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When preparing for the SAT Math section, you’ll often come across probability questions that test your ability to reason through outcomes and likelihoods. SAT probability questions require you to calculate the likelihood of an event by dividing the number of favorable outcomes by the total number of possible outcomes. Common SAT questions test simple probability, the use of the addition and multiplication rules, and probabilities with table data. Keep reading to explore strategies, formulas, and practice examples to help you solve these problems with confidence.
Here are the topics we’ll cover:
- What is Probability?
- The Basic Probability Formula
- The Special Addition Rule of Probability
- The General Addition Rule for Non-Mutually Exclusive Events
- The Multiplication Rule of Probability
- Table Probabilities
- Summary
- Frequently Asked Questions (FAQ)
- What’s Next?
Let’s first discuss some basic information about probability.
What is Probability?
In a nutshell, probability is the chance that a particular event will occur. For example, the chance of choosing a green marble from a jar containing only green and red marbles is the probability of that event.
The study of probability starts with understanding what a sample space is. A sample space consists of all possible outcomes of an experiment. For example, the sample space for rolling a die is the set of outcomes: {1, 2, 3, 4, 5, 6}. The sample space for flipping a coin twice is the set of outcomes: {(H,H), (H,T), (T,H), (T,T)}.
KEY FACT:
Probability is the chance that a particular event will occur.
This leads us to the more formal definition of probability.
The Basic Probability Formula
Here is the basic probability formula:
Probability of an event happening = (number of favorable outcomes) / (total number of possible outcomes)
We see that for rolling a fair die, the sample space has a total of 6 possible outcomes, one for each of the 6 faces of the die. So, to determine the probability of obtaining a 5 on one roll of the die, we would use the probability formula. We would see that the denominator of the fraction is 6, the total number of possible outcomes in the sample space. And the numerator would be 1, because the favorable outcome of “5” occurs only once. Thus, the probability of rolling a 5 would be expressed as:
P(rolling a 5) = 1/6
The basic probability formula is the foundation for essentially every probability problem you will see on the SAT. Thus, it’s worth memorizing so that it will be readily available for quick recall.
TTP PRO TIP:
The basic probability formula is universally used in SAT probability problems.
Let’s take a look at a straightforward basic probability problem.
Probability Example 1: A Single-Event Probability
A jar contains 16 yellow marbles and 12 purple marbles. If 1 marble is drawn at random, what is the probability that it is purple?
- 2/5
- 3/7
- 4/7
- 3/4
Solution:
There are 16 + 12 = 28 marbles in the jar. Thus, the total number of possible outcomes is 28, and every marble has an equal chance of being selected. Of those 28 marbles, there are 12 purple marbles. Since we are interested in drawing a purple marble, the number of “favorable outcomes” is 12. Thus, we can use the basic probability formula to calculate the probability of drawing a purple marble:
P(purple marble) = number of purple marbles / total number of marbles
P(purple marble) = 12 / 28 = 3/7
Answer: B
The Special Addition Rule of Probability
A more sophisticated probability situation involves the calculation of the probability of either of 2 (or more) events happening. For example, we might be asked to calculate the probability of rolling either a 5 or a 6 on one roll of a 6-sided die. In this case, we would use the Special Addition Rule of Probability, stated as follows:
P(A or B) = P(A) + P(B)
The capital letters A and B stand for the 2 events of interest. We sometimes use descriptions of the events instead of the letters. For our example, we might write the probability statement as:
P(5 or 6) = P(5) + P(6)
In English, we would read the above as “the probability of rolling either a ‘5’ or a ‘6’ on one roll of a die is the probability of rolling a ‘5’ plus the probability of rolling a ‘6.’”
The solution is straightforward. We previously calculated that P(5) = 1/6. Similarly, we can see that P(6) is also 1/6.
Thus, we use the special addition rule of probability to obtain the final answer:
P(5 or 6) = 1/6 + 1/6 = 2/6 = 1/3
KEY FACT:
The Special Addition rule of Probability is: P(A or B) = P(A) + P(B).
