Archive for the ‘GRE’ Category

Sometimes, we want to multiply a number by itself several times. For example, if we are trying to guess someone’s password, we may want to know how many 5-letter sequences are possible. Following our lessons on combinatorics (see here), we know that there are 26 \times 26 \times 26 \times 26 \times 26 possibilities. But we may want an easier way to write this down. The solution is to use exponents. In general,

Rule for Positive Exponents

Using this rule, we can rewrite our earlier expression as simply 26^5. We call 26 the base and we call 5 the exponent or power. And in general, for b^n, b is the base and n is the exponent. In English, we say this is “b to the n-th power.” So we would say that 26^5 is 26 to the 5th power.

Now, we can take any number and raise it to the power of any positive integer simply by multiplying the number by itself that many times. So, we have:

We can even do this for irrational numbers, like \pi:

But we can also have negative exponents. When we take a negative exponent, we use the following formula:

Rule for Negative Exponents

    \[b^{-n} = (\frac{1}{b})^n\]

In essence, we flip the base and then take a normal exponent. Here are some examples:

    \[3^{-4} = (\frac{1}{3})^4 = \frac{1}{81}\]

    \[\frac{1}{8}^{-3} = \frac{8}{1}^3 = 8^3 = 192\]

    \[\pi^{-99} = \Big(\frac{1}{\pi}\Big)^{99}\]

Finally, if we raise anything (other than 0) to the 0th power, we get 1. In other words,

Rule for the 0th Power

If b \neq 0, then b^0 = 1.

For example:

    \[29^0 = 1\]

    \[\pi^0 = 1\]

    \[(49+28/3 + \sqrt{2} + \sqrt{3} + \sqrt{4892})^0 = 1\]

Now, there are some rules for how different things with exponents can be combined.

Multiplication and Division Rules for Exponents

    \[a^{b}  a^c = a^{b + c}\]

    \[\frac{a^b}{a^c} = a^{b - c}\]

Exponent Rule for Exponents

    \[(a^b)^c = a^{bc}\]

Practice Problems

Simplify the following expressions:

  1. 5^{20} \times 5^{27}
  2. \frac{5^3}{5^{15}}
  3. \frac{9^3}{3^2}
  4. (4^2)^3
  5. (28^3)^0
  6. (7^{-3})^{-4} + \frac{5^3}{25}


  1. 5^{47} by the Multiplication Rule for Exponents
  2. 5^{-12} or \frac{1}{5}^{12} by the Division Rule for Exponents
  3. We can actually solve this problem two ways. First, we could solve it just by calculating: \frac{9^3}{3^2} = \frac{279}{9} = 81. Second, we could solve it by our formulas: 9 = 3^2 so 9^3 = (3^2)^3 = 3^6 by our Exponent Rule for Exponents. Thus, \frac{9^3}{3^2} = \frac{3^6}{3^2} = 3^4 = 81 by our Division Rule for Exponents.
  4. 4^6 by the Exponent Rule for Exponents
  5. 1 by the Rule for the 0th Power
  6. 7^{12} + 5 by the Exponent Rule for Exponents and either just directly calculating \frac{125}{5} or by using the Division Rule for Exponents: \frac{5^3}{25} = \frac{5^3}{5^2} = 5.

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Linear equations are a specific kind of equation that meets certain limitations. In general, linear equations can have as many variables as you like, but those variables must be raised to only the first power (see here for a guide to powers/exponents). Furthermore, those variables must not be multiplied by any other variables, although they can be multiplied by constants.

Here are some examples of linear equations:

    \begin{align*} x =& 8 \\ 3x + 29 =& 6 \\ 3x - 4y =& 24 + z \end{align*}

Here are some examples of equations that are not linear:

    \begin{align*} x^2 =& 8 \\ xy = & 8 \\ x + y  =& z^2 \end{align*}

In subsequent posts, we will discuss two methods for solving linear equations: elimination and substitution. We will also talk about systems of linear inequalities, which occur when we replace those equality signs with inequalities like < or \geq.

