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Complete Guide to Probability on SAT Math + Practice Questions

Posted by Courtney Montgomery | Jun 24, 2015 8:30:00 AM

SAT Math


Feature_probability.jpgA probability question asks you to identify how likely a particular event is to occur. How likely is it that you’ll pick a red marble out of a bag? How likely is it that a particular person will be chosen out of a lottery? How likely is it that two or more events will BOTH occur? These are just some of the many different types of probability questions you may encounter on the SAT.

This guide will take you through all aspects of probability you’ll need to know for the SAT--exactly what probability means, the typical probability questions you’ll see on the SAT math section, and the steps needed to solve them.


Before You Continue

Probability questions will show up on every SAT, but the vast majority of SAT tests only have one probability question out of the 54 math questions total. So plan your SAT math study prep accordingly.

If you are struggling to understand other fundamental sections of the math test, like integers or single variable equations, you will want to turn your focus there before you tackle this probability guide. The most important part of studying for the SAT is to focus your attention on topics that show up the most. This way, you can maximize your potential point gain per section.

But if you already have a solid grasp of the other fundamental math topics (or you just really want to learn this section first), then let’s get cracking on probability! You'll learn SAT math tips and formulas to work through questions that deal with chance.


Don't worry--I hear the probability of success is higher than you'd think.

What Does Probability Mean?

$\Probability = {\desired \outcome}/{\all \possible \outcomes}$

Remember this SAT math formula! Asking for the probability of an event is the same thing as asking for the “odds” of any particular event happening. And this probability is expressed as a fraction of: the likelihood of the event over all the outcomes possible.

So how likely is it that you’ll get tails if you flip a coin? The chances are 1 in 2.

1 for the number of  outcomes you want (tails) and 2 for the total number of possibilities (heads and tails).

Let’s take a look at another example:

There are ten students in the class. Every day, the teacher selects a random student to erase the board. What are the odds that Student A will be selected to clean the board today?


The probability of Student A being selected is $1/10$. The desired outcome is 1 because Student A is only one student. And there are 10 students total, so there are 10 possible outcomes (students to pick from).

Now what would happen if we had more than one possible choice as our desired outcome?

What are the odds that either Student A or Student B will be selected to clean the board today?


The probability is now $2/10$ (or $1/5$). Why? Because there are now 2 possible students to choose from, but the total number of students is still 10.

Because the probability of any event happening is expressed as a fraction, it means that an event that will absolutely and without a doubt occur has a probability of $1/1$ or 1. There is no higher chance of it happening--this particular event will happen every single time, without fail.

A probability of an absolutely impossible event, however, will be 0 because $0/x = 0$.

You can also think of probabilities as percentages. If I select a red marble from a bag at a probability of $1/5$, it means that there is a 20% chance that I will select a red marble because $1/5 = 0.2$ or 20%.

body_coin_flip.jpgI'm gonna go with tails on this one. 


Either/Or Probability

${\Probability \of \either \event} = [{\outcome \A}/{\total \number \of \outcomes}] + [{\outcome \B}/{\total \number \of \outcomes}]$

(Note: this kind of probability is called “non-overlapping.” This means that the two events cannot both happen at the same time. There is a way to find an either/or probability for overlapping events, but you will never be asked to do this on the SAT, so it is not in this guide)

As we saw above with our example of multiple students selected at random to clean a board, an either/or probability question asks how likely it is that either one of two or more events will occur. This increases the odds of our desired outcome because we do not care which of the two events happen, only that one of them does.

To solve this kind of problem, we must therefore add the probability of each individual event. Their sum is the probability of either event happening.

What is the probability of drawing either an ace or a queen from a deck of cards?


There are 4 aces in a deck of cards and 52 cards total. Therefore, the probability of drawing an ace is $4/52 = 1/13$ (or 7.69%).

There are also 4 queens in a deck of cards. So the probability of drawing a queen is also $1/13$.

