Permutation Combination Probability

The Fundamental Counting PrincipleIn this tutorial we will be going over the Fundamental Counting Principle. It will allow us to count the number of ways a task can occur given a series of events. Basically you multiply the number of possibilities each event of the task can occur. It is like multiplying the dimensions of it. I think you are ready to count away.

Basic Counting PrincipleSuppose that a task involves a sequence of k choices. Let be the number of ways the first stage or event can

occur and be the number of ways the second stage or event can occur after the first stage has occurred.

Continuing in this way, let be the number of ways the kth stage or event can occur after the first k - 1 stages or events have occurred. Then the total number of different ways the task can occur is:

Example 1: A deli has a lunch special which consists of a sandwich, soup, dessert and drink for $4.99. They offer the following choices:

Sandwich: chicken salad, ham, and tuna, and roast beef

Soup: tomato, chicken noodle, vegetable

Dessert: cookie and pie

Drink: tea, coffee, coke, diet coke and sprite

How many lunch specials are there?

Let’s use the basic counting principle:

There are 4 stages or events: choosing a sandwich, choosing a soup, choosing a dessert and choosing a drink.

There are 4 choices for the sandwich, 3 choices for the soup, 2 choices for the dessert and 5 choices for the drink.

Putting that all together we get:

Sand. Soup Dessert Drink # of lunch specials

4 x 3 x 2 x 5 = 120

So there are 120 lunch specials possible.

Example 2: You are taking a test that has five True/False questions. If you answer each question with True or False and leave none of them blank, in how many ways can you answer the whole test?

Let’s use the basic counting principle: There are 5 stages or events: question 1, question 2, question 3, question 4, and question 5.

There are 2 choices for each question. Putting that all together we get:

quest. 1

quest. 2

quest. 3

quest. 4

quest. 5

# of ways to answer test

2 x 2 x 2 x 2 x 2 = 32

So there are 32 different ways to answer the whole test.

Example 3: A company places a 6-symbol code on each unit of product. The code consists of 4 digits, the first of which is the number 5, followed by 2 letters, the first of which is NOT a vowel. How many different codes are possible?

Let’s use the basic counting principle: There are 6 stages or events: digit 1, digit 2, digit 3, digit 4, letter 1, and letter 2.

In general there are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The first digit is limited to being the number 5, so there is only one possibility for that one. There are no restriction on digits 2 - 4, so each one of those has 10 possibilities.

In general, there are 26 letters in the alphabet. The first letter, cannot be a vowel (a, e, i, o, u), so that means there are 21 possible letters that could go there. The second letter has no restriction, so there are 26 possibilities for that one.


digit 1 digit 2 digit 3 digit 4 letter 1 letter 2 # of codes1 x 10 x 10 x 10 x 21 x 26 = 546000

So there are 546000 different 6-symbol codes possible.

These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.

Practice Problems 1a - 1c:

Solve using the counting principle.

1a. One quarter, one dime and one six-sided die are tossed. How many results are possible?

1b. Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?

1c. Six students in a speech class all have to give there speech on the same day. One of the students insists on being first. If this student’s request is granted, how many different ways are there to schedule the speeches?

Permutations

In this tutorial we will be going over permutations. Permutations are an off shoot of the Fundamental Counting Principle. Permutations specifically count the number of ways a task can be arranged or ordered. I think you are ready to go off into the wonderful world of permutations, have fun!

Factorial !

The factorial symbol is the exclamation point: ! So if I wanted to write 7 factorial it would be written as 7!.

In general, n! = n(n - 1)(n - 2)(n - 3)...(1)

Most, (if not all), of you will have a factorial key on your calculator. It looks like this: !

If you have a graphing calculator, it will be hidden under the MATH menu screen and then select your Probability screen - there you should find !

Some calculators don’t have one, so I will show you how to simplify the problems in case you don’t have that key on your calculator.

0! Has a special definition attached with it. 0! = 1

Example 1: Find 7!

If you have a ! key on your calculator you simple press 7 and then ! and in some cases you may have to also press your enter or = key. If you don’t have this key you will have to enter the definition in as follows:

7! = (7)(6)(5)(4)(3)(2)(1) = 5040

Either way 7! = 5040.

We needed to know about factorial because it is used the formula for permutation, which is our next topic

Permutation : An ORDER of arrangements of r objects, without repetition, selected from n distinct objects is called a permutation of n objects taken r at a time, and is denoted as

In other words, when you need to count the number of ways you can arrange items where ORDER is important, then you can use permutation to count.

