Last Updated on December 7, 2023
I work with GRE students every day who struggle with math, and one of their major struggles is with GRE probability questions. However, some of their concerns are more the result of fear than of any inherent difficulty in probability questions. In fact, many of them are surprised at how easily they can solve even the hardest GRE probability problems once they learn a few important probability rules and concepts.
With that in mind, in this article, we’ll cover some probability concepts for the GRE and work through some GRE probability questions with solutions.
While this article can be used for some initial GRE probability test prep or as an intro to basic probability study materials, we have a more in-depth coverage of probability in the Target Test Prep GRE self-study course, so you may want to check out the course after you’re done reading.
Here are the topics we’ll cover in this article:
- What Is Probability?
- The Probability Formula
- The Sum of All Probabilities in a Sample Space is 1
- Complementary and Mutually Exclusive Events
- When We Multiply Probabilities
- Addition Rule for Mutually Exclusive Events
- Addition Rule for Non-Mutually Exclusive Events
- Combinatorics and Probability
- How to Strategically Use Combinations in Probability
- Summary
- What’s Next?
Before jumping into GRE probability practice, let’s first discuss what probability actually is, and do a brief probability review.
What is Probability?
In the most basic sense, probability is the chance that a given event will occur. For example, probability is the chance that a red ball will be selected from a bucket containing blue and red balls or the chance that a child 5 years old or younger will be selected from a kindergarten class.
With this informal definition, let’s discuss one of the most important GRE probability formulas.
The Probability Formula
In basic probability questions for the GRE, we will use the following formula:
Probability of an event happening = (number of outcomes in which the event can happen) / (total number of possible outcomes)
Since this formula is used for virtually all GRE probability questions, it’s worth memorizing, so you can access it at the drop of a hat. We will also be using it often in the probability examples later in this article.
TTP PRO TIP:
Having the probability formula memorized is the first key to crushing GRE probability questions.
So, we can now start our GRE probability practice. It’s helpful to note that the majority of questions on this topic really are GRE probability word problems.
Probability Example 1: Probability Formula
A class roster contains 20 names of brown-eyed students and 30 names of blue-eyed students. If one name is randomly selected from the roster, what is the probability that the selected name is that of a brown-eyed student?
- 1/4
- 2/5
- 3/7
- 3/5
- 2/3
Solution:
There are 20 + 30 = 50 names on the roster. That is, the total number of outcomes is 50, so 50 is the denominator of our probability fraction. Since 20 of these names are those of brown-eyed students, the number of single-draw outcomes in which a brown-eyed student is chosen is 20. So, 20 becomes the numerator of our probability fraction.
P(Brown-eyed student) = number of brown-eyed students / total number of students
P(Brown-eyed student) = 20 / 50 = 2/5
Answer: B
An important part of probability theory is that the sum of all probabilities in any sample space is 1. Let’s discuss.
The Sum of All Probabilities in a Sample Space is 1
The set of all possible results of an experiment is referred to as the sample space. For instance, there are two possible outcomes in the sample space as a result of flipping a coin once: Heads, Tails.
Similarly, if you roll a die one time, there are six different outcomes in the sample space: 1, 2, 3, 4, 5, 6.
One of the most important things to remember about probability is that the sum of all the probabilities of the items in a sample space is 1.
For example, let’s look at the probability of drawing a red marble from a jar that contains four red and five blue marbles. We see there are just two possible outcomes: drawing a red marble or drawing a blue marble.
The probability that a red marble is drawn is 4/9, and the probability that a blue marble is drawn is 5/9. The sum of the two probabilities is 4/9 + 5/9 = 9/9 = 1. As a result, we see that the sum of all probabilities in this scenario is 1. When dealing with occurrences that complement each other, this fact is important.
Let’s discuss such occurrences next.
Complementary and Mutually Exclusive Events
Events are considered complementary when a) there are only two possible outcomes, and b) if one outcome occurs, the other one cannot occur. Also, the sum of the probabilities of complementary events always equals 1.
KEY FACT:
The sum of the probabilities of complementary events always equals 1.
An example of a complementary event is flipping a coin. If you flip a fair coin one time, it will show either heads or tails, but certainly not both at the same time! Also, the probability that the coin lands on heads is 1/2, and for tails it is also 1/2. So, we see that the sum of those probabilities is 1/2 + 1/2 = 1.
We use the term mutually exclusive to describe the relationship between the events that a coin shows heads or that it shows tails. Recall that mutually exclusive events cannot occur at the same time. One flip of a coin shows either heads or tails. In a jar of only blue and red marbles, one draw yields either a blue marble or a red marble.