So the probability of drawing either an ace or a queen is $1/13 + 1/13 = 2/13$ or 15.38%.


Neither/Nor Probability

${\Probability \of \neither \event} = [{\all \outcomes}/{\all \possible \outcomes}] - [{\outcome \A}/{\all \possible \outcomes}] - [{\outcome \B}/{\all \possible \outcomes}]$

(Note: again this type of probability question deals with non-overlapping events, as the SAT will never ask you about overlapping events.)

Other SAT questions may ask you to identify the probability that neither two nor more events will occur. Basically, your desired outcomes are all the potential outcomes minus the ones you highlighted as undesirable.

What are the odds that you will draw neither an ace nor a queen if you draw a card at random from a deck of cards?


There are 52 cards total and 4 of them are aces. $52-4 = 48$. So there is a $48/52$ chance that you will draw anything but an ace.

There are also 4 queens. So $48 - 4 = 44/52$. (Why 48? Because that’s how many possible outcomes you had left after removing your 4 aces)

So the odds of drawing neither an ace nor a queen (in other words, the odds of drawing anything except for an ace or a queen) are $44/52 = 11/13$

You could also say that the odds of drawing any and all possible cards = $52/52$

So $52/52 - 4/52 - 4/52 = 44/52 = 11/13$


♪ You say nee-ther and I say nii-ther 


Combined Probability

${\Combined \probability} = [{\outcome \A}/{\total \number \of \outcomes}] * [{\outcome \B}/{\total \number \of \outcomes}]$

A combined probability question asks: “What are the odds of two or more events BOTH (all) happening?”

Take note that there is a distinct difference between this type of question and one that asks for the probability of either event A or event B happening. An “either/or” question wants to know if one of the events occurs. A “both/and” question requires that multiple events all occur.

An “either/or” question requires you to add your probabilities. A “both/and” question requires you to multiply them.

A good rule of thumb is to note that a question that asks for multiple events (a “both” question) will ultimately have a lower probability than the odds of just one of those events happening. Why? Because two events BOTH occurring will naturally have lower odds than those of just one event. How likely is it that your first AND second coin tosses will BOTH be tails? Lower than the odds of just flipping tails once.

And an either/or probability question will have higher odds than the probability of just one of its events happening. You are combining forces to increase your odds of getting a desirable outcome. How likely is it that you’ll flip either heads or tails for each toss? 100%!

Combined Unconditional Events

Unconditional events are independent of one another. In other words, the outcome of one event does not affect the outcome of the second. In this case, when you are asked for the odds of both events occurring, you can simply multiply them.

Marcus has a pair of dice. He needs to roll a 12 to win the game. What are the odds that he will roll a 6 on both die?

The dice are not affected by one another, so the odds of his rolling a 6 on the first die are $1/6$. And the odds of his rolling a 6 on the second die are also $1/6$.

So the odds of his rolling two 6’s are: $1/6 * 1/6 = 1/36$. He has a 1 in 36 (or a 2.78%) chance of rolling two 6’s.

Combined Conditional Events

Conditional events affect one another. If two outcomes affect one another, they change the number of possible outcomes. For example, if you are continually taking items away from the total number of items, the probability of selecting each remaining item changes.

There are 3 red marbles, 3 green marbles, and 4 blue marbles in a pouch. If Jenna draws out marbles without replacing them, what are the odds that she will draw a red marble AND a blue marble on her first two selections?

The odds of drawing a red marble are $3/10$. Going forward, we are operating under the assumption that Jenna managed to select that red marble.

Now, she goes to select a blue marble. There are 4 blue marbles out of 9 marbles left in the pouch (Why 9? Because she took one marble away when she drew the red marble, so there are now 9 left in the pouch). The odds of drawing a blue marble are $4/9$.

So $3/10 * 4/9 = 12/90 = 2/15$. The odds are 2 in 15 (13.33%) that she will select a red marble and a blue marble on her first two selections.

