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Chapter 1: Introduction to Permutations

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Permutation questions are about taking a group of objects and totaling how many ways we can arrange them in specific ways. Here is an example that we will explain later.

In how many ways can a pet shop line up 3 cats and 3 dogs in 6 cages if the cats must be in the second, fourth, and sixth cages?

I. The Basics: Three Steps to Permutation Clarity

1. Figure out how many places there are to fill.
2. Figure out how many objects potentially can go into each place.
3. Multiply for the answer.

Example
How many outcomes are there when two identical dice are rolled?


Following the steps:

1. Figure out how many places there are to fill
Because there are two dice, there are two places to fill:
__ __

2. Figure out how many objects potentially can go into each place
Because each die has 6 different potential outcomes, we will fill the spaces accordingly:
_6_ _6_

3. Multiply for the answer
_6_ × _6_ = 36

Example 2
In Country X, three digit area codes are to be given to each town. The first digit will be any number from 2-9, inclusive, the second digit can only be either 0 or 1, and the third digit can be any number from 0-9, inclusive. How many different area codes can be issued in Country X?


Following the steps:

1. Figure out how many places there are to fill
Because there are three digits, there are three places to fill: __ __ __

2. Figure out how many objects potentially can go into each place
The question states that the first digit can be any number from 2-9, inclusive. There are therefore 8 potential options. The second digit can be only 0 or 1, therefore, there are 2 potential options. The third digit can be any number from 0-9, inclusive, and there are 10 such numbers. The diagram looks like this:
_8_ _2_  _10_.

3. Multiply for the answer
_8_ × _2_ × _10_ = 160

II. Permutations Without Replacement
Sometimes the number of possibilities decreases instead of remaining the same. With dice, you may role dice as many times as you want, but there will always be 6 possibilities. But sometimes the number of possibilities change in a question.

A student wants to assign 7 different books to 3 spaces, how many different possible possibilities are there?


Would you calculate _7_ × _7_ × _7_= 343 like above?
How could you if ever time you select a book, the number of possibilities decreases?


Use Logic

Logic tells us that there are 7 choices for the first book. Then 6 choices for the second book and then 5 choices for the third book. Possibilities decrease as items are selected.

To calculate the total possibilities for the three spaces we multiply 7 × 6 × 5 = 210



Permutations Without Replacement Formula

There is a more specific formula for this that essentially does the same thing as the logic above.

In this formula, n stands for the distinct objects which you are choosing from, r stands for the number of spaces which those n objects can fit into, and P stands for Permutation, and is not an arithmetic part of the equation. The exclamation point(!) after each letter represents the factorial of that number.

Factorial(!) means multiplying a number by every positive integer below it down to 1.

5! = 5 × 4 × 3 × 2 × 1 = 120
3! = 3 × 2 × 1 = 6

If you want to fit 7 books into 3 spaces, and want to know the possible permutations, you would assign 7 to n (since the books are the distinct objects which you are choosing from), and assign 3 to r (since it is the number of spaces that n can fit into).
Therefore the formula would read:

that would be written out as:

We can cancel out 4 × 3 × 2 × 1 in both the numerator and denominator, so we are left with 7 × 6 × 5 which equals 210.

	800score Tip: It is your choice as a student whether to rely on either the formula or use logic. 800score provides both approaches, but we suggest logic. The GRE isn't interested in your perfect memorization of the permutation formula, the GRE wants you to have a good intuitive sense of how permutations work.

III. Replacement or Non-Replacement

The GRE will test your ability to distinguish problems with or without replacement. So you should be very good at identifying which one it is.

	800score Tip: Keep in mind that the GRE's effectiveness as a test is a function of it's ability not to be "beaten" by standard test preparation. So it is quite logical for the GRE to set up trick questions specifically to penalize over-preparation. Our test prep gnomes who have recently taken the GRE warn us that questions about pulling cards from a deck or marbles may not be what common "cards" or "marble" questions are usually about. So if you have done dozens of generic questions with marbles and you assume it must be a permutations without replacement question... be warned: it may not be that way on test day.

Chapter 2: Problem Variations

Example 1
In how many ways can 5 people sit on a 5 person bench?

In this case, we cannot have repeating values. If someone named Bob is already sitting, he cannot appear again at a different place on the bench!

1. Figure out how many places there are to fill
There are 5 seats on the bench, so there are 5 places:
__   __   __   __   __

2. Figure out how many objects potentially can go into each place
There are 5 people who could sit in the first seat. Once someone actually does sit, though, there are only 4 people who could sit in the second seat. For the third seat, there are only 3 people left, and then 2 and then 1. So the places will look like this:
_5_   _4_   _3_   _2_   _1_

3. Multiply for the answer
_5_ × _4_ × _3_ × _2_ × _1_ = 120

We can also solve Example 3 by using the formula:

nPr = 5P5

Where we have 5 people to choose from (n) for 5 seats. When expanded out this would be:

In permutation problems, there is no divide by zero, so the zero would just be crossed out and we would multiply out 5 × 4 × 3 × 2 × 1 which equals 120.

Example 2
There are six meal options in the cafeteria of a certain school. Assuming that a different meal must be served each day, and each different type of meal must be served once before any type of meal can be served a second time, how many different ordering options are there for a student in the first four days?

