The BriefA Blog about the LSAT, Law School and Beyond
Thus far, we have mostly talked about problems where all of the outcomes are equally likely. So, for example, our coins are equally likely to land heads as they are to land tails, and our spinners are equally likely to land on any of the outcomes, and our die are equally likely to land on any of 1 through 6. But not all GRE problems are like this. Consider the following:
Suppose you have a weighted coin which is twice as likely to land tails as it is to land heads. You are about to flip this coin. What is the likelihood of it landing heads?
In these problems, we need to do a little bit of algebra to find the answer. Feel free to think about it, and if you want to see a solution, click below:
We know that P(Heads) has some value. Let's call that value . So we write: . And, by what the question states, the coin is twice as likely to land on tails as it is to land on heads. So,
Now, we know that P(Tails) + P(Heads) = 1, since it is guaranteed that the coin lands on heads or tails and the two events are mutually exclusive (see our post on P(A or B) if this reasoning is unfamiliar). Thus,
By division, we conclude that
So in questions where the outcomes are unequal, we can use variables as placeholders for the true probabilities and then use the probability rules to derive the actual probabilities of the different outcomes. Here are some practice problems that are basically variations on this theme:
Suppose you have a weighted six-sided die that is four times as likely to land on the number 4 as it is to land on any of the other numbers. What is the probability that it lands on 6?
Let denote the event of the dice landing on 4 (and similarly for all the other numbers). Let Then, We know that since the die has to land on one of the numbers and all of them are mutually exclusive. Thus, we know
Thus, so The probability that the die lands on 6, then, is just
Suppose you have a company raffle. Managers (of which there are 30) get 2 tickets, ordinary employees (of which there are 110) get 1 ticket, and executives (of which there are 5) get 20 tickets. What is the probability that the winner is an executive?
At first, this also looks like a problem where different outcomes have different likelihoods. The chances of a particular executive winning are not equal to the chances of a particular manager winning (since executives get more tickets than managers). But we can think about the problem in a slightly different way and thereby turn it into a problem where the outcomes are equally likely. Thus, we can consider the outcomes for each tickets. There are 100 executive-tickets, 110 ordinary employee tickets, and 60 manager-tickets. Each ticket is equally likely to win, so there is a chance that an executive-ticket wins. In other words, there is a chance that an executive wins.
Suppose you have an 8-sided spinner numbered 1 through 8. Now, your spinner is set up so that you are twice as likely to get 2 as you are to get 1, 3 times as likely to get 3 as your are to get 1, and so on for all the other numbers up until 8. What is the probability that you spin a 1?
Let Then, and and so on for the rest. Thus,
Now, adding Thus, and so
Now, we discuss a trick that helps in some problems. Sometimes, instead of finding the probability of an event, it can be easier to find the probability that the event does not occur. Since we know that either an event will occur or not, we know that the probability of the event occurring plus the probability of the event not occurring must equal to 1. More formally, we write:
Probabilities for Complements
Let A be some event. Then, P(A) + P(A does not occur) = 1.
Here's an example where we apply this rule:
You are about to roll a fair six-sided dice. What is the probability that you roll a prime number?
We know how to do this kind of problem in the traditional way: find the number of outcomes where we get a prime number and divide by the total number of outcomes. But we can also solve this problem by finding the number of outcomes where we do not get a prime number (1, 4, 6) and then calculating:
P(I get a prime number) + P(I do not get a prime number) = 1
Substituting, we get
P(I get a prime number) + 3/6 = 1
Subtracting, we get
P(I get a prime number) = 3/6 = 1/2
which is the same as our previous answer.
Why do we bother with this rule? Sometimes it is actually more convenient to calculate things this way. For example,
You are going to roll a fair six-sided dice 10 times in a row. What is the probability that you get a six at least once?
