GMAT Probability Problems

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Last Updated on November 22, 2023

GMAT probability is one of many math topics you can expect to see on test day. And, many students find probability to be one of the more advanced GMAT quant topics. However, after learning a few basic probability rules and procedures, you’ll be surprised by how easy probability questions can be for you.

In this blog, we will discuss some probability theory and GMAT probability fundamentals, and work through a handful of sample probability GMAT questions. The questions we will solve will all relate to event probability. Probability is nothing more than the numerical likelihood that an event occurs. Examples of events include picking students from a class, flipping a coin, or selecting fish from a lake. Situations like these frequently show up in probability questions, so you’ll find this practice very useful.

Comprehensive coverage of the topic of probability is provided in the Target Test Prep GMAT course.

GMAT Probability Problems

Here are the topics we’ll cover in this article:

Let’s start with the probability formula.

The Probability Formula

When solving any GMAT probability question, we always use the same basic formula:

Probability of an event happening = (number of outcomes in which the event can happen) / (total number of possible outcomes)

You will need this formula for both basic and advanced GMAT probability problems, so make sure to add it to your GMAT probability cheat sheet and have it memorized!

TTP PRO TIP:

Memorize the probability formula!

Let’s now practice using the formula with an example question.

Probability Example 1: Probability Formula

A jar contains 20 blue marbles and 30 red marbles. If one marble is randomly drawn from the jar, what is the probability that the drawn marble is blue?

  • 1/4
  • 2/5
  • 3/7
  • 3/5
  • 2/3

Solution:

There are 20 + 30 = 50 marbles in total in the jar. That is, the total number of outcomes is 50, so 50 is the denominator of our probability fraction. Since 20 of these marbles are blue, the number of single-draw outcomes in which a blue marble can occur is 20. So, 20 becomes the numerator of our probability fraction.

P(Blue Marble) = number of blue marbles / total number of marbles

P(Blue Marble) = 20 / 50 = 2/5

Answer: B

Next, let’s discuss the sample space and the probabilities associated with it.

The Sum of All Probabilities in a Sample Space is 1

A sample space is the set of all possible outcomes of an experiment. For example, when you flip a coin one time, there are two possible outcomes in the sample space: {Heads, Tails}. Or if you roll a die one time, there are 6 possible outcomes in the sample space: {1, 2, 3, 4, 5, 6}.

An important fact about probability is that the sum of all probabilities in a sample space equals 1.

The sum of all probabilities in a sample space is 1.

For example, let’s look at flipping a fair coin. There are only two outcomes that can occur: heads or tails. When tossing a coin, the probability that the coin shows heads is 1/2, and the probability that the coin shows tails is 1/2. The sum of the two probabilities is 1/2 + 1/2 = 1. Thus, we see that the sum of all probabilities in this limited-size sample space is 1. This is important when dealing with complementary events.

Next, let’s discuss complementary and mutually exclusive events.

Complementary and Mutually Exclusive Events

In any experiment, when the sum of two probabilities equals 1, we call those events complementary.

KEY FACT:

Two complementary events’ probabilities sum to 1.

A characteristic of complementary events is that if one event occurs, the other cannot occur. We know that when we flip a coin one time, it will show either heads or tails. It cannot show both heads and tails at the same time. We use the term “mutually exclusive” to describe this relationship. Other examples of mutually exclusive events could be the result of taking a college class (pass or fail), driving to work (wreck or no wreck), or running in a race (win or lose).

KEY FACT:

Mutually exclusive events cannot occur at the same time.

Note, however, that not all mutually exclusive events are complementary. A simple example of this would be drawing one card from a standard deck of 52 cards. The two events “drawing a king” and “drawing an ace” are mutually exclusive because one drawn card cannot be both a king and an ace. But these two events are not complementary because their individual probabilities (4/52 and 4/52) do not add to 1.

KEY FACT:

Two mutually exclusive events whose probabilities do not add to 1 are not complementary.

Let’s consider an example that illustrates both complementary and mutually exclusive events.

Illustration of Complementary and Mutually Exclusive Events

When a candy is randomly drawn from a bowl with 20 sour candies and 60 sweet candies, either a sour candy or a sweet candy will be drawn. That is, picking a sour candy and picking sweet candy are mutually exclusive events. Additionally, these are complementary events because their individual probabilities sum to 1. This is illustrated below.

  • Probability of picking a sour candy = 20/80 = 1/4
  • Probability of picking a sweet candy = 60/80 = 3/4
  • Sum of probabilities = 1/4 + 3/4 = 1

The complementary events formula is powerful because if two events are complementary, knowing the probability that one event will occur allows us to easily calculate the probability that the other event will occur.