Solution
The answer is 360. In this case we cannot have repeating values not because of the nature of the situation, but because we are told that a student cannot be served two of the same type of meal before they go through all the meals first.
Following the steps:

1. Figure out how many places there are to fill:
We want to know how many ordering options a student has on the first four days, so we have four places to fill.
___   ___   ___   ___

2. Figure out how many objects potentially can go into each place:
There are 6 meal options that a student can order from on the first day. Since he/she cannot repeat a meal before all are tried, there are only 5 meal options on the next day. On day three, there are 4 options, and on day four, 3 options.
_6_   _5_   _4_   _3_

3. Multiply for the answer
_6_ × _5_ × _4_ × _3_ = 360

We can also use the formula, nPr = 6P4 where n stands for the 6 meals that there are to choose from and r stands for the 4 days or spaces, that you can fit those meals into.

6P4 = 6 × 5 × 4 × 3 = 360

Example 3
A boy is given a hat with 6 tiles, each numbered with a different digit 1-6. If he will pull out 3 tiles, one tile at a time, and lay them in the order he pulls them out, how many different 3-digit numbers could he possibly create?

1. Figure out how many places there are to fill
3, one for each tile: __   __   __

2. Figure out how many objects potentially can go into each place
Because he is not returning the tiles to the hat, the numbers cannot repeat. There are therefore 6 potential tiles for the first place, then 5, and then 4:
_6_ _5_   _4_

3. Multiply for the answer
_6_ × _5_ × _4_ = 120

Answer = 120

Notice that the places do not have to count down all the way to 1 in order for this to work. This is no longer a complete factorial.

We can also use the formula, nPr = 6P3 where n stands for the 6 tiles that there are to choose from and r stands for the 3 tiles or spaces that the boy will pull out and try to create a number with.

6 × 5 × 4 = 120

For our next example, let’s return to the first problem.

Example 4
In how many ways can a pet shop line up 3 cats and 3 dogs in 6 cages if the cats must be in the second, fourth, and sixth cages?

1. Figure out how many places there are to fill
There are six cages for the animals, so there are six places:
__   __   __   __   __   __

2. Figure out how many objects potentially can go into each place
This is where things change, but just a bit. There are 3 cats and 3 dogs, but the cats must be in the second, fourth, and sixth cages. That means the dogs must be in the first, third, and fifth cages. If we go cage by cage, we can figure out the potential number of options for each cage (remember, there can be no repeating values):
_3_   _3_   _2_   _2_   _1_   _1_

3. Multiply for the answer
_3_ × _3_ × _2_ × _2_ × _1_ × _1_ = 36

We can also use the formula:

nPr   ×   nPr   =   3P3  ×   3P3

In this case we need to solve two permutations, one for the dogs and the other for the cats because the 3 dogs (n) can only go into 3 of the cages (r). Also, the 3 cats (n) can also only go into 3 of the cages (r). Afterwards, we multiply both answers in order to find out how many arrangements are possible.

So, 3P3   ×   3P3 = 3!/0! × 3!/0! = 3 × 2 × 1 × 3 × 2 × 1 = 36 possible permutations

Example 4 (hard)
A company will create different ID numbers for its employees. Senior-level employees will receive 4-digit ID numbers, and junior level employees will receive 5-digit ID numbers. If the first digit of any ID number cannot be zero, and if no digits will be repeated in any ID number, what is the ratio of the total number of senior level ID numbers possible to the total number of junior level ID numbers possible?




This is the same as all the others, we just have to do it twice and make a ratio out of the numbers. There are two tricks in this question:

1) Correctly figure out how to translate the restraints on the numbers into math.
2) To save time, do not multiply until the end. You will see how well that works.

For senior level employees:

1. Figure out how many places there are to fill
4, for each digit of the ID number: __   __   __   __

2. Figure out how many objects potentially can go into each place
The first digit can be any digit except 0. That means there are 9 options. The second one can have any digit including 0, but it cannot have the same digit as the one before it. So there are also 9. When two digits are used, there are 8 options for the third digit, and 7 for the fourth:
_9_   _9_   _8_   _7_

3. Multiply for the answer
_9_ × _9_ × _8_ × _7_
(Don’t multiply yet because the GRE loves cancellation).

For junior level employees:

The logic is the same here, except there are 5 digits, so the final step will look like this:
_9_  × _9_ ×  _8_ × _7_ × _6_

Finally, now, the question wants the ratio of these two permutations. That means one on top of the other:

You can now see why we did not multiply originally. Most of the numbers cancel out! The result is simply 1/6, or 1:6.

Chapter 1: Example of Combinations

When Order Is Not Relevant

Combinations problems are very similar to permutations problems. The key distinction is that placement/order isn’t relevant for combinations, but placement/order is relevant for permutations.

For example, if a committee is being put together for a company and there is a president, vice president, and treasurer, order matters. If that same committee of three people is being put together but nobody has a rank then order does not matter because there is no hierarchy.

To illustrate this, we will show a permutation question and then make a slight change to make it a combinations question.


I. Example of Permutation vs. Combination

Let’s say we play a game of dice where we roll two dice (one red and one blue) and record the results. If the dice roll up the same number, the results aren’t counted and we roll again. The result is that there are no doubles. How many permutations are possible?

1. Figure out how many places there are to fill
Because there are two dice, there are two places to fill: __ __

2. Figure out how many objects potentially can go into each place
Since you cannot get the same result (doubles) on the second die there are only 5 possibilities for the second roll.
_6_ _5_

3. Multiply for the answer
_6_ × _5_ = 30

You eliminate all doubles (shown on the diagonal above) [(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)] because the game says that if get the same number, the results aren’t counted. So the result is 30 different permutations. Just count all of the outcomes above to get 30.


