Topics in quant GRE: Data, Probability and Statistics

Hi there! In this third post of the quant GRE series I wanted to tackle the section that it is commonly regarded as the most challenging, analyzing graphical data, statistics and probability.

Photo by Michał Parzuchowski on Unsplash

This section is a mixture of new ideas combined with several other key points from number properties (like sequences of consecutive numbers) and algebra. Today, I will go through all the various concepts we need to understand in order to become familiar with statistics and feel comfortable solving these problems under pressure on the test day. After studying the concepts and theory, we will work on several exercises of varying difficulty to apply what we learnt today.

If you are not confident on your knowledge of Number Properties I recommend you check out the blog I wrote where I go into detail about the various kinds of exercises and topics on that section we should know. It helped me a lot. After writing the article, and studying with y’all, I felt much more confident about my ability to work out this problems fast. Also, I explain my intentions for this series of blog posts in that one, so if this is your first time reading one of my posts on the GRE it might be a great place to start. Here is the link:

Topics in quant GRE: Number Properties

Disclaimer: I am not selling anything. This is no “Full guide on how to get a 170 on the GRE” or anything of the sort, maybe if I do get a 170 I will change the name of the blog :). This is an honest attempt to democratize information about the exam and a way for me to complement my own studies on such topics. Basically, all of you will be studying with me so if you want please get in touch either by private message or via the comments to chat in further detail about anything in particular or notify me of anything I may have missed I would be thrilled to do so.

Without further ado, let’s get to it! Unlike with previous topics, thanks to the nature of both my majors (Economics and Data Science), I have had plenty of exposure to statistics and data analysis throughout my three years of college so far. Thus, I will make my best effort to layout the main ideas as clearly and as engaging as possible by providing some cool facts and examples whenever I can. I am a data junky (if the term doesn’t exist yet I coin it). I love looking and working with data, hence my decision to double major in Data Science. I believe that data analysis and statistics is more of an art rather than a hard science and, hopefully, by the end of this post you will understand why. First, let’s get through the basics. Probably most, if not all, of you will be very familiar with the topics presented next but it is always good to recap:

Mean: The sum of the elements divided by the number of elements

Median: The number that lies exactly in the middle position of the set of elements.

Range: The last number minus the first number. In other words, it gives you the length of that set of elements.

Variance: The average of the squared deviations (element minus the mean) from the mean. Depending if we are working with the entire data or a subset (sample) of the data we have the population variance:

Or the sample variance:

Standard deviation: How much does your data deviates from the mean. To calculate this we take the square root of variance by determining each data point’s deviation relative to the mean. Note: 68% of your data is explained under 1 sd, 95% under 2 sd and 99% under 3 sd.

Great, hopefully that was a brief recap of the main statistical measurements out there. If you want to analyze these concepts, as well as many others related to statistical analysis, in much more detail then I recommend you check out the following post:

My adventure in Deep Learning — part 3.5

Next, we have quartiles and percentiles. Quartiles are the numbers that divide the given data points into quarters and are defined as:

• Q1: middle value in the first half of the ordered data points
• Q2: median of the data points
• Q3: middle value in the second half of the ordered data points
• Inter Quartile Range (IQR): given by Q3 — Q1

Percentiles are a similar way to refer to the same information. A percentile is defined as the score below which a given percentage of scores in its frequency distribution falls. For example, the 50th percentile is composed of all the scores below which 50% of all the scores may be found. Also, it refers to all the scores below the Quartile 2. It is worth mentioning that a frequency distribution is a table containing the score and the number of times it appears (i.e. its frequency). It looks like this:

or like this:

source: https://www.slideserve.com/Patman/tabulating-numerical-data-frequency-distributions

We get, on average, two or three tables per exam and a couple of questions attached to it. Awesome! We are moving pretty smoothly so far. The concept of tables and frequency distributions set up the starting point for our next topics:

Graphs

There are many types of graphs and each of them are suited for representing a particular format of data. If we develop an intuition for what does each type of graph represent we will be able to save some time during the exam and be more confident of our analysis. Let’s analyze some data!