For now, see if you can identify whether the following equations are linear or not:

Practice Problems:

Identify whether the following equations are linear or not

  1. \frac{x^2}{y} + 4 = z
  2. 4x - 2 = 3
  3. xy - xy + z = 4
  4. x^{-1} + x = 3
  5. \frac{1}{x+y} = 4
  6. \frac{1}{x+y} =z


  1. No, since \frac{x^2}{y} is not a single variable raised to the first power.
  2. Yes, since 4x involves just a single variable raised to the first power. It is multiplied by a constant (namely 4), but that is fine.
  3. Technically no, since xy is not a single variable raised to the first power. But we can see that since xy - xy = 0 no matter what the value of x and y are, our equation has the same solutions as the linear equation z = 4.
  4. No, since x^{-1} is not a linear term.
  5. Technically no, since \frac{1}{x+y} is not a linear term. But provided x + y \neq 0, we can multiply both sides by it in order to get 1 = 4(x+y) which is linear. So again, our equation has the same solutions as a linear equation.
  6. No, and our earlier trick for problem 5 will not even work here since 1 = z(x+y) is still not a linear equation.

Now, systems of linear equations are just groups of 1 or more linear equations. The solutions to the system are all the values for the variables that make every equation true. Note that every solution must specify the value for each variable in our system. So, for example, in the following system:

    \[\begin{cases} 3x + y =& 9 \\ 2x + 2y =& 10\]

The response x = 2 is not a solution. However, x = 2, y = 3 is a solution. And in general, we can give our solutions in the form of ordered pairs. So we would write (2, 3) is a solution, where the first coordinate corresponds to the value of x and the second coordinate corresponds to the value of y.

The reason for requiring that solutions specify values for all variables (not just some) is that sometimes a certain value for one variable can be a solution and sometimes not, depending on the values of the other variables. Consider:

    \[x = y\]

Now, this system has many solutions. For example, x = 3, y = 3 is a solution. But we cannot simply say that x = 3 is a solution, for if y = 4, then that is no solution at all (since 3 \neq 4)! Thus, we require that a solution for a system of equations specify values for every variable in the system.

In our next post, we discuss how to solve systems of linear equations via elimination.

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In this series, we discuss how to solve systems of equations. This is an important skill for the GRE since many of the math problems will involve solving systems of equations (e.g. problems with inequalities, algebra problems, or word problems). Now, this content builds on some other mathematical ideas, like the idea of a variable and the distributive law. If you want a refresher on those ideas, see this post.

What is a “system of equations”?

A system of equations is a group of one or more equations, like this:

    \[x^2 + 6x + 9 = 0\]

Or this:

    \[\begin{cases} 4x + y =& 8 \\ 2x - y =& 4 \end{cases}\]

And we want to find the possible value(s) of x and any other variables in the problem. So, for example, in the second pair of equations we could have:

    \[x = 2, y = 0\]

And then we would get:

    \[\begin{cases} 4x + y = 4(2) +0 &= 8 \\ 2x - y = 2(2)- 0 &= 4\end{cases}\]

making both equations true. But if we had x = 3, y = 2 then we would get:

    \[\begin{cases} 4x + y = 4(3) +2 &= 8 \\ 2x - y = 2(3)- 2 &= 4\end{cases}\]

which would make the second equation true, but not the first. A solution to a system of equations is an assignment of values to every variable such that every equation in the system is true. By what we have said above, x = 2, y = 0 is a solution to the above pair of equations, whereas x = 3, y = 2 is not.

In this series, we will focus on solving two basic kinds of problems:

  1. Linear Systems of Equations: These only have variables raised to the 1st power. For example:
    • x + y = 21
    • \frac{42}{9} x + 2y + 81 = 13
  2. Quadratic Systems of Equations: These include variables raised to the 2nd power. For example:
    • x^2 = 2
    • 4x^2 + 2x + 6 = 13

What Kinds of Problems Involve This Skill?

This skill can come in handy for many different types of GRE problems, such as:

  • Quantity Comparison Problems
    Suppose x > 100. Which of the following is larger?


    x + 9,900
  • Word Problems
    Tom and Liz are baking cookies. Tom can bake 10 cookies in an hour and Liz can bake 5 cookies in an hour. Working together, how long will it take them to bake 40 cookies?
  • Problems with Infinite Series
    Suppose we have the series: 1, ½, ¼, …
    What is its sum?
  • Problems with Inequalities
    Which values of x, y will satisfy the following system of inequalities?

        \[\begin{cases} x + 3y \ge  & 4 \\ 2x - y < & 1 \end{cases}\]

In our next post, we discuss what a linear equation is, a precursor to actually solving them

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In order to solve systems of equations, we need to understand three concepts: variables, how to manipulate equations, and the distributive law.


Variables are letters that stand in for numbers. Often, we will use the letters x, y, z as variables. We can use equations to assign a specific value to a certain variable. For example, x = 3 means that the variable x is equal to the number 3.