II. Example of Combinations

Now let’s do the “combinations” versions of the above question.

We’ll try the game again, but instead of using a red die and a blue die, this time both dice are red. How many different possibilities are there?

First, we can easily identify this as a combinations question because order or placement isn’t an issue. In the first question a (5,2) is different from a (2,5) because they have different colors. This time, they are both the same color, so you can’t tell a (2,5) and a (5,2) apart, so order doesn’t matter. Once order or position doesn’t matter, this becomes a combinations question and not a permutations question.

This chart shows all the possibilities (30) but now you can see that half of the outcomes are redundant. For example, we have counted (4,3) and (3,4) (bolded above) when the two red dice will look the same. One of them can't be counted to the total number of combinations. You have to eliminate the double counted results in combinations questions.

Therefore, we have to reduce the total number of permutations. All the possibilities that are overlaps/double counted in an earlier column we will strike through.

If we count the final results, we get only 15 total combinations (in bold red). Obviously you can’t always rely on using charts like this if the number of possibilities is much larger, so there must be a simpler way to solve these combinations problems.

1. Do the problem as if it was a permutations problem.
2. Divide the answer by (the number of spaces)! (factorial)

You can combine the above two steps into this simple combinations formula:

This formula is very similar to our permutation formula. In this problem, the n stands for distinct objects to choose from, r stands for the spaces into which n objects can fit, and the C stands for stating that this is a Combinations problem. The only difference between is the addition of r on the bottom.

Let’s apply this to our dice problem. We have two dice, so there are two spaces. Following the combinations formula:

6! 2!(6-2)!

=

6 × 5 × 4 × 3 × 2 × 1 2 × 1 (4 × 3 × 2 × 1)

=

(3 × 2) × 5 2 × 1

=

15

Notice that there are only 15 combinations while there are 30 permutations. Having red and blue dice made 15 more possibilities than red and red dice.

	800score Tip: Combinations are always fewer than permutations. The implication of the formula above is that combinations are smaller than permutations of the same numbers because we always start with permutations and then divide. Use logic: combinations mean results that are the same are double counted and therefore don't count, so they need to be excluded. Combinations don't address order and therefore produce fewer possibilities.

Chapter 2: Permutation or Combination?

	800score Tip: Half of the challenge in most combination/permutation questions is correctly identifying if it is a combination or permutation question. Once you know this, most become easy.

1. Does order matter?
“Does it matter if the two items (or however many items you have) positions are changed?” If the answer is yes, it is a permutations problem, if not, it is a combinations problem.

2. If the word “arrangements” is in the problem, it is a permutations problem.
The term “permutations” will rarely be in a GRE problem.


3. If the word “combinations” is in the problem, it is a combinations problem.
Don't think it'll be that easy often!

Here are 7 examples each of permutations and combinations. Read through them and try to think out each one. You will soon see the difference between permutations and combinations.

Drill:
For these 3 statements, answer whether order matters or does not matter.

A. A man is redecorating his apartment and is putting up 2 paintings side by side. How many arrangements are possible if he has 6 paintings to choose from?

B. A woman must pick six apples out of a cart of 15 to purchase. How many different ways can this be done?

C. Alex must read 5 books in the next 3 days. How many ways can he do this?

Solution:
A. Order matters.
If a Monet is put on one side of a wall and a Degas on the other, and then they are switched around, that is a different room design arrangement so therefore, order matters.

B. Order does not matter.
Since whatever 6 apples this woman chooses to purchase represent one large group without assignment, there is no order.

C. Order matters.
If Alex must read 5 books in 3 days, it will matter which book he read on which day. Tolstoy on one day, Satre on another day. The prevailing order is the days of the week.

Example 8
There are 15 available toppings in a pizza restaurant. If Maria will order a 4-topping pizza, how many different pizzas could she order?

Solution
As we’ve already seen, the order of toppings on a pizza does not matter (does it matter if pepperoni is on top or mushrooms?), so this is a combinations question. We want to eliminate all the redundant pizzas and only count the different ones.

Two Steps to Combinations:

1. Do the problem as if it were a permutations problem.

a. Figure out how many places there are to fill
There are 4 spaces because there are four toppings: __ __ __ __

b. Figure out how many objects potentially can go in each place
_15_ _14_ _13_ _12_

c. Multiply
_15_ × _14_ × _13_ × _12_

	800score Combinations Tip: Don’t do the multiplication – we will be dividing and canceling numbers out later! Because combinations usually create fractions, you can cancel out the numbers on top with the numbers on the bottom.

2. Divide the answer by (the number of spaces)!

There are 4 spaces, so that’s 4! That looks like this:


Combinations Formula

You may use logic to solve combinations problems or you may use the combination formula:

nCr = 15 C 4

Since we have 15 toppings to choose from, n is 15 and there are 4 possible spaces, so r is 4. Let’s expand this problem out now.

Example 9
An abstract painter has 15 colors on her pallet to work with. If she decides she will paint with exactly 5 colors, how many different combinations of colors can she choose from?

Solution
1. Do the problem as if it was a permutations problem (remember to try to cancel out!).

a. Figure out how many spaces there are to fill
There are 5 colors so there are 5 spaces: __ __ __ __ __

b. Figure out how many objects potentially go in each place
_15_ _14_ _13_ _12_ _11_

c. Multiply:
_15_ × _14_ × _13_ × _12_ × _11_

2. Divide the answer by (the number of spaces)!

There are five spaces, so divide by 5!

