The Brief
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Initially, our Rule for Permutations may seem like an odd rule, but the following example may help to make it more intuitive. First recall:

Rule for Permutations
Suppose we have n objects that we want to fill k spots where kn. Then, the number of possible ways ("permutations") to do this is:

    \[\frac{n!}{(n-k)!} .\]

Now consider:

Example
Our refrigerator has 5 magnetic numbers on it: 3, 2, 4, 5, and 7. How many ways are there to order the magnets on our fridge into 5-digit numbers?

Solution
By the question, we have 5 objects to fill into 5 spaces (corresponding to the five digits). Thus, by our formula, we get: 5!/(5-5)! = 5!/0! = 5!/1 = 5! = 120.

But to make this answer more intuitive, consider that we have five spaces to fill in with letters:

__               __               __               __               __               

And for the first space, we can pick any number. It doesn’t matter which one. So we have five options for the first space.

Suppose we picked the number A. (A is just a stand-in for whatever number we actually picked; it doesn’t matter which one we picked). Then, we have:

                                                                                                          

(5 options) 

And now, for the second digit, we can pick any number except for A. So we have four options. Suppose our second digit was B. Then we get:

                                                                                                        

(5 options)  (4 options)

And so on for the rest:

                                                                                                        

(5 options)  (4 options)   (3 options)   (2 options)  (1 option)

Thus, the total number of possibilities we could get is: 5 * 4 * 3 * 2 * 1 = 120, exactly what we got by following our rule.

Now, let’s see how permutations and probabilities can interact:

Example
You are forming four-digit numbers from the numbers 1, 2, 3, and 4. Each of your four-digit numbers must use all of those numbers exactly once. What is the probability that a number randomly chosen from your four-digit numbers contains a prime number in the tens place?

Solution
Now, we want to find:

How many four-digit numbers have a prime number in the tens place?

How many four-digit numbers are there?

To find (1), we want to consider numbers like:

                                (prime)                  

There are only two prime numbers here: 2, 3. Thus, all of our numbers must be like:

                                2                 

or

                                3                 

In either case, we have three choices for the remaining numbers (1, 3, 4 in the first case, and 1, 2, 4 in the second). Thus, there are 6 ways of ordering the remaining three in either case, so we have a total of 12 four-digit numbers with a prime in the tens place.

To answer (2), we want to find how many total four-digit numbers there are. This can be done via our rule: we just need to consider how many ways there are to put 4 objects into four slots. Thus, we get \frac{4!}{(4-4)!} = \frac{4!}{0!} = \frac{4!}{1} = 24 ways to do so.

Thus, the answer is that there is a \frac{12}{24} = \frac{1}{2} chance of getting a prime in the tens place.


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In a previous post, we discussed this rule:

Mutually Exclusive Rule for P(A or B)
Let A, B be mutually exclusive events. Then, P(A or B) = P(A) + P(B).

Here, we talk about why this rule makes sense. First, an example:

Example 1
In rolling a fair, six-sided die, what is the probability that you will get a 1 or a 4?

Solution
We know that rolling a 1 and rolling a 4 are mutually exclusive events, since it is impossible for them both to occur. We know that P(rolling a 1) = ⅙ and that P(rolling a 4) = ⅙. Thus, by our above rule, P(Rolling a 1 or rolling a 4) = ⅙ + ⅙ = ⅓.

We may also consider the different possible outcomes that are relevant here:


So we roll a 1 or 4 in 2 of the six possible outcomes. Thus, the probability of doing so is just $\frac{2}{6} = \frac{1}{3}.$

In looking at some other examples, we may see why this more general rule makes sense:

General Rule for P(A or B)
Let A, B be two events. Then, P(A or B) = P(A) + P(B) - P(A and B).

Again, imagine we are rolling our favorite fair six-sided die. We are considering two events: A and B. We know that A occurs in three of the possible outcomes and B occurs in three of the possible outcomes. Can we determine P(A or B)? Now, this rule may seem a little arbitrary at first, but let’s see if we can see why it is true.

The answer is no. To see why, let’s consider two possible cases:

-       A: I roll an even number

-       B: I roll an odd number

Here, A occurs in three possible outcomes (2, 4, 6) and B occurs in three possible outcomes (1, 3, 5). And, in this case, P(A or B) = 1.

We can calculate this by considering how many of our outcomes satisfy A or B. So, for example, suppose we roll a 1. Then, “A or B” is true, since 1 is an odd number (so it is true that “I roll an odd number or I roll an even number”). Now suppose we roll a 2. Again, “A or B” is true, since 2 is an even number. Since all the numbers 1 through 6 are odd or even, we conclude that “A or B” will be true for all six outcomes. So, P(A or B) = 6/6 = 1.