We use variables to capture certain relationships between quantities when we do not know the specific values involved. For example, I could tell you that my cell phone is worth twice as much as my laptop (and this is true). Without variables, we would have no way to represent this mathematically, since we do not know the actual value of either my laptop or cell phone. But with variables, we can say: x = the value of my laptop, y = the value of my cell phone. Then, we can represent this information mathematically as: 

    \[y = 2x\]

Thus, by using variables, we can capture relationships between unknown values.

Manipulating Equations

The second idea we need to understand is how to manipulate equations. Often, we are given a complicated equation like:

    \[3x + 8 = 17\]

And we want to turn it into a simpler equation, like x = 3.

Now, the key is that we are allowed to anything we like to one side of the equation provided we do it to the other side as well. So, for example, we can take our initial equation

    \[3x + 8 = 17\]

And we can subtract 8 from both sides to get:

    \[3x = 9\]

And we can divide by 3, provided we do it to both sides, to get:

    \[x = 3\]

So now we know what the value of x is, namely 3.

The reason why this works is that in an equation, you are saying that the left hand side is equal to the right hand side. In other words, that the left hand side represents the very same number as the right hand side. So in our equation,

    \[3x + 8 = 17\]

We are saying that 3x + 8 is just another way of writing the number 17. So of course we can do anything we like to both sides and the resulting numbers will still be equal, since we are doing the same thing to the same numbers.

Remember, though, that we still cannot do mathematically illicit operations, like dividing by 0, even if we do them to both sides.

Distributive Law

Finally, we need to understand how the distributive law works. It is a rule for how multiplication and addition interact and it states:

    \[a \times (b+c) = (b+c) \times a = a \times b +a \times c\]

An example may help in seeing that the distributive law is, in fact, true:

We know that 56 = 8\times 7, and we know that 8\times 7 = 8\times (3+4) because operations in parenthesis are done before any other operations. Thus, we know that

    \[56 = 8 \times 7 = 8 \times (3 + 4)\]

And by the distributive law,

    \[8\times (3+4) = 8 \times 3 + 8 \times 4 = 24 + 32 = 56\]

So it is true that 8\times (3+4) = 8\times 3 + 8\times 4 since 8\times (3+4) = 56 and 8\times 3 + 8\times 4 = 56 (and of course 56 = 56).

Now, one very important application of the distributive law is used in factoring quadratic equations (as we discuss here). We will need to apply to the distributive law to:

    \[(a+c)\times (b+d)\]

By distributing, we get:

    \[(a+c)\times (b+d) = (a+c)\times b + (a+c)\times d\]

    \[= a\times b +c\times b + a\times d + c\times d\]

And in general, we will omit the multiplication symbol \times and simply write:

    \[(a+c)(b+d) = ab + cb + ad + cd\]

You may have also seen this formula called FOIL (First Outer Inner Last), a mnemonic meant to remind people of the four terms in our equation:

However you choose to remember it, this formula is an important part of understanding how to factor equations.

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Sometimes, the order of our objects does not matter. For example, suppose a professor is trying to assign people to different study groups. Then, it does not matter who gets assigned to the study group first; what matters is who is in which study groups, as in the following example:

A professor is trying to split up his class of 9 students into 3 groups of 3. How many ways can he do this?

Or we might have:

An appeals court is convening a 3-judge panel to hear a certain case. There are 26 judges in the courthouse. How many 3-judge panels are possible?

Now suppose that of the 26 judges, 13 are conservative and 13 are liberal. How many possible 3-judge panels will have at least one conservative and one liberal on the panel?

To address these kinds of problems, we will use the following:

Rule for Combinations
If you have n objects and you want to pick k of them to form a group, then there are n!/((n-k)!k!) ways to do so.

Let's try to solve our earlier example with this rule.

By applying our rule, we know that there are \frac{26!}{(26-3)!3!} ways of choosing 3 judges from the total group of 26 judges in order to form a panel. We can directly calculate this:

    \[\frac{26!}{(23!3!)} = \frac{26 \times 25 \times 24}{(3 \times 2 \times 1)}\]

, since 26! = 26 \times 25 \times 24 \times 23! and \frac{23!}{23!} = 1.

    \[\frac{26!}{(23!3!)} = 26 \times 25 \times 24 = 2600 .\]

So there are 2600 ways to pick such a panel.

But why does this rule make sense? Well, let’s start with our rule for permutations: if I want to form a line of three judges from this group of 26, then there are 26*25*24 ways of doing this. This is because we have three spots to fill:

____  ____  ____

And for the first spot, we can pick any of the 26 judges; for the second, we can pick any of the remaining 25 judges, and for the last spot, we can pick any of the remaining 24 judges.