Now, we are trying to find P(I get a 6 at least once in 10 rows). By our above rule, we know that
P(I get a 6 at least once in 10 rows) + P(I do not get a 6 at least once in 10 rows) = 1
Directly calculating P(I get a 6 at least once in 10 rows) is somewhat tricky. But P(I do not get a 6 at least once in 10 rows) is not nearly as hard to calculate. If I do not get a 6 at least once in 10 rows, then I must have gotten some number 1 through 5 on each roll. Now, the probability that I got some number 1 through 5 on the first roll is 5/6. And similarly, the probability that I got some number 1 through 5 on the second roll is 5/6. And so on for the third, fourth, …, tenth rolls.
Since each roll of the dice is independent of all the other rolls, we get:
P(I do not get a 6 at least once in 10 rows) =
Thus, we conclude that:
P(I get a 6 at least once in 10 rows) + P(I do not get a 6 at least once in 10 rows) = 1
P(I get a 6 at least once in 10 rows) +
P(I get a 6 at least once in 10 rows) =
which gives us the answer.
Now, it will often be easier to just directly calculate the probability of a given event. But if that calculation looks absurdly difficult or tedious, take a minute to step back and consider: can I calculate the probability that the event doesn’t occur? Would that be easier? Sometimes that can save you a lot of hassle.
You are rolling a fair spinner with seven sections, numbered 1 through 7. What is the probability that, in 10 spins, you get at least one prime number?
Let E = the event that you get at least one prime number in 10 spins. By our rule above, 1 = P(E) + P(not E). What is P(not E)? That's just the probability that E does not occur. What does it mean for E to not occur? That means that, in 10 spins, you never get a prime number. What is the probability of that? Well, on each spin, there is a 3/7 chance that you do not get a prime number (since 1, 4, 6 are not prime). Since each spin is independent of each other spin, we get: P(not E) = Thus, P(E) = 1 -
One random employee will be selected to win a company-sponsored vacation. There are 500 employees, 300 of which are female. Also, 50 employees are managers, and half of the managers are female. What is the probability that the winner is not a female manager?
By our rule, we know P(Winner is not a female manager) + P(Winner is a female manager) = 1. What is the probability that the winner is a female manager? We know there are 25 female managers. So, P(Winner is a female manager) Thus, P(Winner is not a female manager) =
The weather reporters say that there is a 60% chance of rain for each of the next seven days. If that is true, what is the probability that it rains at some point over the next seven days?
Let E = the event that it rains at some point over the next seven days. By our rule, P(E) + P(not E) = 1. It is easier to find P(not E), since that is just the probability that it does not rain at any point over the next seven days. Each day, there is a 40% chance that it does not rain. So over the seven days, P(not E) = Thus, P(E) =
Welcome to Law School Success Stories, where we discuss 7Sage applicants who made the most of their GPA and LSAT score.
👤 Who: “Sarah,” an applicant who grew up in China and moved to the United States for college.
- 📈 LSAT: 169
- 📉 GPA: 3.33
- 🏆 Accepted at the University of Michigan Law
- 💵 $35,000 merit scholarship
🥅 Goals and Strategy
Sarah knew she wanted to take her law degree back to China, and the cachet of a T-14 school was important to her. Her parents, however, had a limited ability to pay for her education, and as a Chinese citizen, she wasn’t eligible for federal loans, so she was also hoping for a merit scholarship. Continue reading
Welcome to Law School Success Stories, where we discuss 7Sage applicants who made the most of their GPA and LSAT score. Please note that we changed certain details to protect this applicant’s anonymity, but we did not change his numbers or results.
👤 Who: “Neil,” a recent college grad of Southeast Asian descent
- 📉 LSAT: Under 149
- 📈 GPA: Over 3.8
- 🗞 Two-year résumé gap
- 🏆 Accepted at a T-14 school
- ✍️ Handwritten note from the dean: "I loved your essays" (and more)
🌘 The Strategy: A Shot at the Moon
Welcome to Law School Success Stories, where we discuss 7Sage applicants who made the most of their GPA and LSAT score.