But now consider this case:

-       C: I roll an even number

-       D: I roll a prime number

Again, C occurs in three possible outcomes (2, 4, 6) and D occurs in three possible outcomes (2, 3, 5). Now, in how many cases will “A or B” be true? Well let’s go through the cases. Suppose I roll a 1. Now, 1 is not even and it is not a prime number. So “C or D” will not be true if I roll a 1. Now consider 2; it is even, so “C or D” will be true if I roll a 2. And 3 is a prime number, 4 is even, 5 is prime, and 6 is even; so “C or D” will be true for all of those numbers.

Thus, P(A or B) = 5/6.

Thus, even when we know that A occurs in three of the possible outcomes and B occurs in three of the possible outcomes, we cannot determine P(A or B). (We are only able to do this if we get the additional information that A and B are mutually exclusive.)

To see why this is true, let’s picture each of the different outcomes as corresponding to some box.

Now, remember that for “A or B” to be true, we only need one of the two to be true. Let’s look at which boxes make A true.

And let’s now add which boxes make B true:

So, the boxes that make “A or B” true will be all of the boxes.

But now let’s look at C and D. So, the boxes that make C true will be:

And the boxes that make D true will be:

Thus, the boxes that make “C or D” true will be only 5 of the 6 boxes. Looking at it this way, it becomes clear what went wrong: C and D overlap on one of the boxes. So we are “double-counting” the square 2. C counts it and D counts it. In some sense, we are wasting one of our boxes on 2.

That’s why our formula for calculating P(A or B) looks the way it does. First, we add the boxes where A is true. Second, we add the boxes where B is true.

But, now we’ve double-counted the boxes where A and B are both true – we added them once as part of A, and another time as part of B! That’s why our formula then has us subtract P(A and B). That way, we are cancelling out this double-counting.

So, to re-iterate, the formula for disjunctive probabilities is:

P(A or B) = P(A) + P(B) – P(A and B).

And from this general formula, we can derive the formula for Mutually Exclusive Disjunctions. So you only really need to learn this formula. How do we derive it?

Well, when A and B are mutually exclusive, we know that P(A and B) = 0. That’s just what it means to be mutually exclusive! Thus,

P(A or B) = P(A) + P(B) – P(A and B) = P(A) + P(B) – 0 = P(A) + P(B)

which was exactly our formula.

 


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In a previous post, we discussed this formula:

Probabilities of Compound Events
Let A and B be independent of one another. Then, P(A and B) = P(A)P(B)

An example will help to illustrate why this formula makes sense.

Example
Suppose I roll a fair six-sided die and flip a fair coin. What is the probability that the coin lands heads and the die lands on six?

Answer

In general, listing out all the different possible outcomes will be infeasible. But it helps to get a sense for why this formula works. Here is an example where we wouldn’t bother to list out all the different outcomes:

Example
Suppose I roll three fair six-sided die in a row. What is the probability that the first die lands on an even number, and the second and third die both land on odd numbers?

Answer

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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:

Example 1
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:

Answer

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:

Practice Problems

Question 1
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?

Answer

Question 2
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?

Answer

Question 3
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?

Answer

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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:

Example 1
You are about to roll a fair six-sided dice. What is the probability that you roll a prime number?

Answer

Why do we bother with this rule? Sometimes it is actually more convenient to calculate things this way. For example,

Example 2
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?

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.

Practice Problems

Question 1
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?

Answer

Question 2
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?

Answer

Question 3
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?

Answer

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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

Results:

  • 🏆 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

Featured image: Steps by Free-Photos is licensed under Pixabay License.

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On today's episode, J.Y. talks about Main Conclusion (aka Main Point) questions and why they're foundational to logical reasoning.

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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

Results

  • 🏆 Accepted at a T-14 school
  • ✍️ Handwritten note from the dean: "I loved your essays" (and more)

🌘 The Strategy: A Shot at the Moon

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Featured image: Person Standing on Hand Rails With Arms Wide Open Facing the Mountains and Clouds by Nina Uhlíková is licensed under Pexels License.

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On today's episode, J.Y. talks about how to score your preptests to make them better predictors of your actual LSAT score.

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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

Results:

  • 🏆 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.

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Featured image: White Dump Truck Near Pine Tress during Daytime by 500photos.com is licensed under Pexels License.

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