But here is the problem: we don’t actually care about the order in which we pick the judges. Suppose we have Judge Abby, Judge Ben, and Judge Cynthia. Then, we want to treat this possibility:

Abby Ben Cynthia

the same as this possibility:

Ben Abby Cynthia

which is in turn the same as

Cynthia Abby Ben

Cynthia Ben Abby

Abby Cynthia Ben

Ben Cynthia Abby.

We want to count all of these possibilities just once, since for us, a panel with Abby, Ben and Cynthia on it is just the same as a panel with Cynthia, Abby, and Ben on it. They’re not actually different!

Now, whenever we pick three judges from the 26 total judges, there are 3! ways to order those judges. This is because, for the first slot, we can pick any one of the three judges, for the second slot, we can pick any one of the remaining two judges, and for the last slot, we are stuck with the final judge.

So our permutation 26*25*24 actually counts each of these panels 3! times, when we only want to count each panel once. Suppose a panel has judge A, B, and C on it. Then our permutation counts:







So since our permutation is counting these panels 3! times when it should be counting them just once, we divide by 3! to get our final answer.

And, in general, the formula for how many combinations of k items we can pick from n objects is just equal to the number of permutations of k items we can pick from n objects, divided by (k!). Dividing by k! is how we make sure we are not counting all the different orders a particular combination can be put in.

Practice Problems

Question 1
A professor is trying to split up his class of 8 students into two groups of 4 students each. How many ways can he do this?


Question 2
Congress has decided to randomly pick a group of 4 congresspeople to lead a congressional committee. There are 535 congresspeople. How many groups are possible? What is the probability that any particular congressperson will be on the committee?


Question 3
Congress has decided to randomly pick a committee of 4 congresspeople, but they decide to pick as follows: 2 seats are randomly chosen from the Representatives while the remaining 2 are randomly chosen from the Senate. How many committees are possible? What is the probability that any particular Representative will be on the committee? What is the probability that any particular Senator will be on the committee?


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And finally, we want to address the case when we have permutations with repeated objects. What if we have something like:

The DMV is wondering how many new license plates it can form from just the letters: T, T, and R. How many such license plates are there?

Now, this question is something of a hybrid between our permutations and combinations. It is true that the order of the objects matters: TTR is not the same license plate as RTT. But we can’t just treat it as a standard permutation where we calculate n!/(n-k)!. If we tried that, we would get:

3!/(3-3)! = 3! = 6

But, if we list out all the possibilities, we find that we cannot form 6 different possibilities. The only possible license plates are:




You can try to think of another ordering, but there isn’t one. So treating it as a normal permutation leads us to the wrong answer. What is going on here?

To see what’s going wrong, let’s label our two T’s as T1 and T2. Now, let’s see how many ways we can order them:

T1 T2 R

T1 R T2

T2 T1 R

T2 R T1

R T1 T2

R T2 T1

Now we get the six possibilities we were expecting. But, remember, as a license plate, it doesn’t matter whether one uses T1 or T2. At the end of the day, the letter that shows up on the plate is just T. So, for example, combinations like R T1 T2 are just the same as R T2 T1. They both result in license plates that look like:


So, like in our combination problem, our permutation is over-counting things. It is treating as distinct some possibilities that are, in fact, the same. Here, our permutation gives us 6 possible outcomes. But, in fact, the ones with the same color are the same:

T1 T2 R

T1 R T2

T2 T1 R

T2 R T1

R T1 T2

R T2 T1

And we see that, in fact, we only have 3 different possibilities, which is what we found in listing out the possible license plates.

So our rule for these cases:

Permutations with Repeated Objects
Suppose you want to order k objects where one of the objects repeats n times. Then, the number of possible orderings is: \frac{k!}{n!}.

is again a way of correcting for our over-counting. In our case, we can think of this rule as saying: there are 6 permutations. But within those permutations, the order of the T1 and T2 doesn’t matter. There are 2! ways to order T1 and T2, so the total number of distinct ways to order R, T, and T is just: 6/2! = 3.

Here is an application of this rule that involves larger numbers:

You are playing Scrabble. Your hand has: T, R, S, S, S, E, and F. How many possible ways are there to order all of your tiles?

Following our rule, we want to find a way to order 7 objects (i.e. our 7 letters) where one of the objects is repeated 3 times. Then, there are 7!/3! ways to do this. And again,

7! = 7*6*5*4*3!