Let’s now consider an example problem.
Probability Example 2: The Special Addition Rule
At an animal shelter, there are 6 poodles, 11 schnauzers, and 3 spaniels. If 1 dog is to be selected at random, what is the probability that it is either a poodle or a spaniel?
- 9/40
- 3/10
- 9/20
- 17/20
Solution:
We can use the Special Addition Rule to determine the probability that, of the 20 dogs, we select 1 dog that is either a poodle or a spaniel.
P(poodle or spaniel) = P(poodle) + P(spaniel) = 6/20 + 3/20 = 9/20.
Answer: C
Next, let’s discuss the types of events for which we cannot use the Special Addition Rule.
The General Addition Rule for Non-Mutually Exclusive Events
Two events are said to be mutually exclusive when they cannot occur at the same time. For example, 1 roll of a die cannot yield both a 5 and a 6. So, what do we do when 2 events can take place at the same time? We use the General Addition Rule, which is stated as follows:
P(A or B) = P(A) + P(B) – P(A and B)
Note that the final term says to perform subtraction. Let’s look at how and why we must subtract.
If we want to calculate the probability of drawing 1 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, 1 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 its probability once, to offset the double-counting. Thus, the probability is calculated as follows:
P(queen or heart) = P(queen) + P(heart) – P(queen and heart)
P(queen or heart) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13
KEY FACT:
We use the General Addition Rule when events are not mutually exclusive.
Let’s practice with an example.
Probability Example 3: General Addition Rule
We randomly select 1 number from the set of numbers from 1 to 20, inclusive. What is the probability that the number is either an even number or a number greater than 15?
- 1/2
- 3/5
- 345
- 4/5
Solution:
For the set of numbers from 1 to 20, inclusive, let’s let E be the event “even number” and F be the event “a number greater than 15.”
The even numbers between 1 and 20, inclusive, are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20. Thus, P(E) = 10/20. The numbers greater than 15 are 16, 17, 18, 19, and 20. Thus, P(F) = 5/20.
We now must decide if the events E and F are mutually exclusive. We see that 3 numbers (16, 18, and 20) are both even numbers and greater than 15. Thus, P(E and F) = 3/20. Since one number can be both even and greater than 15, we verify that events E and F are not mutually exclusive. Thus, we must use the General Addition Rule to perform the calculation, as follows:
P(E or F) = P(E) + P(F) – P(E and F)
P(E or F) = 10/20 + 5/20 – 3/20
P(E or F) = 12/20 = 3/5
Answer: B
The Multiplication Rule of Probability
We use the Addition Rule when we want to calculate the probability of either event happening. The key words to look for in the question are “either” or “or.”
But if a question uses the word “and,” it is a signal that we will need to use the Special Multiplication Rule.
P(A and B) = P(A) x P(B)
For instance, let’s say a die is rolled twice and we want to calculate the probability that it shows “6” both times. This means that it must show a 6 on the first roll and a 6 on the second roll. To get the answer, we multiply the probability that it shows a 6 on the first roll (which is 1/6) by the probability that it shows a 6 on the second roll (which is also 1/6). We calculate as follows:
P(6 on first AND 6 on second) = P(6 on first) x P(6 on second)
P(6 on first AND 6 on second) = 1/6 x 1/6 = 1/36
Remember, the multiplication rule is used when we see the word “and.”
KEY FACT:
We multiply probabilities when we want all events of interest to happen.
Independent Events and Dependent Events
In probability, 2 events are called “independent events” if the occurrence of 1 event does not affect the probability of the occurrence of the other event.
For example, when we roll a balanced die, if our first outcome is a “6,” the probability of getting a “6” on the next roll is still 1/6. The first outcome does not affect the probability of the second outcome. Thus, we say that getting a 6 on the first roll and getting a 6 on the second roll are independent events.
We use the Special Multiplication Rule when events are independent.