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 selecting a marble from a bowl containing 9 blue, 3 green, and 4 red marbles. The two events “selecting a blue marble” and “selecting a green marble” are mutually exclusive because one marble cannot be both blue and green. But these two events are not complementary because their individual probabilities (9/16 and 3/16) 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
Let’s go back to our earlier marble example. We were given that there are four red and five blue marbles in a jar. When selecting a marble from the jar, either a red marble or blue marble will be drawn, which of course means these events are mutually exclusive. These events are also complementary because their probabilities add up to 1. Let’s explain this further.
The probability of selecting a red marble is 4/9, and the probability of selecting a blue marble is 5/9. We see the sum of the probabilities is 4/9 + 5/9 = 9/9 = 1. The key fact is that there are just red and blue marbles in the jar.
Another interesting aspect of complementary events is that the probability of one event can be used to determine the probability of the other event.
For example, since the probability of selecting a red marble was 4/9, the probability of selecting a blue marble was 1 – 4/9 = 9/9 – 4/9 = 5/9.
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.
Let’s practice with an example.
Probability Example 2: Complementary Event Probabilities
The probability that I will miss the bus tomorrow morning is 15 percent. What is the probability that I will not miss the bus tomorrow morning?
- 0
- 0.15
- 0.70
- 0.85
- 1.00
Solution:
I will either miss the bus or not miss the bus. Thus, there are only 2 outcomes in this experiment. So, “miss bus” and “not miss bus” are complementary events, and their probabilities sum to 1. The probability that I miss the bus is P(miss bus) = 15% or 0.15. Thus, the probability that I do not miss the bus is P(not miss bus) = 1 – 0.15 = 0.85.
Answer: D
Now, let’s discuss the multiplication rule for probabilities.
When We Multiply Probabilities
If a probability question mentions two events and asks the probability that both happen, we multiply the two probabilities. If the question contains the word “and,” or even if it just implies the word “and,” we multiply the probabilities.
For instance, let’s say a coin is tossed twice and we want to calculate the probability that it lands on heads both times. This is the same as saying it must land on heads on toss one and on toss two. Thus, to determine the answer, we multiply the probability that it lands on heads on the first flip (which is 1/2) by the probability that it lands on heads on the second flip (which is also 1/2).
Doing so gives us the answer 1/4, since (1/2) x (1/2) = 1/4. Why? Because the probability of getting heads on any one coin flip is 1/2.
Next, let’s discuss some GRE probability tips related to the difference between dependent and independent events.
Independent Events and Dependent Events
In probability, two occurrences are considered independent if there is no effect on the likelihood of either occurring depending on whether the other event took place.
For example, if we were to roll a fair 6-sided die two times, the probability that the die yields a “1” on the first roll is 1/6. The probability that the die yields a “1” on the second roll is also 1/6. The probability of getting a “1” doesn’t change from one roll to the next.
To put it another way, the fact that the die lands on “1” on the first roll has no bearing on the likelihood that it will land on “1’ on the second (or any subsequent) roll. Thus, getting a “1” on the first roll and getting a “1” on the second roll are independent events.
Dependent Events
On the other hand, the occurrence of a prior event does impact the probability of a later event when the events are dependent events. This concept gives rise to what is called conditional probability, in which the probability of an event changes because another event has happened. Consider the events “rain” and “going to the beach.” If it rains, the probability of going to the beach decreases. These two events are dependent events.
Here’s another example. Let’s calculate the probability of picking 2 dogs from a group of 10 dogs and 12 cats. First, we see that the probability of picking the first dog is 10/22.
The probability changes for the selection of the second dog. There are now only 9 dogs from which to choose. Therefore, the probability of picking the second dog is 9/21. (Note that the number of eligible animals has decreased from 22 to 21 as a result of selecting the first dog.
Let’s look at some GRE probability practice questions illustrating these concepts.
Probability Example 3: Multiplication Rule
The probability that Mimi locks her keys in her car on any given day is a constant 5 percent. What is the probability that she locks her keys in her car on two successive days?
- 0.0025
- 0.025
- 0.05
- 0.10
- 0.25
Solution:
The probability that Mimi locks her keys in the car on any given day is 5%, or 0.05. We are told that this is a constant value. Therefore, the probability that she will lock her keys in the car on two successive days is the product of the individual probabilities:
0.05 x 0.05 = 0.0025
Notice that each day’s occurrence is independent of the day before or after it. The event of interest is “Mimi locks her keys in the car.” We know that the probability of this event does not change from day to day because the question stem tells us that the probability is a constant 5%.