7!/3! = 7*6*5*4 = 210

There are 210 possibilities

Practice Problems

Question 1
You are trying to come up with anagrams. You get the letters: A, Z, P, P, E, D. How many 6-letter combinations can be formed?


Question 2
You are playing Scrabble. You get the letters: S, S, S, S, T, E, P, X. How many 8-letter combinations can be formed?


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So far we have talked about calculating probabilities of events where we tell you all the possible outcomes. For example, when we flip a coin, the possible outcomes are known: heads or tails. In such examples, you know how many possible outcomes there are. If we roll a six-sided die, there are clearly just six outcomes. But what if we are not given the number of possible outcomes? Consider this problem:

There is a bag of 5 marbles; 3 blue and 2 red. Jane picks one marble out of the bag, looks at its color, sets it to the side (and does not return it to the bag) and picks another marble. What is the probability that she picks two blue marbles?

In this case, it isn’t clear how to apply our earlier rules. Remember our first rule:

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.

How do we apply this rule to Jane’s situation? How many possible outcomes do we have?

This is where combinatorics comes in: it will allow us to determine how many possible outcomes we have. Now the word “combinatorics” is a little intimidating, but it’s just a fancy word for the mathematics of counting. In this case, we will use it to count the number of possible outcomes for Jane.

Let’s try to list out all the possible pairs of marbles that Jane could pick. To do this, let’s number our blue marbles 1 through 3, and number our red marbles 4 through 5. Now, Jane has to pick two different marbles. So, in other words, Jane ends up with some pair of marbles from the following list:

So these are all the possible outcomes Jane could get. She could pick any of the 5 marbles first, and then she could pick any of the 4 remaining marbles second. Now, the question doesn’t say this explicitly, but we will assume (and it is generally fair game to assume) that Jane is equally likely to pick any of the marbles from the bag. After all, the marbles are presumably the same size and feel the same, so she’s just as likely to pick one as to pick any other.

Now, we have our set of equally likely outcomes. And these are all the outcomes. So, in line with our above rule, to find P(Jane picks 2 blue marbles) now we just need to pick out the outcomes where Jane picks two blue marbles. And that’s easy enough to see on our table, we just need to pick the outcomes where two blue marbles are chosen.

Adding up these outcomes, we get 6/20 = 3/10. So that’s the likelihood of Jane picking two blue marbles.

In general, it will not be practical to make our table of possibilities and add things up. For if Jane had a bag of 10 marbles, we would have had 90 possible outcomes (10 choices for the first marble multiplied by 9 choices for the second). So, we introduce the following rule:

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:

    \[\frac{n!}{(n-k)!} .\]

The n! is read as “n factorial” and it means n * (n-1) * (n-2) * … * (1). Basically, we take n and we multiply it by all the positive numbers less than or equal to it. Here are some examples:

5! = 5 \times 4 \times 3\times 2 \times 1 = 120

2! = 2

1! = 1

98! = 98 \times 97 \times 96 \times … \times 3 \times 2 \times 1

And, by a special convention, we say that 0! = 1.

Now, to solve probability questions, we want to find: how many permutations (or possible outcomes) will satisfy our event, and how many permutations in total are possible (i.e. how many possible outcomes there are). So in our above problem, we want to find:

  1. How many permutations involve two blue marbles being drawn? (How many outcomes involve two blue marbles being drawn?)
    We have three blue marbles, and we are drawing two marbles. So we want to use three objects to fill two spaces. Following our rule, we get:

        \[\frac{3!}{(3-2)!} = \frac{3!}{1!} = 3! = 6 .\]

  2. How many total permutations are there? (How many possible outcomes are there in total?)
    We have five total marbles and we are drawing two of them. So we want to use five objects to fill two spaces. Following our rule, we get:

        \[\frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20 .\]

Thus, we get: P(Draw two blue marbles) = \frac{6}{20} = \frac{3}{10}, which matches our previous answer.

Practice Problems

Question 1
Suppose you have three friends: Anne, Beth, and Charles. How many ways can they form a three-person line?


Question 2
Mark is trying to descramble a series of letters: AQCFXDE. How many possible five-letter combinations can be formed from those letters?


Question 3
Jane is drawing four cards from a standard 52-card deck. What is the probability that she first draws the King of Hearts, and then the Ace of Spaces, and then the Jack of Clubs, and finally the Three of Hearts?


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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:

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?

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:

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?

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:


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




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?

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.

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?


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:

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?


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