Dependent Events
Now consider a different scenario in which the occurrence of the first event changes the probability of the second event. An example might be the 2 events “studying” and “getting an A.” If we study, the probability of getting an A increases. These 2 events that are tied to each other are called dependent events.
Here’s another example. Let’s calculate the probability of picking 2 blue marbles in succession from a jar containing 5 blue marbles and 5 green marbles. We see that the probability of choosing the first (blue) marble is 5/10 = 1/2.
The probability changes for the selection of the second marble. The first marble drawn was blue, so there are now only 4 blue marbles remaining in the jar. Thus, the probability of picking the second blue marble is 4/9. (Note that after the first blue marble was drawn, only 9 marbles remained in the jar.)
In experiments like the marble example, we refer to the situation as “sampling without replacement” because we do not replace the first item after it has been selected. This changes the probability that the second event will happen.
Note that when we perform sampling without replacement, we still multiply the 2 probabilities. But we have to keep in mind that the second probability changes after the first outcome has happened. (Do note that the multiplication rule changes slightly, but a discussion of this is beyond the scope of our discussion here.)
KEY FACT:
When we sample without replacement, probabilities change.
Let’s look at some SAT probability practice questions illustrating independent and dependent events.
Probability Example 4: Multiplication Rule: Independent Events
Harold has 2 bins of ping pong balls. The first bin has 4 balls, numbered 1, 2, 3, and 4. The second bin has 5 balls, numbered 1, 2, 3, 4, and 5. Harold randomly selects 1 ball from each of the bins. What is the probability that both balls are odd?
- 1/10
- 1/5
- 1/4
- 3/10
Solution:
The probability that the ping pong ball from the first bin has an odd number is 2/4 = 1/2. The probability that the ping pong ball from the second bin has an odd number is 3/5. We note that drawing a ball from the first bin and drawing a ball from the second bin are independent events, so we can multiply the 2 probabilities.
P(odd on first and odd on second) = P(odd on first) x P(odd on second)
P(odd on first and odd on second) =1/2 x 3/5 = 3/10
Answer: D
Probability Example 5: Multiplication Rule: Dependent Events
In a classroom are 16 girls and 12 boys. We randomly select 2 students to help the principal with a project. What is the probability that we first pick a boy and then a girl to help the principal?
- 24/91
- 12/49
- 7/24
- 1/14
Solution:
There are 12 boys and 16 girls, so there are 28 total students in the room. Thus, the probability of first selecting a boy is 12/28 = 3/7. After we select the boy, the number of girls is 16, but the number of students remaining in the classroom is only 27. Thus, the probability of picking a girl after the boy has been selected is 16/27. So, the probability of picking a boy first and a girl second is:
P(boy first and girl second) = 12/28 x 16/27 = 3/7 x 16/27 = 24/91.
Answer: A
Table Probabilities
You are almost certain to encounter at least 1 table probability question on the SAT. You will be presented with a table of data, sometimes called a 2-way table or a contingency table. The table is a way of organizing and displaying data that shows the relationship between 2 categorical variables. The rows represent the categories of 1 variable, and the columns represent the categories of the second variable. Each cell in the table displays the frequency of cases that fall into the combination of those two categories.
Let’s consider the following table that summarizes the pet preferences of junior high and high school students.
| Cats | Dogs | Total | |
|---|---|---|---|
| Junior High | 12 | 8 | 20 |
| High School | 14 | 22 | 36 |
| Total | 26 | 30 | 56 |
A typical question might ask the probability that an individual selected at random prefers dogs. We see that the total number of students preferring does is 30, out of a total of 56 individuals who were surveyed. Thus, the probability would be 30/56 = 15/28.
A more challenging problem might be this: Of those who prefer dogs, what is the probability that the individual is in junior high? In this case, we limit our consideration to the 30 students who prefer dogs, and we see that 8 of them are in junior high. Thus, the probability would be 8/30 = 4/15.
KEY FACT:
On the SAT, you will likely encounter a table probability question.
Let’s look at two SAT math probability questions with tables.