👤 Who: “Mark,” a Caucasian male in his mid-forties switching careers from the trucking industry
- 📉 LSAT: 166
- 📈 GPA: 3.9
- 🏆 Accepted at Northwestern
- 💵 Significant merit scholarship
🚚 Starting the Journey
Mark worked in the trucking industry for twenty years before he began a new kind of long-haul journey toward his JD. He didn’t know any law school applicants beyond the 7Sage community and had no idea where to begin.
In our last post, we talked about the idea of an experiment, outcome, and event. If you're not familiar with those concepts, it may be a good idea to look at that post. Here, we will talk about some of the basic features of probability. First, a definition:
Definition: The probability of an event is a number that measures the likelihood of the event occurring.
And because it is tedious to always write out things like "the probability that a fair coin lands leads is ½", we will adopt an abbreviation. We use letters to represent events:
E = A fair coin lands heads
And then, we just write:
P(E) = ½
which we read as:
The probability that “A fair coin comes up heads” is ½.
And in general, for any event E, we use P(E) to denote the probability that event E occurs. This shorthand will save us much space in the rest of the series.
Now, a probability measures the likelihood of an event. This brings us to:
5 Basic Facts About Probability
1. A probability of 0 means that an event is impossible.
So if you find that P(E) = 0, that means that E will not occur. As an example, when rolling a six-sided die, the event that we roll a 7 is impossible -- it does not occur in any of our outcomes. Thus, P(Roll a 7) = 0.
2. A probability of 1 means that an event is certain.
So, when rolling a six-sided die, the event that we roll some number is a certainty -- it occurs in all of our outcomes. Thus, P(Roll a number) = 1.
3. An event with a higher probability is more likely to occur.
So, if the probability that it snows is 20% while the probability that it rains is 80%, then it is more likely to rain than it is to snow. And, on the flip side, events with a lower probability are less likely to occur.
4. Probabilities are always between 0 and 1.
This makes sense, since if an event had a probability greater than 1, then it would be more likely to occur. But events with a probability of 1 are already certain to occur! How could any event be more likely than a certainty? Similarly, if an event had a probability less than 0, then it would be less likely to occur, but events with a probability of 0 are already impossible! How could an event be less likely than an impossibility?
This also gives us a helpful way to check our answers: if we get a probability greater than 1 or less than 0, we have made a mistake somewhere.
5. The probabilities of our different outcomes must sum to 1.
E.g. if we have 4 different outcomes, then
P(Outcome 1) + P(Outcome 2) + P(Outcome 3) + P(Outcome 4) = 1.
This is because, when we do an experiment, something is bound to happen. So the probabilities of our outcomes must sum to 1.
Now, for the GRE, there are three main types of probability problems:
- The probability of a single event occurring: P(A)
- The probability that two events both occur: P(A and B)
- The probability that one or another event occurs: P(A or B)
You are about to do an experiment with four possible outcomes: A, B, C, and D. The stated probabilities are as follows:
P(A) = .5
P(B) = .3
P(C) = .38
P(D) = .1
Is such an experiment possible? What if P(D) = -.18?
No, such an experiment is not possible since P(A) + P(B) + P(C) + P(D) = 1.28 which is not equal to 1. And changing P(D) would make P(A) + P(B) + P(C) + P(D) = 1, but then we would have P(D) = -.18, which is a negative number. Since probabilities cannot be negative, this experiment is again impossible.
Give an example of an experiment not discussed above, and give an example of an event with a probability of 0 for that experiment, and another event with a probability of 1 for that experiment.
There are many possible examples; here is one: investing in the stock market. It is a certainty that "The value of my portfolio will either increase, decrease, or stay the same." It is an impossibility that "The value of my portfolio will both increase and stay the same." It may do one, or the other, but to do both is impossible.
Translate P(A) + P(B) = ½ * P(C) into a natural language (like English, French, Chinese, etc.).
