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# GMAT Probability Problems and Questions

by | Jul 21, 2024 | GMAT

#### Fatma Xhafa

Fatma is our Director of Operations and has been with Apex for over 3 years. With a background in computer science and business, she has the skills and expertise to manage day-to-day operations and lead the team effectively.

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.

In the GMAT exam, these questions are found in the Quantitative Reasoning section.

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.

## Understanding the Fundamentals:

Before diving into specific problem-solving strategies, it’s essential to grasp the three fundamental combinatorics tasks that form the basis of GMAT probability questions:

Ordering: Lining up elements spatially or temporally

Permutations: Selecting and ordering a subgroup of elements

Combinations: Selecting a subgroup of elements without regard for their order

Recognizing which task(s) a probability question involves is crucial to applying the appropriate problem-solving approach.

The ordering task is “lining up” elements spatially or temporally (in time).

Spatial: How many ways can 7 cars be lined up in 7 parking spaces?

Temporal: In how many different orders can 7 contestants perform in a talent show?

There is no mathematical difference between these two scenarios. Both can be represented with the same standard formula.

### The Permutations Task: Selecting and Ordering

The permutations task is “lining up” a subgroup of elements spatially or temporally. We may refer to permutations as “selecting and ordering.”

We will order elements in time or space, but only the subgroup of elements that have been selected out of the larger group. As with the ordering task, spatial and temporal instances of the permutations task are mathematically identical.

Spatial: How many ways can 4 cars from a collection of 7 cars be lined up in 7 parking spaces?

Temporal: The judges of a talent show will create a performance schedule for 4 finalists selected from among 7 semifinalists. How many different schedules can be created?

### The Combinations Task: Selecting Without Ordering

The combinations task is selecting a subgroup of elements out of a larger group, without regard for their order. We may refer to combinations as “selecting without ordering,” or simply “selecting.” In combinations, the spatial/temporal distinction disappears altogether.

Since nothing is being “lined up,” there is no time or space where this “lining up” occurs. Selecting a subgroup out of a larger group is not a spatial or temporal operation.

Spatial: How many ways can 4 cars be selected from a collection of 7 cars?

Temporal: How many ways can 4 finalists be chosen from among 7 semifinalists in a talent show?

With all three tasks in view, it makes sense to think of permutations as “selecting, then ordering.”

Permutations entail performing the combinations task and then performing the ordering task on the selected subgroup. In the coming articles, we will see how the standard formulas for each task relate and support this understanding.

## Breaking Down the Problem

Let’s examine a sample GMAT probability question to demonstrate how to break it down into manageable parts:

“Xavier, Yvonne, and Zelda are solving problems independently. The probability of Xavier solving a problem is 1/4, Yvonne solving it is 1/2, and Zelda solving it is 5/8. What is the probability that Xavier and Yvonne solve a problem, but Zelda does not?”

Step 1: Determine the probability of Zelda not solving the problem.

If Zelda solves a problem 5 out of 8 times, she doesn’t solve it the other 3 out of 8 times (3/8).

Step 2: Treat each person’s probability independently.

Since each person’s probability is independent of the others, we can calculate the probabilities separately and then multiply them together.

Visualizing the Solution:

To better understand the problem, let’s visualize the solution using a tree diagram or a table.

For every four attempts by Xavier, he solves the problem once (1/4). If and only if Xavier solves the problem, we move on to Yvonne’s probability. Yvonne solves the problem 1 out of 2 times (1/2). Therefore, the probability of both Xavier and Yvonne solving the problem is 1/4 × 1/2 = 1/8.

Finally, we multiply the probability of Xavier and Yvonne solving the problem (1/8) by the probability of Zelda not solving it (3/8). This gives us the final probability: 1/8 × 3/8 = 3/64.

Interpreting the Result:

The final probability of 3/64 means that out of every 64 attempts, there will be 3 instances where Xavier and Yvonne solve the problem, but Zelda does not.

Mastering GMAT probability questions requires a solid understanding of the fundamental combinatorics tasks, the ability to break down complex problems into manageable parts, and the use of visualization techniques to clarify the solution process.

