Permutatation Comb Notes

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Chapter 1:

ntroduction to

Permutations

Print out chapter

Permutation questions are about taking a group of objects and totaling how many ways we can arrange them

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?


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Because there are three digits, there are three places to fill: __ __ __


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 potentialoptions. 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_.


_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

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 choice

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.

If want to fit 7 books into 3 spaces, and want to know the possible permutations, you would assign 7 ton (sinc

is the objects which you are choosing from), and assign 3 to r(since it is the number of spaces that n can fit i

Therefore the formula would read:


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In this formula, n stands for the distinct objects which you are choosing from, r stands for the number of space

which those r objects can fit into, and Pstands for Permutation, and is not an arithmetic part of the equation.

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

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 GMAT isn't interested in your

perfect memorization of the permutation formula, the GMAT

wants you to have a good intuitive sense of how

permutations work.

III. Replacement or Non-Replacement

The GMAT will test your ability to distinguish problems with or without replacement. So you should be very go

at identifying which one it is.

Replacement

Potential outcomes that are replaced or constant

Non-Replacement

(Potential outcomes that decrease with each selectio


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Rolling a set of 6-sided diceHow many possibilities in 10 dice rolls

Set of 6-sided dice on paintHow many possibilities in 10 dice rolls on wet red

paint. If a painted side shows, then the dice must bre-rolled.

Pulling from a bag of marbles and then puttingthem back

A student pulls a marble from a bag andthenputs it back in the bag. If he does this four timesand the bag contains 9 green marbles and 9 blue

marbles, how many different possibilities arethere?

Pulling from a bag of marblesA student pulls 4 marbles from a bag that contains green marbles and 9 blue marbles, how many differe

possibilities are there?

Combination of safe or lockHow many possibilities for a combination lock with

40 numbers that requires 3 selections.

Sum of the combination of a safe where numbecan't be repeated

How many possibilities for a combination lock with 4numbers that requires 3 selections and cannot hav

the same number twice.

Pulling from a repeatedly shuffled deck ofcards.

4 cards are pulled from a deck where the dealershuffles the deck and replaces the cardimmediately after each card is pulled.

Pulling from a deck of cards.4 cards are pulled from a deck of cards.

800score Tip:

Keep in mind that the GMAT's effectiveness as a test is

a function of it's ability not to be "beaten" by standardtest preparation.

So it is quite logical for the GMAT to set up trick questions

specifically to penalizeover-preparation. Our test prep

gnomes who have recently taken the GMAT 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
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hapter 2: Problem

ariations

Print out chapter

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 app

again at a different place on the bench!


There are 5 seats on the bench, so there are 5 places:

__ __ __ __ __


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_


_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 woul

multiply out 5 4 3 2 1 which equals 120.
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Example 2

There are six meal options in the cafeteria of a certain school. Assuming that a different meal must be e

each day, and each different type of meal must be eaten once before any type of meal can be eaten 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 situa

but because we are told that a student cannot eat two of the same types of meal before they go throug

the meals first.


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 4 three options.

_6_ _5_ _4_ _3_


_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 fromr stands for the 4 days or spaces, that you can fit those meals into.

6P4 = 6 5 4 3 = 360

Example 3A boy is given a hat with 6 tiles, each numbered with a different digit 1-6. If he will pull out 3 tiles, one ti

a time, and lay them in the order he pulls them out, how many different 3-digit numbers could he possib

create?


3, one for each tile: __ __ __


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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_


_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 lon

complete factorial.

We can also use the formula, nPr = 6P3 where n stands for the 6 tiles that there are to choose from 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, lets 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 secon

fourth, and sixth cages?