Does this work if she were to select a blue marble first? Absolutely! The odds of selecting that first blue marble would be $4/10$. And the odds of selecting a red marble next would be $3/9$.

$4/10 * 3/9 = 12/90 = 2/15$.

Whether she selects red then blue or blue then red, the odds remain the same.



Life would be better if there were a much higher probability of this actually happening. 


Typical SAT Probability Questions

Probability questions generally fall into one of two categories--word problem questions or geometry questions. Let’s look at each category.


Word Problem

A word problem question will not give you any figures to work from, but instead will describe either a figure, a scene, or simply provide you with a numerical problem.


We have two sets of three numbers (j and k) and we are looking for their products to find the odds of particular types of products (multiples of 5) occurring. This means that we must also find the total number of outcomes possible (in this case products between the two sets).

We know that the total number of products will be $3 * 3 = 9$ (the numbers of each set multiplied together). If we listed out each product, it would look like this:

$(4)(10) = 40$

$(4)(11) = 44$

$(4)(12) = 48$

$(5)(10) = 50$

$(5)(11) = 55$

$(5)(12) = 60$

$(6)(10) = 60$

$(6)(11) = 66$

$(6)(12) = 72$

We have found our 9 total products. Now count the ones divisible by 5, which leaves us with:

40, 50, 55, 60, 60

So out of 9 total products, there are 5 that are divisible by the number 5. This means our final answer is $5/9$.

Geometric Probability

Sometimes you will be given a geometric probability question. A problem like this will usually ask you to find the likelihood of randomly hitting a point in a particular region, or to find the odds of randomly selecting an item from a particular section in a drawn figure.

Think of these kinds of problems as a standard geometry question with a small probability twist. They are essentially asking you to find the fraction of the highlighted region over the entire figure.

Because probability is desired outcome/all possible outcomes, this is exactly the same as saying:

${\highlighted \section \of \figure}/{\entire \figure}$



Again, this problem is basically asking you to identify what fraction the shaded region is of the whole figure. Because the dart is thrown at random, it has an equal chance of hitting any and all points on the figure. So it will hit anywhere in the shaded region at the same rate as the shaded region is a fraction of the whole.

So if the shaded region is $1/2$ of the entire figure, a dart will hit the shaded region $1/2$ the time, etc.

So we must first determine what percentage (fraction) the shaded region is of the entire figure. We can do this many different ways and if this is an unfamiliar type of problem to you, check out our guides on polygons (coming soon!) and plugging in numbers.

For this example, I have chosen to use the technique of plugging in numbers to better show those of you who have not tackled the geometry guides yet (or who feel uncomfortable with geometry right now) the easiest way to approach this kind of problem.

So let’s choose our own numbers and say that the length of the rectangle is 12 and the width is 4. (I chose these numbers because the question tells us that we are working with halfway points and even numbers are easily to divide into two, but you can pick any numbers you would like.)


So now let’s divide the rectangle into half and work with one half at a time.


The larger rectangle divided into the two halves that we'll work with.


We begin by looking at the left half of the larger rectangle. 


Because we are working with half the larger rectangle, the length of this rectangle (which is half of the larger) will be 6, while its width will still be 4. So its total area will be $(6)(4) = 24$. The shaded region cuts this rectangle in half, which we can see visually, and which we know if we know our geometry laws.

But if you wanted to make absolutely sure that the shaded region is one half of this figure, we can also confirm it by finding the area of the shaded triangle.

The area of a triangle is: $1/2bh$. So $1/2(6)(4) = 12$ and this is half of 24.

So the shaded region of this rectangle is $1/2$ of its total area.

But wait, we can't stop here! We are trying to find the percentage the shaded region is of the whole figure. Because we divided the larger rectangle into two halves, and this shaded region is half of a half, this shaded region is $1/2 *1/2 = 1/4$ of the total area of the rectangle.