Answer: A
Probability Example 4: Multiplication Rule
In a room of 8 girls and 6 boys, what is the probability of first picking a boy and then a girl from the room if each child can be picked only once?
- 12/49
- 24/91
- 7/24
- 17/43
- 1/2
Solution:
There are 6 boys and 8 girls in a room of 14 children. Thus, the probability of first selecting a boy is 6/14 = 3/7. After the boy has been selected, we see that the number of girls is 8, but the number of children remaining is only 13. Thus, the probability of picking a girl is 8/13. Notice that the probability of picking the girl was affected by the fact that one child had already been selected, reducing the pool of available selectees from 14 to 13. Thus, the probability of picking a boy first and a girl second is:
P(boy first and girl second) = 3/7 x 8/13 = 24/91.
Answer: B
Next, let’s discuss some addition rules.
Addition Rule for Mutually Exclusive Events
When discussing some earlier probability strategies, we learned that mutually exclusive events cannot happen at the same time. So, if events A and B are mutually exclusive, the probability that event A or event B occurs is:
Probability of event A or event B = probability of event A + probability of event B
We should notice the keyword “or” is used to indicate that we have to add the probabilities.
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 a queen or a jack is found by adding the probability of drawing a queen and the probability of drawing a jack:
P(queen or jack) = P(queen) + P(jack)
Since we cannot draw a single card that is both a queen and a jack, we know that these two events are mutually exclusive, so we can add their individual probabilities. Thus, we have:
P(queen or jack) = P(queen) + P(jack) = 4/52 + 4/52 = 8/52 = 2/13
Remember, the above rule does not just have to be for two events. The formula works for any number of events.
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 a bakery counter, there are 6 doughnuts, 11 muffins, and 3 cookies. If one item is to be selected at random, what is the probability that it is a doughnut or a cookie?
- 9/200
- 9/40
- 6/20
- 9/20
- 17/20
Solution:
The events doughnut, muffin, and cookie are mutually exclusive. Therefore, we use the addition rule for mutually exclusive events to determine the probability that, of the 20 bakery items, we select one item that is either a doughnut or a cookie:
P(doughnut or cookie) = P(doughnut) + P(cookie) = 6/20 + 3/20 = 9/20.
Answer: D
Next, let’s discuss events that are not mutually exclusive.
Addition Rule for Non-Mutually Exclusive Events
Two events are said to be mutually exclusive when they cannot occur simultaneously. So, what do we do when two events can take place at the same time? Well, we have a very simple process we can follow. The likelihood of either outcome A or B is calculated by applying the following formula:
probability of event A or B = probability of A + probability of B – the probability of A and B
We see that because the two events (A and B) can happen at the same time, we subtract the probability of their overlap.
Let’s consider an example that illustrates this concept.
If we want to calculate the probability of drawing one card that is either a king or a diamond from a standard deck of 52 cards, we must first realize that the events “king” and “diamond” are not mutually exclusive. That is, one card can be both a king and a diamond (the king of diamonds). Thus, when we calculate the probability of drawing a king or a diamond in one draw, we double-count the king of diamonds as both a king and a diamond. But since the king of diamonds is only one card, we must subtract out its probability once, to offset the double-counting. Thus, the probability is calculated as follows:
P(king or diamond) = P(king) + P(diamond) – P(king and diamond)
Let’s practice with an example.
Probability Example 6: Addition Rule #2
We randomly select one number from the set of numbers from 1 to 25, inclusive. What is the probability that the number is either an odd number or a number greater than 21?
- 1/5
- 2/5
- 3/5
- 17/25
- 19/25
Solution:
For the set of numbers from 1 to 25, inclusive, let’s let O be the event “odd number” and G be the event “a number greater than 21.”
The odd numbers between 1 and 25, inclusive, are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, and 25. Thus, P(O) = 13/25. The numbers greater than 21 are 21, 22, 23, 24, and 25. Thus, P(G) = 5/25 = 1/5.
We now must decide if the events O and G are mutually exclusive, and we quickly see that 3 numbers (21, 23, and 25) are both odd numbers and greater than 21. Thus, P(O and G) = 3/25. Since one number can be both odd and greater than 21, we know that events O and G are not mutually exclusive. Thus, we will use the addition formula for non-mutually exclusive events to perform the calculation, as follows:
P(O or G) = P(O) + P(G) – P(O and G)
P(O or G) = 13/25 + 5/25 – 3/25
P(O or G) = 15/25 = 3/5
Answer: C
Let’s now take a look at the final probability topic: GRE probability and combinatorics.
