic example that can help us understand probability.

There are two possible outcomes when we flip a coin: heads or tails. If we want the coin to land on heads, then we divide 1 (the chance that the coin will land on the desired outcome, heads) by 2 (all possible outcomes, heads and tails), and the result is ½ or 0.5 (50%), meaning that there is a 50% chance that the coin will land on heads.

Although this is an elementary example, it demonstrates the fundamental concept behind all probability problems–a ratio between a part and a whole expressed as a fraction or percentage.

Probability of Independent Events

The probability of x discrete events occurring is the product of all individual probabilities.

For example, imagine that we toss a coin twice. Each toss is independent of the other, meaning that each toss has an equal chance of landing on either heads or tails (0.5). If we want to calculate the chance of getting heads twice in a row, we need to multiply the probability of getting heads the first time by the probability of getting heads the second time.

Or, represented as an equation:

½ x ½ = ¼

We get a 0.25 or 25% chance that the coin will land on heads twice.

Probability of Getting Either A or B

Keep in mind that the sum of all possible events is equal to 1 (100%).

If we continue with the coin toss example, we know that the probability of landing on heads is 0.5, and that the probability of landing on tails is also 0.5. Therefore:

0.5 + 0.5 = 1

The possibility of landing on either heads or tails is equal to 1, or 100%. In other words, every time we flip a coin, we can be certain that it will land on heads or tails.

Probability Of An Event Not Occurring

Following the concept that the sum of all possible events is 1, we can conclude that the probability of event A not happening (A’) is 1 – A, or equal to the probability of event B occurring.

The chance that the coin will not land on heads is equal to the chance that the coin will land on tails:

1 – 0.5 = 0.5

This method is most useful in situations with many favorable events and fewer unfavorable ones. Since time management is essential on the GMAT, it’s better to avoid solution paths that require more calculations. Subtracting the number of unfavorable events from the whole is quicker and simpler, and thus, less likely to result in mistakes.

Pay Attention to Keywords

Read each problem’s wording with great care to determine exactly which operations to use.

For example, if the problem uses the word “and,” you need to find the product of the probabilities. If the question uses the word “or,” you need to solve for their sum.

If we flipped one coin and we wanted to know the chances of landing on either heads or tails, we would calculate it like this:

0.5 + 0.5 = 1

Similarly, if we were to toss two coins and we wanted to find the probability of landing on both heads and tails, we would use this equation:

0.5 x 0.5 = 0.25

Avoid Common Errors

Minor errors, such as missing possible events, can lead to incorrect answers.

These pointers will help you avoid some common mistakes on probability questions:

• List all possible events before starting any calculations;
• Sum up the probabilities of all possible events to make sure they add up to 1;
• If there are several different arrangements possible (for example, picking different colored balls from a box), find the probability of one of the events and multiply it by the number of different possible arrangements.

If you enjoyed this article make sure to check out our other How To articles like: Efficient Learning &

Contributor: Altea Sulollari
Date: 20th November 2020

Posted on
06
Aug 2020

Probability GMAT Problem

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.

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

If you enjoyed this GMAT probability problem, try your hand at these other types of challenging problems: Combinatorics & Algebra