So this shaded region is $1/4$ of the entire figure.


Now let’s look at the other half of the rectangle.


We know that this rectangle is again going to be $(6)(4) = 24$.

We can also see that the NON-shaded region makes a triangle with an area of:

$1/2bh$ => $1/2(2)(6) = 6$ (Why did we use 2 for the base? Well the line cuts halfway through the larger rectangle’s width of 4, which means that the shorter leg of the triangle will be 2.)

Because the NON-shaded region has an area of 6, it means that our shaded region has an area of $24 - 6 = 18$. So the area of the shaded region is 18 and the fraction of the shaded region to the whole of this rectangle is $18/24 = 3/4$.

But remember! We are again trying to find the percentage the shaded region is of the whole figure. Because we divided the larger rectangle into two halves, and this shaded region is half of a half, this shaded region is $3/4 * 1/2 = 3/8$ths of the whole rectangle.

Now, we must put the two regions together to get the shaded region of the whole, larger rectangle.

$1/4 + 3/8 => 2/8 + 3/8 = 5/8$

Because the shaded region is $5/8$ths of the whole figure (or 62.5%), that means that a dart thrown at random will have a $5/8$ths (or 62.5%) chance of landing in the shaded region.


body_mockingjay.jpgI somehow don't think the odds are that much in my favor in this game....


How to Solve a Probability Question: SAT Math Strategies

You will know if you are being asked for a probability question on the SAT because, somewhere in the problem, it will ask you for the "probability of," the "chances of," or the "odds of" one or more events happening. When you see those words, follow these steps to solving a probability question:


1) Make sure you examine exactly what the question is asking.

It can be easy to mistake an either/or question with a both/and (or an unconditional both/and and a conditional both/and), so make sure you carefully examine the question.

Mary has had three children, each of them a boy. What are the odds that her next child will be a girl?


You may be tempted to say that the number of desired outcomes (your numerator) is influenced by the existing number of children, or even that the odds for a girl will be high because statistically she’s “due” for a girl.

But in all actuality, Mary’s chance for a girl is $1/2$. Why $1/2$? The number of desired outcomes--having a girl--is 1. The number of possible outcomes--boy or girl--is 2. So there is a $1/2$ or 50% chance of Mary having a girl for her next child.

Now let's look at a slightly different question:

What are the odds that Mary would have three boy children in a row?


This question may sound similar to the previous one, but it is actually asking you for the probability of multiple combined events, which makes it a both/and question (you can rephrase it mentally as “What are the odds that BOTH her first two children will be boys? And what are the odds that her third child will ALSO be a boy?”). So if we take what we learned about combined probabilities from above, we would now have:

$1/2 * 1/2 * 1/2 = 1/8$

Each birth has a 50% chance for producing a boy, but the odds of having three boys in a row are only 1 in 8 or 12.5%.

2) Think logically about when odds will increase or decrease

The odds of either two or more events occurring will be greater than one of the events alone. The odds of both (or multiple events) all occurring will be less than the odds of one of those events alone.

Always take a moment to think about probability questions logically so that you don’t multiply when you should add or vice versa.

3) Simplify the idea of a probability

Once you get used to working with probabilities, you’ll find that probability questions are often just fancy ways of working with fractions and percentages.

If the question uses geometric figures, solve it the way you would a geometry problem and then find the fraction of the highlighted area to the whole. If the question is an equation, solve the equation as normal and then find the fraction of the indicated variable to the whole.

Don’t forget to think about the problem as a whole in terms of probability, but don’t get intimidated by the word “probability” either.


body_roulette.jpgWith odds of 37 to 1, how could you lose? (Answer: easily)

Test Out Your Knowledge with SAT Math Practice Questions






3) The local high school has 340 students with an equal number of freshmen, sophomores, juniors, and seniors. Once a year, one student is selected at random to be the student body president. What are the odds that the student selected this year will NOT be a freshman?