There are six cages for the animals, so there are six places:

__ __ __ __ __ __


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 our the potential number of options

for each cage (remember, there can be no repeating values):_3_ _3_ _2_ _2_ _1_ _1_


_3_ _3_ _2_ _2_ _1_ _1_ = 36

We can also use the formula:


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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

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-dig

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

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. Ther

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:


4, for each digit of the ID number: __ __ __ __

2. Figure out how many objects potentially can go into each placeThe 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_


_9_ _9_ _8_ _7_

(Dont multiply yet because the GMAT 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 othe


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You can now see why we did not multiply originally. Most of the numbers cancel out! The result is simp1/6, or 1:6.

GMAT Combinations

hapter 1: Example of

ombinations

Print out chapter

When Order Is Not Relevant

Combinations problems are very similar to permutations problems. The key distinction is

that placement/order isnt relevant for combinations, but placement/orderis 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

Lets 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 arent counted and we

roll again. The result is that there are no doubles. How many permutations are possible?


Because there are two dice, there are two places to fill: __ __


Since you cannot get the same result (doubles) on the second die there

are only 5 possibilities for the second roll.

_6_ _5_


_6_ _5_ = 30
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(Red,Blue) (Red,Blue) (Red,Blue) (Red,Blue) (Red,Blue) (Red,Blue)

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

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

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

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

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

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

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 arent counted. So

the result is 30 different permutations. Just count all of the outcomes above to get 30.

II. Example of Combinations

Now lets do the combinations versions of the above question.

Well try the game again, but instead of using a reddie 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

isnt 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 cant tell a (2,5) and a

(5,2) apart, so order doesnt matter. Once order or position doesnt matter, this becomes

a combinations question and not a permutations question.(Red,Red) (Red,Red) (Red,Red) (Red,Red) (Red,Red) (Red,Red)

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

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

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

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

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

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

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


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are overlaps/double counted in an earlier column we will strike through.

(Red,Red) (Red,Red) (Red,Red) (Red,Red) (Red,Red) (Red,Red)

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

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

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

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

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

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

If we count the final results, we get only 15 total combinations (in bold red). Obviously

you cant 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, rstands for the spaces into which n objects can fit, and

the C stands for stating that this is a Combinations problem. The only difference betweenis the addition ofron the bottom.

Lets apply this to our dice problem. We have two dice, so there are two spaces.

Following the combinations formula:

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.


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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?

Chapter 2:

Permutation or

Combination?

Print out chapter

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

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 GMAT 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 You will soon see the difference between permutations and combinations.
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Permutations

(Things that require order, positioning).

Combinations

(Groups without order, rank position).

Committees with position titles(it matters if one person is the president and theother a vice president, or the other way around).

Generic committees(There is no rank so the committee is just one larg

union).

Layer of colors used in a painting or project(It matters if blue goes first and then red, or the

other way around).

Different colors used in a painting or project(If the colors are not layered, then they are all on t

canvas just the same).

Combination of safe or lock(1435 is different from 1345).

Sum of the combination of a safe or lock(1 + 4 + 3 + 5 is the same as 1 + 3 + 4 + 5).

Different outfits to wear from a full wardrobe.(If you wear a red shirt and blue pants that isdifferent from a blue shirt and red pants).

Different clothes to take on vacation from alarger selection.(If a red shirt, blue shirt, red pants, and blue pant

are all put in a suitcase it does not matter which ois first or last).

Order of winning 1st, 2nd, 3rd in a race (alsoGold, Silver, Bronze).

(As in the Olympics, if one person wins the goldand another the silver that is different than the

other way around).

Different prizes to take home from a largerselection.

(If you bring home a stuffed animal and a hat youcan only bring both home one way).

Arranging people in a row

(Steve, Maria and John in a row are different thanMaria, Steve, and John).

Putting people in a class

(If Steve, Maria and John are all in a class togethethere is no first or last).

Saying Hello(You can say hello to a friend and he/she can say

hello back, and they are two different events).

Handshakes(If you and a friend shake hands you are both doithe same act at the same time, so there can be n

order).

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 arrangementspossible 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?


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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

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, t

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, Sa

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 differe

pizzas could she order?