Again, there are many possible solutions. Not confident in my French, I'll stick with giving the English translation: "The probability of A plus the probability of B is equal to half of the probability of C."
Some GRE questions ask about the likelihoods of different events. For example:
You are about to roll two fair six-sided dice. What is the probability that they sum to 7?
E. None of the above.
In this series, we cover the strategies you need for probability questions on the GRE.
Before continuing, you should know that probability questions make up only about 5% of all math questions on the GRE. That totals to about 2 questions per test. So if you are weak in other areas that appear more frequently (e.g. algebra, fractions/ratios, reading graphs), it might be wiser to look at those topics first and return here later.
What Is Probability?
Sometimes, we are not sure what will happen. It may rain tomorrow or not. I may win the lottery or lose. Federer may win Wimbledon again, or not. Probability is a way to handle this uncertainty. Even if we cannot know exactly what will happen, we can at least determine how likely the different possibilities are. So, if I roll a fair die, I can't know if it will land on 6 or not. But I know that it is more likely to land on an even number than to land on 6 specifically.
Talk about probability is commonplace. We might say it's "pretty likely" to rain later today, or that some team has "no chance" of making it to the playoffs. The mathematical theory of probability gives us a way to make that kind of talk precise, by turning it into formulas and numbers.
Probability begins with the idea of an experiment:
Definition: An experiment is an action that, when performed, leads to exactly one of many possible outcomes.
All sorts of things can be experiments. If I don't study for the final, that is an experiment; it could lead to my acing the class (unlikely), barely passing (more likely), or failing (very likely). In the land of GRE problems, standard experiments include things like flipping a coin (which either leads to the outcome of heads or the outcome of tails), rolling a die (which leads to some number 1 through 6), or picking a winner for a raffle (where the "outcome," or in other words the winner, could be anyone who bought a ticket).
Now, let's consider rolling a fair six-sided die. That experiment has 6 possible outcomes:
And there are lots of questions we could ask about rolling a dice. For example: am I more likely to get an even number or an odd number? What are the odds of getting a prime number? To answer these questions, we need the idea of an event:
Definition: An event is a set of some (or none, or all) of our experiment's outcomes.
Here are some examples of events:
- My die lands on an even number
- My die lands on a number which, in English, is made of 4 letters
- My die lands on 6
Note that events do not need to be possible; we can also consider events that simply will not occur:
- My die lands on 7
- My die lands on a number that is neither even nor odd
- My die lands on a number which, in English, is made of 47 letters
Now we are ready to get at the heart of probability: the whole point of probability is to figure out what the likelihoods of different events are. These GRE questions will give you some setup and some event, and ask you to find the probability of that event. And the rest of this series will be devoted to doing exactly that -- figuring out those likelihoods -- across a series of increasingly complicated contexts.
And here are some practice questions to test your understanding of the above.
I roll a 21-sided die. How many outcomes are there?
There are 21 outcomes; each of the 21 sides of the die corresponds to a different outcomes. (Within the land of GRE problems, we assume things like the die never lands on a corner, teetering between two numbers, and that no one catches our die midair, and so on).
Give an example of an experiment that was not discussed above.
Many examples would work; here is one: eating days-old leftover food. The possible outcomes are: you get sick, or you do not.
In rolling a 6-sided die, I list seven outcomes: I roll a 1, I roll a 2, I roll a 3, I roll a 4, I roll a 5, I roll a 6, and I roll an even number. Are there actually seven outcomes?
No. Recall that an experiment is an action which leads to exactly one of several possible outcomes. But it is possible to roll a die and get a 6 (one of the outcomes I list) and get an even number (another outcome I list). After all, 6 is even! So this is an improper listing of outcomes for our experiment.
In our previous posts, we've talked about the basic concepts of probability and some fundamental facts about probabilities. Here, we'll show how to calculate the probability of a single event when all the outcome are equally likely. This is, in a sense, the simplest case that we will cover, and it is crucial for everything we'll do later (e.g. in finding the probability of two events occurring).