We can also use the combinations formula to solve for this:

nCr = 15 C 5

Since we have 15 colors to choose from, n is 15 and she is going to paint with only 5 colors, so r is 5 since we are choosing only 5 of the 15 colors:

Chapter 3: Groups/Pairings

Now that you understand combinations problems and the method to solve them, there are two important combinations question types you should know.

Variation 1: Combinations from Multiple Groups
In this situation, combinations are being drawn from several groups to form a complete set. Figure out the combinations from each group and then multiply them together.

Example 10
At Sam’s Pizza Parlor, there are 8 meats, 7 vegetables, and 5 cheeses to choose from. Jonathan would like to make a pizza with 4 meats, 3 vegetables, and 3 cheeses. How many different pizzas could he order?

Solution
This problem is different from example 8. In example 8, Maria wanted to order a pizza mixing all the toppings together. In this case, Jonathan wants his pizza a specific way. But there are many ways he could have meat on his pizza, many ways he could have vegetables, and many ways he could have cheese. To answer this correctly, you need to solve each combinations problem individually and then multiply the answers together for the correct answer.

1. Do the problem as if it were a permutations problem.

2. Divide the answer by (the number of spaces)!

Meats Vegetables Cheeses

3. Multiply the answers together for the final answer.
70 × 35 × 10 = 24,500

We can also use the combinations formula to solve for this:

nCr × nCr × nCr = 8C4 × 7C3 × 5C3

For this problem, we need to set up three separate combinations equations and then multiply them all together to get the final answer. Our first equation represents the 8 meats Jonathan can put into the four spots on the pizza. The second represents the 7 vegetables for the 3 spaces available, and the last equation represents the 5 cheeses that can go into the 3 available spaces on the pizza. Now let’s expand out the equation:

And now we multiply each result to get the final answer, and we get:

70 × 35 × 10 = 24,500

Example 11
A committee of 7 members will be chosen from 3 groups:
3 from Green Group, which has 6 people
3 from Red Group, which has 8 people
1 from Purple Group, which has 5 people
How many different committees can be created?

1. Do the problem as if it were a permutations problem.

2. Divide the answer by (the number of spaces)!

3. Multiply the answers together for the final answer.
20 × 56 × 5 = 5600

We can also use the combinations formula to solve for this:

nCr × nCr × nCr = 6C3 × 8C3 × 5C1

Once again, we need to set up three separate combinations equations and then multiply them all together to get the final answer. Our first equation represents the 6 people we can choose for the 3 spots from the Green group. The second represents the 8 people from the Red group available for the 3 spots on the committee, and the last equation represents the 5 people from Purple that can go into the 1 space on the committee.

Now let’s expand out the equation:

And now we multiply each result to get the final answer, and we get:

20 × 56 × 5 = 5600

Variation 2: Pairings
The title says it all: these are pairing questions – so they must take place in twos.

Example 12
6 people in a room each shake hands with one another. If no one shakes hands with any other person more than once, how many handshakes take place?

Let’s approach this with a discussion. How many spaces should there be? Many people want to put six. But actually, there are only 2. Why? Because there are only 2 people involved in any handshake!

Solution
In this case, there are 6 people who could be the first person, and then five people to shake that person’s hand. So following our approach for combinations:

1. Do the problem as if it was a permutations problem.
_6_ × _5_

2. Divide the answer by (the number of spaces)!
There are two spaces, so divide by 2!

6 × 5	= 15
2 × 1

You can also solve for this by using the combinations formula:

nCr = 6C2

Where n stands for the 6 people that we are choosing from to shake hands, and r stands for the 2 people who are actually shaking hands.

Example 13
7 basketball teams with five players each are at a tournament. If each player shakes hands with every other player NOT on his own team, how many handshakes take place?

Explanation
This one is MUCH trickier than the last one. But the logic remains the same. How many people take part in a handshake? Two. So there must be two spaces. But in this case we have to use the logic of the problem to answer it correctly. There are 35 people who could be in the first spot, but that person cannot shake hands with anyone on his own team. So that person has only 30 people who’s hands he can shake! That will be reflected in the combination. So let’s approach it using the method and this logic:

1. Do the problem as if it was a permutations problem.
_35_ × _30_

2. Divide the answer by (the number of spaces)!
There are two spaces, so divide by 2! And don’t forget to cancel!

Chapter 1: Simple Probability

Probability is often seen as the real evil of the GRE. But the truth is that by mastering one simple fraction, you can make probability approachable and even easy.

Let’s say that you have a single six-sided die. If you role it, what is the probability you would roll a 5?

There are 6 sides to the die, and each one could come up. The 5 side is one of those sides, right? So there are 6 possible outcomes, and the five is one of them. Therefore, there is a 1 in 6 probability that the five will come up.

	800Score Technique: The Bottom and the Top If you want to make probability approachable, just think of it as a fraction to solve Bottom to Top. The bottom number is the total number of possibilities that could happen, and the top number is the number of possible ways to achieve the desired result.

Or, more simply:

Probability =	what we want
	all of what’s possible.

Apply that to the above example:
Bottom: 6 outcomes are possible
Top: 1 outcome we want
Probability: 1/6

Let’s change the above example just a bit to see this in action. Now, we still have one die, but we want to know the probability of rolling a 5 or a 6?