First, what are the types of charts or graphs we could encounter on the exam? Well, we could get the following:

• Histograms
• Bar graphs
• Pie charts
• Box plots
• Frequency distributions, which we already covered :)

It is often useful to start from the beginning, so let’s do exactly that. Histograms are very useful at visualizing numerical data, specially frequency distributions, because you plot the frequency on the y-axis and the range of values (called bins) on the x-axis. The higher the bar, the more frequent those values appear. These are not to be confused with bar graphs which, even though they look very similar, are used for different purposes. Bar graphs are used to represent nominal data, meaning the measure the frequency of the values of a given class (i.e. the number of male or female students in a classroom). In this case, the bins are specific categories rather than a range of values like in histograms. Another difference is that bar graphs could be arranged in descending order (with the most frequent class first) while histograms cannot be rearranged. I understand these are weird definitions that may be difficult to grasp in text so here are two beautiful illustrations of each type of graph. Can you recognize some differences?

Histogram

Bar graph

Are you able to quickly recognize what differentiates one from the other? A good rule of thumbs is that if the x-axis has text (names or categories) then it will always be a bar graph. What type of animal do you think people bought the most during the pandemic?

Next in line we have a very controversial type of chart, pie charts. I say controversial because most statisticians discourage the use of pie charts to present information as they are considered “misleading” most of the time, yet this is only because people don’t know how to use them and not because the type of chart itself is flawed. Regardless of the popular opinion we need to be able to work with them so, what are they and what are they used for? Well, pie charts get their name for their shape and the way proportions are represented with them. They are used for… did you catch the hint I just gave you? Yes, they are used to visualize proportions. Another definition from chartio is “A pie chart shows how a total amount is divided between levels of a categorical variable as a circle divided into radial slices.” Also, like bar graphs, these are used to represent categorical variables where each “slice” represents a category (dog, rabbit, cat, goldfish or hamster if we follow our bar graph example). In summary, if we are shown a pie chart we immediately know we are dealing with proportions and/or ratios. Here is how they look like:

Although on the exam we would probably get something more like this:

Awesome! I hope I didn’t make you hungry with all those pies. Last but not least we should be comfortable working with Box plots. This are arguably the most important type of chart you will have to know throughout your career (assuming you will following some sort of STEM discipline). It is very efficient as it presents a lot of information about the data in a single plot. Remember quartiles and percentiles which we studied a few moments ago? I hope you do. If not, go back and re-read it, and check the graph I used to explain those concepts visually while you are at it. That is a box plot! Literally a box plotted in an xy-plane that shows you the mean, median, 1st and 3rd quartiles and even the outliers of your data. It is super useful. Besides their usefulness, I have not seen many exercises involving box plots in all the quizzes and practice tests that I have taken so far but it is very important for us to be familiar with them. Here some more pictures of box plots. Can you spot its different components?

They could also be presented horizontally, like this:

The latter picture offers a very nice illustration of skewness. Do you know what that means? Basically, it is a term that refers to how to data is distributed. In statistics jargon, it is a measure of symmetry. If the mean is higher than the median then it is positively skewed (or skewed to the right), and vice versa. If the mean and median are equal we get a normal distribution which is symmetrical on both sides (it has no skewness). We are not going to be tested on this, but if this is making your brain itch here is the formula to calculate skewness:

source: https://corporatefinanceinstitute.com/resources/knowledge/other/skewness/

This concludes our review on statistics and data analysis for the quant GRE, but don’t go anywhere yet because there is one more little thing we need to cover:

Counting and Probability

There are not many probability problems in the GRE (maybe around 3 or 4 on average) but counting exercises come in many forms and is very important, as well as complex, to develop an intuition for these topics. But do not fear, we will go through them together and I am certain that by the end of the article they won’t give us much trouble. If you’d like, take a break and grab a coffee or something. I’ll see you in 5 minutes :)

Hopefully you recharged. In order to master counting and probability we should start by looking at the building blocks of every mathematical object: Sets. A set is a collection of elements, where the elements can be anything you could think of: numbers, names, pictures, suits… They are represented in between curly braces and the name of the set is capitalized. For example, the set X containing the suits of poker cards would be:

X = {Diamonds, Hearts, Spades, Clubs}

Moreover, you can perform special operations with sets, namely union, intersection and difference. The union of two sets yields a new set containing the elements found on both sets, their intersection yields a new set containing the elements which those sets have in common and their difference would yield a set containing all the elements in the first set but not on the one being substracted. For example, If A = { 1, 3, 5, 7, 9 } and B = { 2, 3, 5, 7, 9} then their union A∪B = ​​{1,2,3,5,7,9}, their intersection A∩B​={3,5,7} and their difference A - B = {1}. It is often useful to visualize sets using a tool called Venn diagrams, this is how these basic operations look like (each circle represents a set):