## GMAT Probability – Fundamental Rules & Formulas

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 in these types of questions is mastering the Probability formula:

Probability = 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)

### Complementary and Mutually Exclusive Events

#### Complementary Events:

Complementary events are pairs of events where one event is the opposite of the other. In other words, if event A occurs, then its complement, event A′ (read as “A prime”), cannot occur, and vice versa. The probabilities of complementary events always sum up to 1.

Example:

Event A: Drawing a red card from a standard deck of 52 cards.

Event A′: Drawing a card that is not red (i.e., drawing a black card).

The probability of drawing a red card (event A) is 26/52 = 1/2, and the probability of drawing a black card (event A′) is also 26/52 = 1/2. Notice that P(A) + P(A′) = 1/2 + 1/2 = 1.

In GMAT probability questions, you may be given the probability of an event and asked to find the probability of its complement. To do this, simply subtract the given probability from 1.

#### Mutually Exclusive Events:

Mutually exclusive events are events that cannot occur at the same time. In other words, if event A occurs, then event B cannot occur, and vice versa. The intersection of mutually exclusive events is always an empty set.

Example:

Event A: Rolling an even number on a six-sided die.

Event B: Rolling a prime number on the same die.

The possible outcomes for event A are {2, 4, 6}, and the possible outcomes for event B are {2, 3, 5}. Notice that the events share only one common outcome, the number 2. However, in a single roll of the die, you cannot obtain an even and a prime number simultaneously. Therefore, events A and B are mutually exclusive.

When dealing with mutually exclusive events, the probability of either event occurring is the sum of their individual probabilities. In the example above, P(A or B) = P(A) + P(B) = 3/6 + 3/6 = 1.

It’s important to note that complementary events are not always mutually exclusive, and mutually exclusive events are not always complementary.

For example, when drawing a card from a standard deck, the events “drawing a heart” and “drawing a black card” are mutually exclusive but not complementary, as there are other possibilities (e.g., drawing a red diamond).

## 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 you to determine what type of events you are 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.

## One Problem, Two Approaches: When Probability Meets Combinatorics

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 more difficult 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?

1. 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 with 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 the 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 canceling 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 the 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 the 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.

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?

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

## GMAT Probability Cheat Sheet

This cheat sheet provides a breakdown of key concepts, formulas, and problem-solving strategies to help you tackle GMAT probability questions with confidence.

Probability Formula:

• Probability = number of desired outcomes / total number of possible outcomes
• The probability of an event not happening = 1 – probability of the event occurring

Dependent (Disjoint) Events:

• Two or more events with a probability of simultaneous occurrence equaling zero
• Probability P of events A and B = P(A) + P(B)
• If events A and B are not disjointed: P(A and B) = P(A) + P(B) – P(A and B)

Independent (Discrete) Events:

• Two or more events that do not have any effect on each other
• Probability P of events A and B = P(A) × P(B)

Complementary Events:

• Pairs of events where one event is the opposite of the other
• P(A) + P(A′) = 1, where A′ is the complement of A

Mutually Exclusive Events:

• Events that cannot occur at the same time
• P(A or B) = P(A) + P(B), where A and B are mutually exclusive

Problem-Solving Strategies:

• Understand the three fundamental combinatorics tasks: ordering, permutations, and combinations
• Break down complex problems into manageable parts
• Use visualization techniques (e.g., tree diagrams or tables) to clarify the solution process
• Identify dependent and independent events using keywords like “or” and “and”

Keywords:

• “OR” indicates dependent events; add the probabilities
• “AND” indicates independent events; multiply the probabilities

Visual Representations:

• Draw visual representations of events to determine if you should include or exclude the intersect

Two Main Approaches to Solving Probability Problems:

1. Combinatorics Approach:
• Involves counting the number of favorable outcomes and total possible outcomes
• Relies on understanding permutations and combinations
• Useful when the problem provides information about the number of ways events can occur
1. Pure Probability Approach:
• Involves using probability formulas and rules directly
• Relies on understanding the relationships between events (dependent, independent, complementary, or mutually exclusive)
• Useful when the problem provides probabilities of events rather than the number of ways they can occur

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 a tailored GMAT curriculum are ideal for gaining more test confidence and understanding the underlying purpose of each question, which can bridge your future GMAT score and your desired business school admission.