Suppose we flip a fair coin. What is the probability we get heads? Intuitively, the answer should be 1/2. And that's exactly what the following rule would say:
Probability of Equally Likely Outcomes:
If you have n possible outcomes, all of which are equally likely, then the probability of any particular outcome occurring is 1/n.
So when we flip a fair coin, there are 2 possible outcomes (heads and tails). So n = 2 and the probability of one outcome (e.g. heads) occurring is 1/n = 1/2. And if we roll a six-sided die, there are 6 possible outcomes. So the probability of any particular outcome (e.g. rolling a 4) is 1/6. And if we held a raffle where there were 109 different entrants, the probability of any one of them winning would be 1/109.
Note that this rule only applies when all the outcomes are equally likely. In most GRE problems, the outcomes will be equally likely, and the question will signal that by saying that the outcome is "random" or that the outcomes are "equally likely." So, the question might say things like: "a name is chosen at random" or that "each outcome is equally likely." When the outcomes are not equally likely, all bets are off, and you will have to be more careful in how you approach the problem.
Now, we want to find the probability of some event occurring. Suppose I am going to roll a six-sided die, numbered 1 through 6. What is the probability that I get an even number? To calculate this, we use the following rule:
Probability of Single Events (for equally likely outcomes)
Suppose you have n equally likely outcomes. Then, the probability of some event E occurring is:
where the # of total outcomes = n.
So to find the probability of rolling an even number, we need to find the number of outcomes where we roll an even number. If we roll an even number, then we must have rolled a 2, 4, or 6. Then, we divide by the number of total outcomes, in our case 6. So, P(Roll an even number) = 3/6 = 1/2.
Here’s another example in a similar spirit:
Suppose you randomly choose a number from 1 to 50. What is the probability that you chose a prime number?
There are 50 possible outcomes to your random choice. Now, we need to know: in how many of those outcomes do you choose a prime number? In other words, how many of the numbers 1 to 50 are prime? Here, we just have to go through the list: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. So there are fifteen such prime numbers. Thus: P(Pick a prime number) = 15/50 = 3/10
Now we know how to find the probability of a single event when the possible outcomes are equally likely. Our next step is to learn how to combine these probabilities in order to get the probabilities of more complex events.
130 people line up to buy raffle tickets. Every 10th person who buys a ticket gets a teddy bear as a promotional item. What is the probability that a randomly chosen person from the line will receive a teddy bear?
There are 130 people in line. Since every 10th person gets a teddy bear, we know 13 people got teddy bears. To find the probability that a randomly selected person gets a teddy bear, we just need to calculate: # of people who get teddy bears/# of people total. Thus, we get 13/130 = 1/10.
You have 50 friends. 12 of them have blue hair. You randomly pick one of your friends to invite to dinner tomorrow. What is the probability that you invite a person with blue hair?
We are looking for P(I invite a person with blue hair). Now, if I randomly pick a friend, that means there are 50 possible outcomes: I get dinner with friend 1, with friend 2, ..., with friend 50. Since the question tells us that the outcome is randomly selected, we know that they are all equally likely. So we can apply our rule. In how many of the outcomes will you get dinner with a blue-haired friend? In 12 of them. And we already know how many total outcomes there are. Thus, P(I invite a person with blue hair) = 12/50 = 6/25.
You still have 50 friends. 12 of them still have blue hair. What is the probability that you do not invite a person with blue hair?
This is just like our previous question, except now we want to find P(I do not invite a person with blue hair). If there are 12 people with blue hair, then there are 50 - 12 = 38 people without blue hair. So there are 38 outcomes where I do not invite a person with blue hair. Again, since the outcomes are chosen randomly, we can apply our principle to get P(I do not invite a person with blue hair) = 38/50 = 19/25.