Probability Example 6: Table Probabilities #1
A random sample of Taylor Swift concert-goers was taken. They were asked their age and the type of Taylor Swift music they prefer. The results are summarized in the table below.
| Country | Rock/Pop | Total | |
|---|---|---|---|
| Age 16 or less | 16 | 84 | 100 |
| Age 17 or greater | 30 | 74 | 104 |
| Total | 46 | 158 | 204 |
What is the probability that a randomly-chosen concert-goer is age 17 or greater?
- 23/102
- 25/51
- 26/51
- 79/102
Solution:
This is a straightforward probability calculation. We see that there are 104 concert-goers who are age 17 or greater, out of a total of 204. Thus, the probability that a randomly selected concert-goer is 17 or greater is 104/204 = 52/102 = 26/51.
Answer: C
Let’s consider another table probability question. We’ll use the same scenario but change the type of question.
Probability Example 7: Table Probabilities #2
A random sample of Taylor Swift concert-goers was taken. They were asked their age and the type of Taylor Swift music they prefer. The results are summarized in the table below.
| Country | Rock/Pop | Total | |
|---|---|---|---|
| Age 16 or less | 16 | 84 | 100 |
| Age 17 or greater | 30 | 74 | 104 |
| Total | 46 | 158 | 204 |
What is the probability that a randomly selected individual is both 16 or less and prefers rock/pop?
- 21/51
- 25/51
- 42/79
- 79/102
Solution:
The person selected will be both in the age category of “16 or less” and in the category “rock/pop.” We see that there are 84 concert-goers who fit both criteria, out of the total sample size of 204. Thus, the probability is 84/204 = 42/102 = 21/51.
Answer: A
Summary
In this article, we concentrated on how to solve SAT probability questions.
SAT probability test preparation can take time and concentration. You will need to invest time and effort into mastering the basic probability formulas and rules. When you’re finished, you’ll have a leg up on doing well on any probability question they throw at you.
In this article, we learned the following key concepts:
- The basic probability formula: Probability of an event happening = (number of favorable outcomes) / (total number of possible outcomes)
- The special and general addition rules for probability are used for an “or” probability question.
- Mutually exclusive events cannot happen at the same time; non-mutually exclusive events do happen at the same time.
- When we want both of 2 events to happen, we use the multiplication rule. A hint to use the multiplication rule is the presence of the word “and.”
- Two events are called “independent events” if the occurrence of 1 event does not affect the probability of the occurrence of the other event.
- Two events are called “dependent” events if the occurrence of 1 event does affect the probability of the occurrence of the other event.
- A data table is often used to test your ability to calculate probabilities.
Frequently Asked Questions (FAQ)
How many probability questions are on the SAT?
There is no specific number of probability questions that will appear on every SAT exam. However, there is a very good chance that you will encounter at least 1 probability question.
What are the most common SAT probability formulas?
In this article, we presented you with the basic probability formula, the special and general addition rules, and the multiplication rule. It’s a good SAT probability tip to memorize these formulas.
How do you solve SAT probability word problems?
If you are having trouble with hard SAT probability questions, especially word problems, your best SAT probability strategies are to read the question carefully and then translate the words into numbers. If this is not easy for you, look at some SAT probability examples with solutions provided. Then, follow the solution, line by line, to get the needed insight for solving the problem. Over time, with practice, you’ll be a word problem ninja!
Is probability hard on the SAT?
Generally, SAT probability questions are relatively straightforward. While you may encounter advanced SAT probability problems, you will generally need only the formulas presented in this article. You won’t see sophisticated concepts such as Bayes’ Theorem or Monte Carlo simulations.
What’s Next?
Probability is a major branch of Statistics. You can get more insight into the types of Statistics problems on the SAT by reading this article.
In this article, we learned several probability formulas. Read this article to learn additional important formulas on the SAT.
While it’s important to master concepts for the SAT, it’s also important to know pacing strategies on the digital SAT. This will ensure that you have enough time to do your best on this challenging exam.
Good luck!