Now, there are still six different possible outcomes, right? Of those, we want a 5 or a 6. Therefore, two outcomes give us what we want. Think Bottom to Top:
Bottom: 6 outcomes are possible
Top: 2 outcomes we want
Probability: 2/6 = 1/3

Here are several other examples to help you understand the basics of probability:
In a deck of 52 standard playing cards, what is the probability that pulling a single card from the deck will produce a 4?

Answer: Think Bottom to Top
Bottom: 52 outcomes are possible
Top: 4 cards give us a 4
Probability: 4/52 = 1/13

Example 2
In a deck of 52 standard playing cards, what is the probability that pulling a single card from the deck will produce a black card?

Answer: Think Bottom to Top
Bottom: 52 outcomes are possible
Top: 26 cards are black
Probability: 26/52 = 1/2

Example 3
In a deck of 52 standard playing cards, what is the probability that pulling a single card from the deck will produce the Ace of Spades?

Answer: Think Bottom to Top
Bottom: 52 outcomes are possible
Top: 1 Ace of Spades
Probability: 1/52

Chapter 2: Probability of Multiple Events (Advanced)

The only vocabulary you’ll ever need: “And” and “Or”
There are two ways events can happen together in the same probability problem: either they could both happen separately or they must happen together.

A) Scenario 1 – “Or”:
If both events do not necessarily have to occur together, an “or” may be used as in:

I will be happy today if I win the lottery OR have email.

“OR” means that we add probabilities together to get a higher overall probability.


Example 4
John will win $100 if, from a deck of 52 standard playing cards, he chooses either a 7 or a 9 when pulling a single card from the deck. What is the probability that John will win $100?

Answer:
Start by noticing the word “or” in the question. How can John win? He can win by pulling out either a 7 or a 9. His chances of doing that are higher than if he could win only by pulling out a 7. In that case, he’d only have 4 cards that would make him win $100 (because there are 4 7's in a standard deck), now he has 8 cards. To find the total probability, we need to figure out the probability of each event and then add them together.

So, what is the probability of each? Think Bottom to Top:

Since this is an "or" question, ADD the possibilities.

Total probability of choosing a 7 or a 9: 1/13 + 1/13 = 2/13.

Example 5
A fair die is tossed once. What is the probability the die will land on a 2 or a 5?

Answer:

Total probability of landing on a 2 or a 5: 1/6 + 1/6 = 2/6 = 1/3

B) Scenario 2 – “AND”: If two events have to occur together, generally an "and" is used. Take a look at this statement: "I will only be happy today if I get email and win the lottery." The "and" means that both events are expected to happen together. He had better odds with "Or".

In the case of “and,” we multiply probabilities together to get a lower overall probability.

	800score Logic In general, the odds of getting many things to happen is generally lower than just requiring one or two things to happen out of many. The odds of getting 4 foul shots in a row is smaller than making one in four. The reason for this is that probability is expressed as a fraction. When you multiply fractions you get smaller fractions as in "AND" scenarios. When you add fractions you get larger fractions as in "OR" scenarios.

Example 6
If a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the second toss the coin lands tails?

Answer:
First note the "and" in this problem. That means we require both events to occur together. If this coin does what the problem says it will do, the coin toss will look like this:

H T

The probability that the coin will land on heads, you should now understand, is 1/2. The probability that the coin will land on Tails is also 1/2.

Expect the answer to be less than the individual probabilities of either event A or event B, so less than 1/2. Since we want them to happen together, we multiply individual probabilities. 1/2 × 1/2 = 1/4.

Example 7
What is the probability that rolling two identical dice together will result in two fives?

Answer:
Note: not all “and” questions include the word “and”. They may just imply it. In this case, to get the result of two fives, the first die must be a five and the second one must be a five. We have the “and” without seeing it.

Probability first roll is 5: 1/6
Probability second roll is 5: 1/6
Probability both are 5: 1/6 × 1/6 = 1/36

Chapter 3: Independent and Dependent Events

Whenever two or more events happen at the same time, i.e., when we use the word “and,” you will have to decide if the events are independent or dependent.

Examples 6 and 7 are examples of Independent probabilities. The outcome of the first event does not affect the probability of the second. Coin tosses are independent. They cannot affect each other's probabilities; the probability of each toss is independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards are independent events ONLY if you put the cards back.

Dependent events are the opposite. The probability of the second event is affected by the first event. An example of a dependent event is drawing a card from a deck but not returning it. By not returning the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of whatever kind of card you drew. If you draw an ace of spades, then there are 1 fewer aces and 1 fewer spades. This affects our simple probability.

Note: that dependent probabilities always coincide with “and” problems, so they will always be multiplication problems.

Example 8
Two cards are pulled from a deck of 52 cards. They are pulled one after the other, and the first is not returned to the deck. What is the probability that both cards will be spades?

Answer
Since both cards must be spades, this is an “and” question. We need to calculate the individual probability of each card and then multiply.

The first card is like many of the other examples:
Bottom: 52 cards total
Top: 13 cards are spades
Probability: 13/52 = 1/4

For the second card, we have to pull it out after the first one has already been pulled. There are no longer 52 cards, but rather 51. And, assuming the first card was a spade, there are now no longer 13 spades, but only 12. So the probability for the second card is:

Bottom: 51 cards left
Top: 12 cards are spades
Probability: 12/51 = 4/17

Finally, we multiply them together:
Probability of two spades: 1/4 × 4/17 = 4/68 = 1/17

Note again that 1/17 is smaller than both 1/4 and 4/17, since it is harder to pull out two spades in a row.