Solution

As weve already seen, the order of toppings on a pizza does not matter (does it matter if pepperoni is on to

mushrooms?), so this is a combinations question. We want to eliminate all the redundant pizzas and only c

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:

Dont 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.


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

There are 4 spaces, so thats 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. Lets expand

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 colo

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_


There are five spaces, so divide by 5!


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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

are choosing only 5 of the 15 colors:

Chapter 3: Problem Variations

Chapter 3:

Groups/Pairings

Print out chapter

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 Sams Pizza Parlor, there are 8 meats, 7 vegetables, and 5 cheeses to choose from. Jonathan would lik

make a pizza with 4 meats, 3 vegetables, and 3 cheeses. How many different pizzas could he order?

SolutionThis problem is different from example 8. In example 8, Maria wanted to order a pizza mixing all the topping

together. In this case, Jonathan wants his pizza a specific way. But there are many ways he could have me

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

the correct answer.
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MEATS VEGETABLES CHEESES

_8_ _7_ _6_ _5_ _7_ _6_ _5_ _5_ _4_ _3_


Meats Vegetables Cheeses

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


nCr nCr nCr = 8C4 7C3 5C3

For this problem, we need to set up three separate combinations equations and then multiply them all toget

to get the final answer. Our first equation represents the 8 meats Jonathan can put into the four spots on th

pizza. The second represents the 7 vegetables for the 3 spaces available, and the last equation represents5 cheeses that can go into the 3 available spaces on the pizza. Now lets expand out the equation:


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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?


Green Red Purple

_6_ _5_ x _4_ _8_ _7_ _6_ _5_



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3. Multiply the answers together for the final answer.

20 56 5 = 5600


nCr nCr nCr = 6C3 8C3 5C1

Once again, we need to set up three separate combinations equations and then multiply them all together t

get the final answer. Our first equation represents the 6 people we can choose for the 3 spots from the Gre

group. The second represents the 8 people from the Red group available for the 3 spots on the committee,

the last equation represents the 5 people from Purple that can go into the 1 space on the committee.

Now lets 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


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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 mor

than once, how many handshakes take place?

Lets approach this with a discussion. How many spaces should there be? Many people want to put six. Bu

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 persons

hand. So following our approach for combinations:

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

_6_ _5_


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 w

are actually shaking hands.

Example 13

7 basketball teams with five players each are at a tournament. If each player shakes hands with every othe

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

handshake? Two. So there must be two spaces. But in this case we have to use the logic of the problem to


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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 whos hands he can shake! That will be

reflected in the combination. So lets approach it using the method and this logic:

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


There are two spaces, so divide by 2! And dont forget to cancel!

GMAT Probability: Simple Probability

Chapter 1: Simple

Probability

Print out chapter

Probability is often seen as the real evil of the GMAT. But the truth is that by mastering one simple fraction,

can make probability approachable and even easy.

Lets 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

6 possible outcomes, and the five is one of them. Therefore, there is a 1 in 6 probability that the five will com

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:
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Probability =what we want

all of whats possible.

Apply that to the above example:

Bottom: 6 outcomes are possible

Top: 1 outcome we want

Probability: 1/6

Lets change the above example just a bit to see this in action. Now, we still have one die, but we want to k

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:


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 willproduce a 4?

Answer: Think Bottom to Top


Top: 4 cards give us a 4


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?



Top: 26 cards are black


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?



Top: 1 Ace of Spades

Probability: 1/52


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Probability of Multiple Events

apter 2: Probability ofltiple Events

dvanced)

Print out chapt

The only vocabulary youll 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 pulli

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

His chances of doing that are higher than if he could win only by pulling out a 7. In that case, hed only

cards that would make him win $100 (because there are 4 7's in a standard deck), now he has 8 cards

find the total probability, we need to figure out the probability of each event and then add the together.