From above, if we know that the probability of selecting a sour candy is 1/4, then the probability of selecting a sweet candy must be 1 – 1/4 = 3/4.

TTP PRO TIP:

If two events are complementary, knowing the probability of one event allows you to easily calculate the probability of the other.

Let’s practice with an example.

Probability Example 2: Complementary Event Probabilities

The probability that it will rain in town X tomorrow is 25 percent. What is the probability that it will not rain in town X tomorrow?

  • 0
  • 0.25
  • 0.50
  • 0.75
  • 1.00

Solution:

It will either rain or not rain. Thus, there are only 2 outcomes in this experiment. So, “rain” and “not rain” are complementary events, and their probabilities sum to 1. The probability of rain tomorrow is P(rain) = 25% or 0.25. Thus, the probability that it will not rain is P(not rain) = 1 = 0.25 = 0.75.

Answer: D

Now, let’s discuss addition and multiplication rules for probabilities.

When We Multiply Probabilities

Anytime we hear the word “and” in a probability question, or even if “and” is implied, we will multiply probabilities.

For example, if a fair coin is tossed two times, and we need to determine the probability that it lands on heads twice, that is the same thing as saying it needs to land on heads on the first flip and on the second flip. Thus, we multiply those two probabilities together.

So, since the probability of heads on any flip of a coin is 1/2, and we want to compute the probability that the coin lands on heads in two successive flips, we multiply the two probabilities, obtaining (1/2) x (1/2) = 1/4.

Now, let’s look at two quick examples in which we need to multiply probabilities. For the first, the probabilities remain constant because we are dealing with independent events. For the second, the probabilities do not remain constant because we are dealing with dependent events.

Independent Events and Dependent Events

In probability, independent events are events for which the probability of one event does not change based on whether the other event occurred. For example, if we toss a coin two times, the probability that the coin lands on heads on each toss does not change, as it’s 1/2 for each toss. In other words, the fact that the coin landed on heads for the first toss does not affect the probability that it will land on heads on the second (or subsequent) tosses.

On the other hand, with dependent events (which give rise to what is called conditional probability), the prior event’s occurrence does affect the probability of the subsequent event. For example, if we select 2 dogs from a group of 10 dogs and 12 cats, we see that the probability of selecting the first dog is 10/22.

When we select the next dog, the probability changes. It changes because there is 1 less dog in the group. So, the probability of selecting the second dog is 9/21. (Note that there are fewer animals from which we can select because the first dog’s selection reduced the number of available animals from 22 to 21.)

Let’s look at a couple of practice questions illustrating these concepts.

Probability Example 3: Multiplication Rule

The probability that a certain archer can hit the target’s bullseye with an arrow is a constant 30 percent. What is the probability that the archer can hit the bullseye four times in a row?

  • 0.0081
  • 0.0039
  • 0.012
  • 0.3
  • 0.5

Solution:

The probability that the archer will hit the bullseye in one shot is 30 percent, or 0.3. We are told that this is a constant value. Therefore, the probability that he will hit the bullseye four times in a row is

0.3 x 0.3 x 0.3 x 0.3 = 0.0081

Notice that each shot that the archer takes is independent of the shot before or after it. The successful outcome is hitting the bullseye. We know that the probability of hitting the bullseye does not change with each shot the archer takes because the problem statement tells us that the probability is a constant 30%.

Answer: A

Probability Example 4: Multiplication Rule

In a room of 6 girls and 6 boys, what is the probability of picking two boys in a row from the room if each boy can be picked only once?

  • 1/6
  • 5/22
  • 1/4
  • 11/23
  • 12/23

Solution:

There are 6 boys in a group of 12 people. Thus, the probability of picking the first boy is 6/12 = 1/2. After the first boy has been picked, there are only 5 boys to be selected out of 11 available people. The probability of picking the second boy is then 5/11. Notice that the probability of picking the second boy was affected by the fact that one boy had already been selected. The probability of picking both boys is:

P(first boy and second boy) = 1/2 x 5/11 = 5/22.

Answer: B

Next, let’s discuss some addition rules.

Addition Rule for Mutually Exclusive Events

When two events are mutually exclusive (meaning that they can’t happen at the same time), the probability that either event A happens or event B happens can be determined using the following formula:

probability of event A or B = probability of event A + probability of event B

We use this addition rule when we encounter the word “or” in a probability question and the events are mutually exclusive.