Combinatorics and Probability
If you are familiar with combinations and permutations, then I have some good news for you. Those concepts can be used with GRE math probability questions! Before getting into how they can be used, let’s do a brief review of combinations.
We use combinations when determining how many ways an item can be selected from a larger group of items. Specifically, when determining the number of ways of selecting r items out of n items, we use the following formula:
nCr = n! / [r! X (n-r)!]
Let’s use this formula to solve a combinations question, and then we will use it again to solve a very similar probability question.
Combinations Example
There are 6 fitted hats and 4 adjustable hats in a bin. If 4 hats are selected, in how many ways can 2 fitted hats and 2 adjustable hats be selected?
- 80
- 90
- 105
- 120
- 135
Solution:
First, we determine the number of ways to select two fitted hats from six fitted hats.
Thus, n = 6 and r = 2:
6C2 = 6! / [2! X (6 – 2)!] = (6 x 5 x 4 x 3 x 2 x 1) / (2 x 1 x 4 x 3 x 2 x 1) = (6 x 5)/2 = 15 ways
Next, we determine the number of ways to select two adjustable hats from four adjustable hats.
Thus, n = 4 and r = 2:
4C2 = 4! / [2! X (4 – 2)!] = (4 x 3 x 2 x 1) / (2 x 1 x 2 x 1) = 6 ways
Thus, the total number of ways to select two fitted and two adjustable hats is:
15 x 6 = 90 ways
Answer: B
Now, let’s see how we can use what we did above in probability questions.
How to Strategically Use Combinations in Probability
Previously, we solved a combination question related to selecting fitted and adjustable hats. This problem can be converted to a probability question with a simple tweak. Also, we can use combination methods to solve the problem! Let’s look at the adjusted example.
Combination Probability Example
There are 6 fitted hats and 4 adjustable hats in a bin. If 4 hats are selected, what is the probability that 2 fitted hats and 2 adjustable hats are selected?
- 1/7
- 2/7
- 3/7
- 4/7
- 5/7
Solution:
We should notice right away that the only difference between the previous question and the one above is what we have in bold. Even better, we will be able to use the work we did in the previous question to solve for the favorable outcomes in our probability ratio.
Favorable Outcomes
First, we determine the number of ways to select two fitted hats from six fitted hats.
Thus, n = 6 and r = 2:
6C2 = 6! / [2! X (6 – 2)!] = (6 x 5 x 4 x 3 x 2 x 1) / (2 x 1 x 4 x 3 x 2 x 1) = (6 x 5)/2 = 15 ways
Next, we determine the number of ways to select two adjustable hats from four adjustable hats.
Thus, n = 4 and r = 2:
4C2 = 4! / [2! X (4 – 2)!] = (4 x 3 x 2 x 1) / (2 x 1 x 2 x 1) = 6 ways
Thus, the total number of ways to select two fitted and two adjustable hats is:
15 x 6 = 90 ways
Again, the work above is from the previous combinations example.
Total Outcomes
Next, to finish the problem, we just need to determine the total number of outcomes, which is selecting four hats from ten.
Thus, n = 10 and r = 2:
10C4 =
10! / [4! X (10 – 4)!]
= (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / ( 6 x 5 x 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1)
= (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)
= 10 x 3 x 7
= 210
With all this, we can use the basic probability formula to determine the probability that two fitted hats and two adjustable hats are selected:
Probability of an event happening = (number of favorable outcomes) / (number of possible outcomes)
Probability (two fitted and two adjustable hats) = 90 / 210 = 9 / 21 = 3/7
Answer: C
Summary
GRE probability questions can be challenging. But, if you invest time and effort into learning the basic probability formulas and rules, you can make a huge dent in probability.
In this article, we learned the following key concepts:
- The basic probability formula: number of favorable outcomes / number of all possible outcomes.
- Complementary events: events whose probabilities sum to 1.
- The multiplication rule of probability: used when you want to calculate the probability of two or more events happening.
- Independent events: If the occurrence of one event does not change the probability of the other event, the events are independent.
- Dependent events: If the probability of one event is changed because the other event has occurred, the events are dependent.
- Addition rule of probability: used to calculate the probability that either of two (or more) events occurs. P(A or B) = P(A) + P(B) when two events A and B are mutually exclusive.
- If two events are not mutually exclusive, the addition rule is P(A or B) = P(A) + P(B) – B(A and B).
- Probability questions can sometimes be answered by using combinatorics.
What’s Next?
While it’s important to master concepts for the GRE, it’s also important to have a proper mindset about doing well on GRE quant. You’ll find that no one is terrible at GRE math as long as they have a positive attitude and commit to some hard work!