Example 9
There are 6 black marbles and 4 red marbles in a jar. Two marbles are pulled out in succession without replacing them in the jar. What is the probability that both will be black?

Answer:
Notice that this is again a dependent probability situation. Once the first marble is pulled out, there are less marbles for the second pull.

First Pull:
Bottom: 10 marbles total
Top: 6 marbles are black
Probability: 6/10

Assuming the first marble was black, there will now be only 5 black marbles left, and 9 marbles left all together.

Second Pull
Bottom: 9 marbles left
Top: 5 marbles are black
Probability: 5/9

Total probability: 6/10 × 5/9 (don’t forget to reduce before you multiply) = 3/5 × 5/9 = 1/5 × 5/3 = 5/15 = 1/3

Example 10
After each throw of a red die, the face that shows is marked with a blue stripe. What is the probability that after 6 throws all faces of the die will be marked blue?


Thinking through this problem logically is going to get you the right answer. It is a dependent probability problem, because after each throw the number of sides the die can land on diminishes by 1. Remember to think bottom to top for each step.

We now have: (remember to cancel)

Example 11
If a coin is tossed twice what is the probability that it will land either heads both times or tails both times?
A)1/8
B)1/6
C)1/4
D)1/2
E)1

This question brings both “and” and “or” (independent or dependent) into the same problem. For the “or” part, we have two scenarios that could both happen (both heads or both tails). But within each option, there are two probabilities, and both must happen. (both must be heads or both must be tails). Let’s attack it systematically:

Both Heads: This means heads on the first toss and heads on the second toss. Both tosses have a probability of coming up heads, and 1/2 × 1/2 = 1/4.
Both Tails: Same as heads, but reversed. Probability = 1/4.
Both Heads or Both Tails: 1/4 + 1/4 = 2/4 or 1/2.
The correct answer is (D).

Example 12
A fair coin is flipped three times. What is the probability that heads will come up only once?

Solution
This is an “or” question, even though the “or” isn’t written. Why? Think about what it means to have heads come up only once. What are the scenarios? Let’s write them out:
H T T or T H T or T T H

There are three situations, all of which are valid, that would achieve the result we’re looking for. Again, let’s go one by one:

Probability of H T T = 1/2 × 1/2 × 1/2 = 1/8
Probability of T H T = 1/2 × 1/2 × 1/2 = 1/8
Probability of T T H = 1/2 × 1/2 × 1/2 = 1/8

Add 1/8 + 1/8 + 1/8 = 3/8
Answer: 3/8

Example 13
The first basket contains 4 blue and 5 red marbles; the second basket contains 3 blue and 4 red marbles. One marble is randomly extracted from the first basket and put into the second. After that, a marble is extracted from the second basket. What is the probability that this marble is blue?
A. 1/3
B. 15/36
C. 27/72
D. 4/9
E. 11/18

Solution
Among all possible scenarios there are two that suit us:

1. A blue marble is put into the second basket and then a blue marble is extracted from the second basket;

2. A red marble is put into the second basket and then a blue marble is extracted from the second basket.

The probability of the first scenario: probability that a blue marble is taken from the first basket 4/9 (4 blue, 5 red). The probability that a blue marble is then extracted from the second basket is now 4/8 because there are now 4 blue onesout of 8 total marbles.
4/9 × 4/8 = 16/72.

The probability of the second scenario: probability that a red marble is taken from the first basket 5/9 probability that a blue marble is then extracted from the second basket = 5/9 × 3/8 = 15/72.

The probability of EITHER first OR second scenario: 16/72 + 15/72 = 31/72.

Answer: C

Example 14
In a 1 mile race, 5 athletes compete from each of three different schools: Washington High, Duke High, and Cherry Hill High. What is the probability that a student from Cherry Hill will take first place, a student from Duke will take second place, and another student from Duke will take third place?

Solution
Multiply the Individual Probabilities
Probability that:
Cherry Hill student will take first: 5/15
Duke student takes second: 5/14
Duke student takes third: 4/13

Total Probability:

Example 15
Brian rolled two identical six-sided dice. What is the probability that the sum of the dice equaled 7?

Solution
For this problem, like the others, we need to think systematically. Let’s start by thinking about all the situations that could yield a 7:
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

As we saw in Example 7, each situation has a 1/36 chance of happening. Since any one of them would yield 7, this is an “or” question. To solve, we’ll just add up all the probabilities: 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 6/36 = 1/6
Answer: 1/6

Chapter 4: Working Backwards

Let’s go right back to the first example in this chapter:

Let’s say you have a single six-sided die. If you role it, what is the probability you will roll a 1?

If you remember, the answer was 1/6.

Now let’s put some money on this. You’re in Vegas, and you’re going to win $500 if the die lands on 1. When it does, you just won $500! Let’s call that a success. You had a 1/6 chance to achieve that success.

But you decide to let it ride! You’re convinced it can happen again. You go double or nothing, betting again on 1. This time, the die comes up on something else! You’ve failed. And since 5 out of the 6 outcomes would have caused that failure, the probability to fail is 5/6.

Now, let’s turn it around. What if someone were to offer you $500 if, when you roll one die, you rolled at least a 2? Think about that for a second. What outcome would win you $500? If you’re just thinking of 2, think again. In fact, if you rolled a 2, or a 3, or a 4, or a 5, or a 6, you would win! That’s pretty good. What are the odds of that happening? You could figure out that there are 5 ways to win out of 6, so the answer would be 5/6. The probability of success would be 5/6.