A. $1/4$

B. $1/3$

C. $1/2$

D. $2/3$

E. $3/4$


4) body_math_q_figure.png

A box is divided into 9 compartments. Each of the equal rectangles G, H, and I has an area that is three times that of the equal squares A, B, C, D, E, and F. If a marble is dropped into the box at random, what is the probability that it will land in compartment G?

A. $2/3$

B. $3/5$

C. $1/3$

D. $1/5$

E. $1/9$


Answers: C, C, E, D


Answer Walk-Throughs:

1) This is a variable equation problem with a probability twist.

There are 10 positive integers less than or equal to 10. These are: 1, 2, 3, 4, 5, 6, 7, 8, 9, & 10. (0 is NOT a positive integer).

If you solve your given equation:

$5n + 3 ≤ 14$

$5n ≤ 12$

$n ≤ 2.2$

There are 2 positive integers less than or equal to 10 that are also less than 2.2 (1 and 2). This means that our probability of the equation being true is $2/10$, or $1/5$.

So our answer is C, $1/5$.

2) This is a both/and question. We are being asked for the probability that EACH of them will BOTH be assigned a cubicle with an X. This means that this is also a conditional problem, as one woman’s desk assignment depends on the desk assignment of the other woman (they cannot both be assigned to the same office space).

So let’s first take the probability of Karen being assigned to one of the desks with an X. There are four desks total and two of them have an x.

So the probability of Karen being assigned to one of the desks is $2/4$ or $1/2$.

Now let’s take a look at the odds of Tina being assigned to a desk with an X.

We already said that Karen took an X-marked desk, so one of those desks is now no longer an option for Tina. This leaves Tina with one X-marked desk out of three desks total.

So the probability of Tina being assigned to the remaining X-marked desk is $1/3$.

Because we also know this is a combination (both/and) question, we now must multiply these odds together.

$1/2 * 1/3 = 1/6$

This means our final answer is C, there is a 1 in 6 chance that both women will be assigned to one of the X-marked desks.

3) This is a neither/nor question, as we are being asked to take out a fraction of the total population in order to find the probability of all the other options.

If there are 340 students divided into 4 equal parts, then the freshmen make up 1/4 of the total student body (85 students total).

This means that, to select a sophomore, junior, or senior at random, the odds are:

$1 - 1/4 = 3/4$ (1 because the entire student body is 100% of the student population--${340 \students}/{340 \students}$)

So the answer is E, there is a 3 in 4 chance that the random student chosen with NOT be a freshman.

4) This is a probability geometry problem, which means we are essentially being asked to find the fraction of the area of the selected portion of the figure to the whole figure.

We know that each of the rectangles G, H, and I are three times as big as the smaller squares. So we could say that the box was instead made up of 15 small squares, rather than 9 differently sized compartments.

This means that our larger rectangle, G, is made up of 3 small squares out of the 15 smaller squares total in the box.

So the probability that you will drop a marble into box G at random is $3/15$, or $1/5$.

This means our final answer is D, $1/5$.


body_confetti.jpgWhoo! Take your new probability knowledge and celebrate!


The Take-Aways

Probability questions can seem much trickier than they actually are. By taking the time to analyze what is being asked of you--are they asking for the probability of both events happening, neither, either?--and understanding that probabilities are simply fractional relationships of desired outcomes over all potential outcomes, you’ll be able to tackle SAT probability questions in no time.

What’s Next?

You’ve stacked the odds in your favor by mastering SAT probability. Now that you’re done, it’s probably a good idea to take a look at all the topics covered on the SAT math.

Don’t know how you could possibly finish a math section on time? Look no further than our article on how to buy yourself time and complete your SAT math problems before it’s pencils down.

Want to get a perfect score? If you’ve already mastered your timing and score, it may be time to look at our article on how to get a perfect 800 on SAT math, written by a perfect-scorer.

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Courtney Montgomery
About the Author

Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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