Set A is a subset of a set B if every element of A is also an element of B. So, in our previous examples sets A and B are not subsets of each other, but if A = {2, 4, 6, 8} and B = {2, 8} then B is a subset of A. Finally, a universal set is the set consisting of all elements under consideration. In our first example, the universal set would consist of U = {1, 2, 3, 5, 7, 9}. Awesome, this lays an important foundation for what we are going to study next, which is counting. If you, like me, had probability theory classes in college but feel like you learn nothing because your professor’s approach was to give you a problem and then the formula which you are supposed to use but never took the time to help you develop an intuition for these problems, or at least help you understand why the heck we are suppose to use those formulas, then we are riding the same train. Let’s help each other develop this intuition and master probabilistic thinking once and for all. First, I am going to define the terms and work through a problem without the use of any formulas, just reasoning. I will attempt to write down my thought process as detailed and clear as possible. Then, I will present different ways to think about counting and probability, and finally I will show you the formulas. I am hoping that after hammering down different ways of thinking about it the formulas will make intuitive sense.

Let’s start with the basics, there are two ways of counting one is called permutations and the other combinations. Permutations are counting problems where the order matters. In other words, you do care about how the different possibilities are arranged. In other words, we count each possible arrangement as a different scenario. Consider the following GRE example:

There are 5 doors to a lecture room. In how many ways can a student enter the room through a door and leave the room by a different door?

Let’s work through it step by step. First, we are told that there are 5 doors, and we are asked in how many ways he can enter and leave the room without repeating the door. So, the order matter because we cannot go in through door 1 and leave through door 1 as well. ({door1, door1} would be a combination though, which we will look at next). Cool, how many possibilities could we count? Well, we could go in through door1 and leave through any of the four remaining doors, or enter through door2 and leave through the other four doors. Do you notice a pattern? For every door we go in we have four possible doors from which to choose to get out. It follows that, since we have to choose one path OR the other — we are not interested in counting the different ways someone could take one path and then another one, meaning we don’t care about the sequence of events combined. We are analyzing each path independently — we have to sum the possibilities per path, which would result in 4 + 4 + 4 + 4 + 4 = 20, which is our final answer YAY! We will touch upon this idea of summing or multiplying counts when we cover probability at the end of this post, for now I am just trying to work on building a probabilistic intuition.

Combinations, on the other hand, are problems where you are asked about all the different combinations you could arrange from a given set of data (like how many pairs could you form with the four suits of poker cards? Or, following from our permutations exercise, in how many ways can a student enter the room? Note that here he could go in and out through the same door.). The main difference with permutations is that order does not matter here. Let’s work through an example together:

In how many ways can 3 students be selected from a group of 12 students to represent a school in the inter school essay competition

We are told that there are 12 students, and we are asked to calculate in how many ways can I arrange groups of three students. We are not asked to consider the order or anything, just the amount of ways in which I could arrange 3 students. This is a typical combinations problem. How would we go about solving this? Well, we can check how many options I have per spot. In other words, from how many students can I select to place in spot 1, then spot 2 and finally spot 3 in order to form a 3 student group. So, for the first spot I could choose any student so I have 12 options, for the second spot I have 11 because I already chose one, and for the third one — I bet you figured it out already — 10. Thus, we have 12 * 11 * 10 = 1320. Good math but wrong reasoning! It is important to note, and this took me a while to realize, that the fact that the order does not matter means that sets {A,B,C}, {B,C,A} or {C,B,A}… are all the same! So we must count all six permutations as ONLY ONE combination. With our first approach we were double counting a bunch. So, how could we account for this? To build up this intuition let’s take a subset of the data, how many groups of 3 students could we create from a class of 4 students? First we have 4 options, then 3 and finally 2 which results on 24 total possibilities. Great, we are at the same spot as before. Now, the first thing that comes to my mind when I need to “eliminate” something in math terms is division (and I mean eliminating numbers not your math professor or Mr. GRE). So, let’s try dividing 24 by something. What could that something be? Well, the other possible permutations that could’ve taken into account. I chose {A,B,C} from {A,B,C,D}, but I could’ve chosen {A,B,D} or {A,C,B} or {A,C,D}. Think of this pattern carefully, I think I just had an EUREKA moment. You have element A fixed and you have two spots left, and 3 letters to choose for that spot! So, 1 spot fixed, then we have three letters to choose from for spot 2, and two letter for the last spot, 1 * 3* 2= 6, which is the number of different ways I could rearrange a group of 3 elements. Also, you could think about it such as you have three options for the first spot, 2 for the second and 1 for the last one, same thing. I think we should divide by 6 to get our final answer 4. I believe we are getting somewhere. Let’s apply this for our original problem. We have 12 students and already calculated that we have 1320 total possibilities. By what number do we need to divide it to avoid double counting? Let’s use the same reasoning as before. We are still working with groups of 3 elements so there are 6 different ways in which we could rearrange them. Divide 1320 over 6 to get 220, which is our final answer. If you already know the formula for combinations double check this with your calculator.