Let’s illustrate this addition rule with a straightforward example:

From a standard deck of 52 cards, we draw one card. The probability that it is either an ace or a king is found by adding the probability of drawing an ace to the probability of drawing a king:

P(ace or king) = P(ace) + P(king)

Since we cannot draw a card that is both an Ace and a King, we know that these two events are mutually exclusive, so we can add their individual probabilities. Thus, we have:

P(ace or king) = P(ace) + P(king) = 4/52 + 4/52 = 8/52 = 2/13

Of course, this addition formula can be extended to more than two events. For example, given that events A, B, C, and D are mutually exclusive, the probability that event A, B, C, or D will occur is:

probability of event A or B or C or D = probability of A + probability of B + probability of C + probability of D

TTP PRO TIP:

When events A and B are mutually exclusive, the probability of event A or B = probability of A + probability of B.

Let’s practice applying this rule with an example.

Probability Example 5: Addition Rule #1

At an animal shelter, there are 6 cats, 11 dogs, and 3 birds. If one animal is to be selected at random from the shelter for a TV commercial, what is the probability of selecting a cat or a bird?

  • 9/200
  • 9/40
  • 6/20
  • 9/20
  • 17/20

Solution:

The events cat, dog, and bird are mutually exclusive. Therefore, we use the addition rule for mutually exclusive events to determine the probability that, of the 20 animals, we draw one animal that is either a cat or a bird:

P(cat or bird) = P(cat) + P(bird) = 6/20 + 3/20 = 9/20.

Answer: D

Next, let’s see what happens when we encounter two events that are not mutually exclusive.

Addition Rule for Non-Mutually Exclusive Events

We know that two events are mutually exclusive when they can’t both happen at the same time. So, how would we deal with events that can happen at the same time? In that case, we determine the probability of A or B by using the following formula:

probability of event A or B = probability of A + probability of B – the probability of A and B

Notice that because these two events can happen at the same time, we must subtract the probability of their overlap.

Let’s consider an example that illustrates this.

If we want to calculate the probability of drawing one card that is either a queen or a heart from a standard deck of 52 cards, we must first realize that the events “queen” and “heart” are not mutually exclusive. That is, one card can be both a queen and a heart (the queen of hearts). Thus, when we calculate the probability of drawing a queen or a heart in one draw, we double-count the queen of hearts as both a queen and a heart. But since the queen of hearts is only one card, we must subtract out its probability to offset the double-counting. Thus, the probability is calculated as follows:

P(queen or heart) = P(queen) + P(heart) – P(queen and heart)

Let’s practice with an example.

Probability Example 6: Addition Rule #2

If a number is to be selected at random from the numbers from 1 to 25, inclusive, what is the probability that the number is either a perfect square or an even number?

  • 1/5
  • 2/5
  • 3/5
  • 17/25
  • 19/25

Solution:

For the set of numbers from 1 to 25, inclusive, let’s let S be the event “perfect square” and E be the event “even number.”

The perfect squares between 1 and 25, inclusive are 1, 4, 9, 16, and 25. Thus, P(S) = 5/25. The even numbers between 1 and 25, inclusive, are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, and 24. Thus, P(E) = 12/25.

We now must decide if the events S and E are mutually exclusive, and we quickly see that 2 numbers (4 and 16) are both even numbers and perfect squares. Thus, P(S and E) = 2/25. Since one number can be both a perfect square and an even number, we know that events S and E are not mutually exclusive. Thus, we will use the addition formula for non-mutually exclusive events to perform the calculation, as follows:

P(S and E) = P(S) + P(E) – P(S and E)

P(S and E) = 5/25 + 12/25 – 2/25

P(S and E) = 15/25 = 3/5

Answer: C

Let’s now take a look at the final probability topic: using combinatorics to answer probability questions.

Combinatorics and Probability

You may recall that we use the combination formula when we are selecting r items out of n items. The formula is:

nCr = n! / [r! X (n-r)!]

For example, let’s say that we have 4 red marbles and 5 green marbles in a jar. We’ll choose 5 marbles from the jar, and we want to know the probability of selecting exactly 2 red marbles and 3 green marbles.

We need to determine the following, which is a restatement of the basic probability formula that we used earlier:

P(2 red and 3 green) = (number of ways to select 2 red and 3 green) / (total number of ways to select 5 marbles out of 9)

First, let’s calculate the numerator of this probability fraction.