But, by using the phrase “at least,” the problem created more ways to succeed than to fail. That means more work. On the GRE, we want to avoid hard work. Try to figure out how you could fail, and then reverse it? The only way NOT to win $500 is to roll a 1. And what’s the probability that you’d roll a 1? 1/6. The probability of failure is 1/6, so the probability of success must be 5/6.

To simplify it, look at this formula:

P(success) + P(failure) = 1
or
P(success) = 1 – P(failure)

Example 15
A jar contains 10 red marbles and 6 black marbles. If three marbles are pulled from the jar one after another without replacing them back into the jar, what is the probability there will be at least one red marble among them?

Answer: 27/28
Can you see why this would be tough? There are so many ways to pull out “at least one” red marble. The red one can come first, second or third. Or, you could pull out two red marbles and one black one, in any order, or you could pull out three red marbles. Figuring out the total number of ways to do that would take a very long time.

But, think about the failure in this case. What is the one way to fail? You can fail by not pulling out any reds at all! That’s right – in this case the failure is the case where you pull out three black marbles!

So let’s figure that out. What is the probability of pulling out three black marbles? This is a dependent probability problem, as you learned before. Just multiply the individual probabilities:
6/16 × 5/15 × 4/14 = 1/28

Since the probability of the failure, i.e., the probability of all black, is 1/28, the probability of the success, i.e., the probability of at least one red, is 27/28.

Example 16
Two identical six-sided dice are rolled. What is the probability that the sum of the dice will be at least 5?

Answer: 5/6
Pay attention to the phrase “at least,” and think about what it means for this problem. “At least five” means anything five or above. If we just counted all the ways the dice could come up at least five, it would take several minutes. Rather, let’s figure out the failure, and reverse the answer. What is the opposite of “at least five”? Anything less than five, or, in other words, up to four. So we have:

(1,1) = 2 (1,3) = 4 (2,2) = 4
(1,2) = 3 (2,1) = 3 (3,1) = 4

So there are 6 ways to come up with 4 or less. As you have already seen, each one has a probability of 1/36, so together the failure has a probability of 6/36 or 1/6.
The probability of success would then be 1 – 1/6 = 5/6.

Chapter 5: A Different Method (Advanced)

NOTE: Please read this section after you have mastered the previous four sections.


Example 17
What is the probability that rolling two identical dice together will result in a 3 and a 4?

Answer: Think Bottom to Top
Bottom: We have two dice being rolled, with 36 total outcomes possible.
Top: There are two outcomes that can happen: (3,4) and (4,3).
Probability: 2/36 = 1/18

Example 18
What is the probability that rolling two identical dice together will result in the sum of the dice adding to 7?

Answer: Think Bottom to Top
Note: We saw this question above in example 14. Try solving it a different way.
Bottom: Total outcomes = 36
Top: How many of those outcomes yield a 7? Just list them:
(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

There are 6 in total.

Probability: 6/36 = 1/6

Example 19
Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, numbered 5 - 9. What is the probability that the sum of a card pulled randomly from Bowl X and a card pulled randomly from Bowl Y will equal 8?

Answer: Think Bottom to Top
Bottom: How many different pairs of cards can be pulled? There are 4 cards in Bowl X and 5 cards in Bowl Y. Using permutations: 4 × 5 = 20. So there are 20 total outcomes possible.
Top: Which pairs work for this question?
(1,7) (2,6) (3,5)
3 outcomes.
Probability: 3/20

Example 20
Bowl X has 4 cards in it, numbered 1-4. Bowl Y has 5 cards in it, numbered 5-9. What is the probability that the product of a card pulled randomly from Bowl X and a card pulled randomly from Bowl Y will be even?

Answer: Think Bottom to Top
Bottom: 20 total outcomes
Top: Which pairs will multiply to an even number?

14 total pairs.
Probability: 14/20 = 7/10

Alternative Answer
Think about success and failure. What is a success in this problem? An even outcome. How can two numbers multiply together to make an even? There are lots of ways – all we need is an even number. What if we reverse this one? Since there are only two outcomes (odd and even), it should be clear that there are less odd pairs than even ones. Odd outcomes are failures. Let’s calculate the probability of failure, and then subtract from 1 to find the success:

(1,5) (3,5)
(1,7) (3,7)
(1,9) (3,9)

Bottom: 20 total outcomes
Top: 6 odd outcomes
Probability of Odd Outcome (failure): 6/20 = 3/10
Probability of Even Outcome (success): 1 – 3/10 = 7/10

Example 21
A fair coin is flipped three times. What is the probability that heads will come up only once?

Answer: Think Bottom to Top
Note: We saw this question above in example 11.
Bottom: How many different outcomes are possible? Each time the coin is tossed, there are two outcomes, and we’re tossing it three times: 2 x 2 x 2 = 8
Top: How many outcomes have only a single instance of heads? There are only 8 possibilities, as we learned above, so there can’t be too many. Work it out:

H T T
T H T
T T H

3 total outcomes
Probability: 3/8

Chapter 6: Extra Questions

1. A jar has 10 marbles, a combination of black and white. 2 marbles are randomly chosen from the jar. If q is the probability that both will be black, is q > 1/3?