Wow, that was the first time I ever reasoned through a combinations problem like that. I already feel like I developed a much deeper understanding of the topic. How about you?

Here are the formulas, do you feel like we pretty much reconstructed them from scratch while working on the latter exercise? I think we nailed it!

Permutations -> P(n, k) -> n! / (n — k)!

Combinations -> C(n, k) -> n! / k! * (n — k)!

where n is the total number of elements in the universal set (12 students in our exercise) and k is the number of elements we choose (3 for our students problem). Let’s double check our intuition. The second problem was a combinations exercise. If we were to use the formula we would have gotten:

12! / 3! * (12–3)! -> 12*11*10 *9! / 3! * 9! -> 12*11*10 / 3! = 220

Since we are choosing k students we know that we could rearrange them in k! ways, which was exactly the reasoning we used. I am perplexed. Great work! Let’s end with probability and some harder problems.

Probability

Probability theory is a beautiful field and, in my opinion, one of the most interesting ones. But, it is pretty counterintuitive. It has applications all over the place, from gambling and taking the GRE to Quantum physics and modelling your chances of getting hit by a lightning (which are 1 in 7.5 million if you live in California and 1 in ~260k if you live in Montana. Also, it is more likely to get hit by a lighting than winning the lottery). Anyways, for the exam we don’t need to know those fun facts but we do need to develop an intuition for probabilistic outcomes. As their name suggests, probabilistic outcomes are those that are not discrete, meaning that there is no yes/no answer, there is a chance the answer is yes and a chance the answer is no. In other words, it is a way for us to describe uncertainty (or risk) with numbers. These numbers indicate the likelihood of a certain event or group of events occurring and can be written as decimals or fractions. The formula is pretty simple: just dividing favored outcomes (which are the outcomes you are interested in) over the total amount of outcomes.

Great, but how the heck do we calculate the favorable outcomes or the total outcomes? Counting! We will be able to solve any probability problem from the GRE by utilizing the two concepts we just learnt: permutations and combinations. Very well. If you read my second post of this series, then you know that I promised you a poker game. So, let’s learn some poker.

A deck of cards has 52 cards of 4 different suits and 13 different cards per suit. If you never seen a deck of cards before here is a picture of one:

I highly recommend you learn these values as there are some probability questions on the GRE related to cards and knowing the basic properties of a deck of cards might save you some precious time.

Also, you probably already noticed that just like there are 4 suits there are 4 groups of each card (one card per suit). So, can you tell what would be the probability of getting a 4? It would be 4/52 because the 4’s are our favorable outcomes and there are 52 cards total. Pretty straightforward. Some basic rules and jargon of probability to get us started:

• Probability of a certain outcome is P(E) = 1
• Probability of an uncertain outcome is P(E) = 0
• Probability of an event is always between 0 and 1
• Probability of an event not occurring is 1 - P(E)
• A random experiment is an event with uncertain outcome
• An outcome is the result of an experiment
• An event is a particular outcome or set of outcomes of a random experiment
• A sample space is the set of all possible outcomes
• Mutually exclusive events are two or more events that cannot occur simultaneously
• Independent is when the occurrence of one event does not affect the probability of the occurrence of another event.
• Dependent is when the occurrence of one event does affect the probability of the other. There is a whole branch of statistics dedicated to these problems. Fortunately, we won’t be tested about conditional nor Bayesian statistics on the GRE but I thought it was worth mentioning it.
• Random selection is the selection of an item from a sample space where all items are equally likely to get selected.