Number of ways to select 2 red marbles out of 4

Here, n = 4 and k = 2, so we have:

4C2 = 4! / [2! X (4 – 2)!] = (4 x 3 x 2 x 1) / (2 x 1 x 2 x 1) = 24/4 = 6

Number of ways to select 3 green marbles out of 5

Here, n = 5 and k = 3, so we have:

5C3 = 5! / [3! X (5 – 3)!] = (5 x 4 x 3 x 2 x 1) / (3 x 2 x 1 x 2 x 1) = 120 / 12 = 10

The 6 ways to select 2 red marbles are multiplied by the 10 ways to select 3 green marbles, giving us 6 x 10 = 60 ways in total to select 2 red and 3 green marbles. Thus, 60 is the numerator of our probability fraction.

Let’s now calculate the denominator of the probability fraction.

Number of ways to select 5 marbles out of 9

Here, n = 9 and k = 5, so we have:

9C5 = 9! / [5! X (9 – 5)!]
= (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (5 x 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1)
= 362,880 / 2,880
= 126

The total number of ways to select 5 marbles out of 9 is 126.

We now substitute our calculated values into the probability formula:

P(2 red and 3 green) = (number of ways to select 2 red and 3 green) / (total number of ways to select 5 marbles out of 9)

P(2 red and 3 green) = (6 x 10) / 126 = 60 / 126 = 0.476

Let’s try an example.

Probability Example 7: Combinatorics

At a veterinary clinic, there are 5 cats and 3 dogs waiting to be seen. The vet technician chooses 3 animals to be seen in the 3 available treatment rooms. Which of the following is closest to the probability that the technician will choose 2 cats and 1 dog?

  • 0.08
  • 0.18
  • 0.22
  • 0.25
  • 0.38

Solution:

We need to determine the following, which is a restatement of the basic probability formula that we used earlier:

P(2 cats and 1 dog) = (number of ways to select 2 cats and 1 dog) / (total number of ways to select 3 animals out of 8)

First, let’s calculate the numerator of this probability fraction.

Number of ways to select 2 cats out of 5

Here, n = 5 and k = 2, so we have:

5C2 = 5! / [2! X (5-2)!] = (5 x 4 x 3 x 2 x 1) / ( 2 x 1 x 3 x 2 x 1) = 120/12 = 10

Number of ways to select 1 dog out of 3

Here, n = 3 and k = 1, so we have:

3C1 = 3! / [1! X (3-1)!] = ( 3 x 2 x 1) / (1 x 2 x 1) = 6 / 2 = 3

The 10 ways to select 2 cats are multiplied by the 3 ways to select 1 dog, giving us 10 x 1 = 10 total ways to select 2 cats and 1 dog. Thus, 10 is the numerator of our probability fraction.

Let’s now calculate the denominator of the probability fraction.

Number of ways to select 3 animals out of 8

Here, n = 8 and k = 3, so we have:

8C3 = 8! / [3! X (8-3)!]
= (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1 x 5 x 4 x 3 x 2 x 1)
= 40,320 / 720
= 56

The total number of ways to select 3 animals out of 8 is 56.

We now substitute our calculated values into the probability formula:

P(2 cats and 1 dog) = (number of ways to select 2 cats and 1 dog) / (total number of ways to select 3 animals out of 8)

P(2 cats and 1 dog) = 10 / 56 = 5 / 28 = 0.18

Answer: B

GMAT Probability Cheat Sheet

Probability questions on the GMAT are challenging. But learning some basic probability rules and practicing a variety of probability problems can help you to get a solid grasp of the topic. 

Use this GMAT probability cheat sheet of key concepts to get started:

  • The basic probability formula lays the foundation for all probability questions. The numerator is the number of favorable outcomes, and the denominator is the number of all possible outcomes. 
  • Complementary events are two events whose probabilities sum to 1. This concept is straightforward and is often the first step in solving more complicated probability questions.
  • The multiplication rule of probability is used when you want to calculate the probability of two or more events happening. The words “and” and “both” are common hints that the multiplication rule will come into play. 
  • Two events are described as independent if the occurrence of one event does not change the probability of the other event. In contrast, two events are described as dependent if the probability of one event is changed because the other event has occurred.
  • We use the addition rule of probability when we want to calculate the probability that either of two (or more) events occurs. The words “or” and “either” are common in questions that require the use of the addition rule.

Probabilities can be calculated by using combinatorics. Using combinatorial mathematics can make answering some tough probability questions very straightforward.

What’s Next?

Probability is only one of 20+ major topics tested on the GMAT. So, while it’s important to master probability, it’s also critical that you perform well on a variety of quant question types that could be thrown your way on test day. Read our top 10 GMAT quant tips to get a leg up on the entire quantitative section!

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