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

Probability of Choosing a 7: Probability of Choosing a 9:

Bottom: 52 outcomes are possible Bottom: 52 outcomes are possible

Top: 4 cards are 7's Top: 4 cards are 9's
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Probability: 4/52 = 1/13 Probability: 4/52 = 1/13

Since this is a "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:

Probability of landing on a 2: Probability of landing on a 5:

Bottom: 6 outcomes are possible Bottom: 6 outcomes are possible

Top: There is only one "two" Top: There is only one "five"

Probability: 1/6 Probability: 1/6

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

statement: "I will only be happy today if I get email and win the lottery." The "and" means that both eve

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 LogicIn 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 arow 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 largerfractions as in "OR" scenarios.


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Example 6

If a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the se

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 do

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

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

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

result of two fives, the first die must be a five and the second one must be a five. We have the a

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

apter 3: Independent

d Dependent Events

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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 eachother'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.

Note: 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


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


Finally, we multiply them together:


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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.

For the 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 blackProbability: 5/9

Total probability: 6/10 5/9 (dont 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 red?


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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)


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Example 10

If a coin is tossed twice what is the probability that it will land either heads both times or tails bothtimes?

A)1/8

B)1/6

C)1/4

D)1/2

E)1

Answer: D

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 orboth tails). But within each

option, there are two probabilities, and both must happen. (both must be heads or both must betails). Lets 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.

Example 11

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 isnt written. Why? Think about what it means to have

heads come up only once. What are the scenarios? Lets write them out:

H T T orT H T orT T H


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There are three situations, all of which are valid, that would achieve the result were looking for.

Again, lets 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 = 1/8Probability 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 12

The first jar 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. 31/72

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 secondbasket.

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


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Example 13

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 14

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. Lets 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, well 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

Chapter 4: Working

Backwards

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Lets go right back to the first example in this chapter:

Lets 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 lets put some money on this. Youre in Vegas, and youre going to win $500 if the die lands on 1.

When it does, you just won $500! Lets call that a success. You had a 1/6 chance to achieve that succ

But you decide to let it ride! Youre convinced it can happen again. You go double or nothing, betting ag

on 1. This time, the die comes up on something else! Youve failed.And since 5 out of the 6 outcomes

would have caused that failure, the probability to fail is 5/6.

Now, lets turn it around. What if someone were to offer you $500 if, when you roll one die, you rolled a

least a 2? Think about that for a second. What outcome would win you $500? If youre just thinking of

think again. In fact, if you rolled a 2, or a 3, or a 4, or a 5, or a 6, you would win! Thats pretty good. Wh

are the odds of that happening? You could figure out that there are 5 ways to win out of 6, so the answ

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 GMAT, 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 whats the probability that youd roll a 1? 1

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) = 1or

P(success) = 1 P(failure)

Example 15

A jar contains 10 red marbles and 6 black marbles. If three marbles are pulled frothe 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. Tred one can come first, second or third. Or, you could pull out two red marbles and one black one, in an

order, or you could pull out three red marbles. Figuring out the total number of ways to do that would ta

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

at all! Thats right in this case the failure is the case where you pull out three black marbles!


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So lets 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 succi.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 thedice 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 me

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, lets figure out the failure, and reverse the answer. What is the opposite o

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 o

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

Chapter 5: A Different

Method (Advanced)

Print out chapter

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

Example 17

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


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

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

Answer: Think Bottom to TopNote: We saw this question above in example 14. Try solving it a different wBottom: Total outcomes = 36Top: 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.


Example 19Bowl 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 pulledrandomly from Bowl X and a card pulled randomly from Bowl Y will equ8?


Bottom:How many different pairs of cards can be pulled? There are 4 cards in Bowl X and 5 cards 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 20Bowl 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 pulledrandomly from Bowl X and a card pulled randomly from Bowl Y will be


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even?


Bottom: 20 total outcomesTop: Which pairs will multiply to an even number?