Alright, that pretty much covers everything we need to know about probability jargon. To conclude this section I will give you two more formulas you should remember and then we will get our hands dirty by doing a lot of exercises to recap everything we’ve seen today :)

Formula for the probability of two independent events occurring:

P = P(A) * P(B)

Formula for the probability of two mutually exclusive events occurring:

P(A or B) = P(A) + P(B)

Where A and B are the two events and P(A) and P(B) refer to the probability of that event. Okay, great but what do these formulas mean? Why do they make sense? As you probably already told I hate using formulas blindly so let’s try to understand why they are the way they are. One approach would be counting all the possible outcomes, for example if I want to calculate the probability of tossing a coin twice and landing two heads then I would have 2 possibilities for both tosses (heads or tails), or four possibilities total {HH, TT, HT, TH}, and only one outcome that interests me {HH}. Therefore, the probability is 1 (favorable outcome) over 4 (total outcomes). Note that 1/4 is the same as 1/2 * 1/2, which correspond to the probability of getting a head on the first toss times the probability of getting head on the second one. Here, the events are independent because no matter what I get on the first toss that outcome will not affect the probability of getting a head on the second toss, which will always be 1/2 (unless its rigged of course). It is very helpful to think in terms of sets and counting. Here is another way I thought about this, since we have two independent events we will have two separate sample spaces, right? First we have the sample space of the first coin toss {H, T}, then we have a second samples space also {H, T}. I want to know the probability of tossing a head in both spaces, therefore it makes sense that the total number of outcomes grows. With coin tosses we have 2 total outcomes per toss so we have 4 total outcomes for two tosses, and 8 outcomes for three tosses. That makes intuitive sense, so how do we represent that mathematically? Well, since we are working with fractions we would have to multiply them since summing fractions with the same denominator just increases the numerator. Cool, I think we nailed it.

What about for mutually exclusive events? Well, let’s try the same thought process as before. Let’s say we want to calculate the probability of getting an A of spades OR a K of hearts. Then, I have only 1 card I am interested in out of 52 cards for both events. Also, note that here, unlike with independent events, we are only working on one sample space (we only have one deck of cards). So, what we are really trying to calculate is how many of the cards in the deck of 52 are either A of spades or K of hearts. Well, there is only one of each so 2 cards in total that are either one of those. (read the last two sentences a couple of times as they are confusing). Therefore, we have 2 (favorable outcomes) out of 52 (total outcomes) cards that satisfy our condition. Hence, the probability is 2/52. To conclude, note that we never added a second deck because we are interested in knowing how likely I am of drawing either one of those cards from the deck we are currently playing with. How do we represent that mathematically? Well, since we are working with one deck, the total number of outcomes (denominator) should not change and the way to achieve that is by summing both fractions: the probability for getting the A of spades is 1/52 and for K of hearts is also 1/52 so for getting one or the other is 1/52 + 1/52 = 2/52.

If you are still unsure or would like some more detailed explanations of these ideas here is an article in which a math professor explains them in a very detailed and clear way.

WHY Do We Add or Multiply in Probability?

Wow, what a journey! This really was a serious study session. I have learnt and understood a ridiculous amount of material and ideas that I didn’t truly grasp before. Hopefully you felt that way too :)

As I always say, there is no better way to get all those new ideas stuck in your head than practice and repetition. So let’s do exactly that. Join me in doing some exercises extracted from previous GRE exam or from Kaplan QBank. I will order them in two levels, medium and advanced. I will work on two medium exercise and two advanced exercises in detail and give you the answers for the others for you to work on your own.

Medium level exercises

There are 12 male students and 12 female students in a class of 24 students.

Quantity A: The probability that a group of 4 students selected at random will contain 2 males and 2 females

Quantity B: 1/2

Answer: First, notice that the order does not matter. I only care about getting 2 males and 2 females. So, we know this is a combinations problem. Thus, let’s calculate the total possible ways in which we can select 2 males from 12 and 2 females, also from 12. The formula looks like 12! / 2!(12–2)! and so, by canceling out a 10!, we get 12 * 11 / 2 * 1 = 66. The formula is exactly the same for both cases (males and females). Hence, our total favorable outcomes is 66 * 66. Next, we must calculate the total possible outcomes which is done in a similar manner. We must select 4 people from 24 so 24! / 4!(24–4)! and so 24 * 23 * 22 * 21 / 4! = 23 * 22 * 21. Finally, to get the probability divide favorable outcomes over total outcomes: 66 * 66 / 23 * 22 * 21 = 66/161, which is less than 1/2. Therefore, Quantity B is the correct answer.