(1) Less than 1/2 of the marbles in the jar are white.
(2) The probability that 1 white marble and 1 black marble will be chosen together is 7/15.

(A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
(B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
(D) Either statement BY ITSELF is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

Solution
This is a dependent probability problem. If you want to find the probability of choosing 2 black marbles, you will need to figure out the probability that the first marble will be black and that the second marble will be black. In this case, the question wants to know if that probability is larger than 1/3.


Statement 1 tells us that less than half the marbles are white, which means that more than half the marbles are black. The best way to approach this is to systematically (but quickly) figure out what the probability of two black marbles is for each scenario. We can do it easily by drawing a chart:

As you can see, when less than half the marbles are white, the probability of choosing 2 black marbles can be higher or equal to 1/3, depending on how many black marbles there are. This is not sufficient.

Statement 2 tells us that the probability of choosing one black marble and one white marble is 7/15. This is a trap. Since the probability given is exact, it may seem that only one scenario of black marbles and white marbles will work. If you work through all the scenarios, you will see that when there are 7 black marbles and 3 white marbles, the probability of choosing one of each is 7/15. However, it would also be true in reverse: If there were 7 white marbles and 3 black marbles, the probability would also be 7/15. Therefore, this is not enough information.

Combining them does give us enough information. From statement 2 we know that there must be 7 of one color and 3 of the other, and from statement 1 we know that there must be more black than white, so we know there must be 7 black marbles and 3 white marbles.
The correct answer is (C).

2. In a class with 12 children, q of the children are girls. Two children will be randomly chosen simultaneously. What is the value of q?

(1) The probability that two girls will be chosen together is 1/11.
(2) The probability that one boy will be chosen and one girl will be chosen is 16/33.

(A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
(B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
(D) Either statement BY ITSELF is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

Solution
To answer this question, you can do the math, or you can rely on the experience you have gained thus far. Let’s work out statement 1 by thinking it through:

Statement 1: We know there are a specific number of girls (q). Since each number of girls would yield a different probability of choosing 2 girls, there must be only one specific number that would yield 1/11. So it must be enough information.

Now, statement 2 requires a little more thought. Let’s work it out by doing the math:
Statement 2: This one may seem to follow the same logic, as they are giving us a specific probability. However, this time we are asked to pick one boy and one girl. Look at the following chart to see why this isn’t enough information:

*Note: we will multiply each probability by 2 -- see below (1/12 × 11/11)/2, because we can choose a boy and a girl, or a girl and a boy, and both will yield the desired result.

As you can see, each probability is repeated for inverse combinations of boys and girls. There are two ways to get 16/33, once with 4 boys and 8 girls, and also with 4 girls and 8 boys. This is not enough information. We do not know what q is.
The correct answer is (A).

3. In a hotel with single rooms and double rooms, what is the probability that a room chosen at random will be a double room painted red?

(1) 1/6 of the rooms in the hotel are painted red.
(2) 2/3 of the hotel’s rooms are double rooms.

(A) Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
(B) Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
(D) Either statement BY ITSELF is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

Solution
Statement 1 tells us the fraction of rooms painted red, but we do not learn anything about double rooms.

Statement 2 tells us the fraction of the rooms that are double rooms, but we do not know anything about red rooms.

Putting the statements together still does not give us enough information because we do not know how many of the red rooms are double rooms. Imagine if we said there were 12 rooms. We would then know that 2 were red, and 8 were doubles. But we do not know if any of the doubles are red or not. There is simply no information connecting the two categories, so we cannot solve the probability (E) is the correct answer.

4. Two identical dice are rolled together. If the sum of the dice is 7, what is the probability that one of the numbers showing is a 4?
(A) 1/36
(B) 1/18
(C) 1/6
(D) 11/36
(E) 1/3

Solution
For this problem, we want to calculate the probability of rolling a 4 knowing that the sum of the dice is 7. We therefore do not have to take into account any other combinations of dice other than those that equal 7:

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1)

There are six pairs that equal seven, and 2 of them have a four in them. Therefore, the probability is 2/6 = 1/3.

The correct answer is (E).

5. At 3 pm, Jennifer went into labor. There is a 70% chance her baby will be born each hour that she is in labor. What is the probability that her baby will be born at 6 pm on the same day?
(A) .027
(B) .063
(C) .147
(D) .27
(E) .343

Solution
This question is an independent probability question in disguise. The probability that her baby will be born each hour does not change. Each hour, there is a 70% chance the baby will be born, which means there is a 30% chance the baby will not be born.

Therefore, we can see that from 3 pm – 4 pm, the baby is not born and the probability of that happening is 30%, from 4 pm – 5 pm, the probability is 30%, and from 5 pm – 6 pm, when the baby is born, the probability is 70%. Since the baby must not be born in the first hour, and must not be born in the second hour, and must be born in the third hour, the probability is (0.3)(0.3)(0.7) = .063
The correct answer is (B).

6. A fair coin is to be flipped four times. What is the probability that the coin will land on the same side on all four flips?
(A) 1/32
(B) 1/16
(C) 1/8
(D) 1/4
(E) 1/2

Solution
The probability the coin will land on the same side is the probability that it will land on all heads or all tails. All heads looks like: H H H H and all tails looks like: T T T T.

The probability of each is 1/2 × 1/2 × 1/2 × 1/2 = 1/16. Since either could happen, we will add the two probabilities together: 1/16 + 1/16 = 2/16 = 1/8.
The correct answer is (C).

The probability, combinations, permutations chapter is complete.