(1,6) (2,8) (4,6)

(1,8) (2,9) (4,7)

(2,5) (3,6) (4,8)

(2,6) (3,8) (4,8)

(2,7) (4,5)

14 total pairs.


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. W

if we reverse this one? Since there are only two outcomes (odd and even), it should be clear that there

less odd pairs than even ones. Odd outcomes are failures. Lets 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 outcomesTop: 6 odd outcomes

Probability of Odd Outcome (failure): 6/20 = 3/10Probability of Even Outcome (success): 1 3/10 = 7/10

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


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Answer: Think Bottom to TopNote: We saw this question above in example 11.Bottom:How many different outcomes are possible? Each time the coin is tossed, there are twooutcomes, and were 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 welearned above, so there cant be too many. Work it out:

H T TT H TT T H

3 total outcomes

Probability: 3/8

Chapter 6: Extra Questions

Chapter 6: Extra

Questions

Print out chapter

1. A jar has 10 marbles, either black or 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 tha

further information would be needed to answer the question.

SolutionThis is a dependent probability problem. If you want to find the probability of choosing 2 black marbles,

will need to figure out the probability that the first marble will be black and that the second marble will b

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 mar

are black. The best way to approach this is to systematically (but quickly) figure out what the probability
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two black marbles is for each scenario. We can do it easily by drawing a chart:

Black Marbles White Marbles P(2 Black)

6 4 6/10 5/9 = 1/3

7 3 7/10 6/9 = 21/45

8 2 8/10 7/9 = 28/45

9 1 9/10 8/9 = 4/5

10 0 10/10 9/9 = 1

As you can see, when less than half the marbles are white, the probability of choosing 2 black marblesbe 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 7/15This is a trap. Since the probability given is exact, it may seem that only one scenario of black marbles

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 oone color and 3 of the other, and from statement 1 we know that there must be more black than white,

we know there must be 7 black marbles and 3 white marbles.

Answer: C

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

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.



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 tha


Solution

To answer this question, you can do the math, or you can rely on the experience you have gained thus

Lets 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


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different probability of choosing 2 girls, there must be only one specific number that would yield 1/11. S

must be enough information.

Now, statement 2 requires a little more thought. Lets 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 th

isnt enough information:

Boys Girls P(1 boy and 1 girl)*

1 11 1/12 11/11 = 1/6

2 10 2/12 10/11 = 10/33

3 9 3/12 9/11 = 9/22

4 8 4/12 8/11 = 16/33

5 7 5/12 7/11 = 35/66

6 6 6/12 6/11 = 6/11

7 5 7/12 5/11 = 35/66

8 4 8/12 4/11 = 16/33

9 3 9/12 3/11 = 9/22

10 2 10/12 2/11 = 10/33

11 1 11/12 1/11 = 1/6

*Note: we will multiply each probability by 2, because we can choose a boy and a girl, or a girl and a bo

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 enoughinformation. We do not know what q is.

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

be a double room painted red?

1) 1/6 of the rooms in the hotel are painted red.


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2) 2/3 of the hotels rooms are double rooms.



C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even thoughNEITHER 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 tha


Solution

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

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

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).

4. Two identical dice are rolled together. If the sum of the dice is 7, what is the probability that one of th

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 equ

(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/

1/3.

Answer: E

5. At 3 pm, Jennifer went into labor. There is a 70% chance her baby will be born each hour that she is

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


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Solution

This question is an independent probability question in disguise. The probability that her baby will be bo

each hour does not change. Each hour, there is a 70% chance the baby will be born, which means the

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 happenin

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 secon

hour, and must be born in the third hour, the probability is (0.3)(0.3)(0.7) = .063.4.

Answer: B

6. A fair coin is to be flipped four times. What is the probability that the coin will land on the same side o

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 ta

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.

Answer: C

The probability, combinations, permutations chapter is complete.

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Permutatation Comb Notes