List X: 4, 5, 8

List Y: 5, 8, 10, 12

One number is selected at random from list X, and one number is selected at random from list Y. What is the probability that the numbers selected from the two lists are different?

Answer: This one is tricky, we are told to calculate the probability that the two numbers selected are different but this could be done in multiple ways so it seems too complicated. We can approach the problem backwards and look at how many possible ways could I get the same numbers. Well, I could get (5,5) or (8,8). What are the probabilities of getting each number? For list X I have a 1/3 chance of getting either a 5 or an 8. On the other hand I have a 1/4 chance of getting a 5 or an 8 from list Y. Thus, the probability of getting (5,5) is 1/3 * 1/ 4 = 1/ 12 (same thing for (8,8)). Now, we know the probability of getting what we DO NOT want. What could we do? Remember that the probability of something is 1 minus the probability of not getting it. So to get the probability of obtaining different numbers we substract 1/12 + 1/12 = 2/12 = 1/6 (remember why we sum instead of multiplying here?) from 1. Thus, 1- 1/6 = 5/6, which is the right answer.

The infinite sequence 24, 25, 26, 27, 28, 29, 30, 24, 25, 26, 27, 28, 29, 30… repeats indefinitely. What is the 886th term of the sequence?

Hint: Recall how to deal with remainders (covered in detail on my number theory post). Also, do you notice a pattern in the given sequence? Think of how you could exploit it. The answer is 27.

Advanced exercises

The only items in a container are 3 blue disks, 3 green disks, and 2 orange disks. If 5 disks are randomly removed from the container, one after the other, and without replacement, what is the probability that 2 blue disks, 2 green disks, and 1 orange disk are removed?

Answer: We must choose 2 from 3 blues, 2 from 3 greens and 1 from 2 oranges from a total of 5 choices from 8 disks, and we don’t care about the order in which the disks are drawn. This reasoning clearly lays out the approach we need to take. The number of possible ways of choosing 2 out of 3 is 3 and to choose 1 out of 2 is 2. Thus, 3 * 3 * 2 = 18 and the total number of ways in which we could select 5 from 8 is 8! / 5!(8–5)! = 56. Hence, we divide both quantities to get our final probability which is 18 / 56 = 9 / 28.

The scores for bowler A are normally distributed about an average (arithmetic mean) of 190 with a standard deviation of 10. The scores for bowler B are normally distributed about an average (arithmetic mean) of 180 with a standard deviation of 20.

Quantity A: The probability that bowler A scores 210 or greater in a given game

Quantity B: The probability that bowler B scores 210 or greater in a given game

Hint: Think about what it means for a result to be under 1 or 2 standard deviations from the mean. Is it more or less likely to happen if it is closer to the mean? The answer is Quantity B.

Team X competes against Team Y in a series of games in which the first team to win four games wins the series. There are no ties; each game is won by Team X or Team Y. Once one team has won four games, the series ends. The probability that Team Y wins any given game is 0.6. If Team X wins two of the first three games, what is the probability that Team Y wins the series?

Answer: First, we are told that Team X has one 2 our of 3 games. Thus, Team Y has won 1. Also, we are told that the probability of Team Y winning any given game equals 0.6, which means Team X has 0.4 probability of winning a game, and that the first to win 4 games wins the series. So, in how many possible ways can Team Y win the series? Well, it could win the following 3 games with a probability of 0.6 * 0.6 * 0.6 = 0.216; or it could lose the next game and win the last four with a probability of 0.4 * 0.6 * 0.6 * 0.6 = 0.0864; or it could win the following game then lose and win the last two with a probability of 0.6 * 0.4 * 0.6 * 0.6 = 0.0864; or it could win the following two games then lose and win the last one with a probability of 0.6 * 0.6 * 0.4 * 0.6 = 0.0864. Remember how we calculate the total probability of one outcome or another one happening? Yes, we sum both their probabilities. Hence, we sum our four probabilities to get our final answer 0.4752.