For example, you may want to know how many ways to pick a 1st, 2nd, and 3rd place winner from 10 contestants. Since you are arranging them in order, you could use a permutation to do this. Or if you were wanting to know how many ways your committee could pick a president, a vice president, a secretary, and a treasurer, you could use permutations.

Example 2: In how many ways can 8 CD’s be arranged on a shelf?

Since we are arranging these CD’s that means order is important. So we can use permutations to help us out here. First we need to find n and r : n is the number of CD's we have to choose from. What do you think n is in this problem?

If you said n is 8 you are correct!!! There are 8 CD's in this problem.

r is the number of CD’s we are using at a time. What do you think r is?

If you said r is 8, pat yourself on the back!! We are arranging all 8 CD's on the shelf. Putting this into the permutation formula we get

*n = 8, r = 8

*0! = 1

*Expand out 8!

If you have a factorial key, you can put it in as 8! divided by 0! and then press enter or =.

If you don't have a factorial key, you can simplify it as shown above and then enter it in. It is probably best to simplify it first, because in some cases the numbers can get rather large, and it would be cumbersome to multiply all those numbers one by one.

Wow, this means there are 40320 different ways to arrange those 8 CD’s, that’s a lot.

Example 3: If a softball league has 10 teams, how many different end of the season rankings are possible? (Assume no ties)

First we need to find n and r :

n is the number of teams we have to choose from. What do you think n is in this problem?

If you said n is 10 you are correct!!! There are 10 teams in this problem.

r is the number of teams we are ranking at a time. What do you think r is?

If you said r is 10, pat yourself on the back!! We are ranking all 10 teams. Putting this into the permutation formula we get:

*n = 10, r = 10

*0! = 1

*Expand out 10!



Wow, this means there are 3,628,800 different ways to rank those 10 teams, that’s a lot.

Example 4: In how many ways can a sorority of 20 members select a president, vice president and treasury, assuming that the same person cannot hold more than one office.

Since we are choosing offices, which is a way to rank members, that means order is important. So we can use permutations to help us out here.


n is the number of members we have to choose from. What do you think n is in this problem?

If you said n is 20 you are correct!!! There are 20 members in this problem.

r is the number of members we are selecting for offices at a time. What do you think r is?

If you said r is 3, pat yourself on the back!! There are 3 offices. Putting this into the permutation formula we get:

*n = 20, r = 3

*Expand 20! until it gets to 17! ( which is the ! in den)

*Cancel out 17!'s



Wow, this means there are 6840 different ways to select the three officers, that’s a lot.

Example 5: How many different arrangements can be made using two of the letters of the word TEXAS if no letter is to be used more than once?


n is the number of letters we have to choose from. What do you think n is in this problem?

If you said n is 5 you are correct!!! There are 5 letters in TEXAS. r is the number of letters we are using at a time. What do you think r is?

If you said r is 2, pat yourself on the back!! We are using 2 letters at a time. Putting this into the permutation formula we get:

*n = 5, r = 2


*Cancel out 3!'s



This means there are 20 different 2 letter arrangements.

Practice Problems 1a - 1c: 1a. A company issues a questionnaire whereby each employee must rank the 5 items with which he or she is most satisfied. The items are wages, work environment, vacation time, job security, supervisors, health insurance, break time, and retirement plan.

The ranking is to be indicated by the numbers 1, 2, 3, 4 and 5, where 1 indicates the item involving the greatest satisfaction and 5 the least. In how many ways can an employee answer this questionnaire?

1b. A key pad lock has 10 different digits, and a sequence of 5 different digits must be selected for the lock to open. How many key pad combinations are possible?

1c. In how many ways can 7 books be arranged on a shelf?

Combinations

In this tutorial we will be going over combinations. When you need to count the number of groupings, without regard to order, then combinations are the way to go. Recall that permutations specifically count the number of ways a task can be arranged or ordered. That is the difference between the two, permutations is with regard to order and combinations is without regard to order. Let's see what you can do with these combinations.

Combination: An arrangement of r objects, WITHOUT regard to ORDER and without repetition, selected from n distinct objects is called a combination of n objects taken r at a time. The number of such combinations is denoted by

The difference between combinations and permutations is in combinations you are counting groups (order is not important) and in permutations you are counting different ways to arrange items with regard to order.

The n and the r mean the same thing in both the permutation and combinations, but the formula differs. Note that the combination has an extra r! in its denominator.

Example 1: In a conference of 9 schools, how many intraconference football games are played during the season if the teams all play each other exactly once?

When the teams play each other, order does not matter, we are counting match ups. For each game there is a group of two teams playing. So we can use combinations to help us out here.

Note that if we were putting these teams in any kind of order, then we would need to use permutations to solve the problem.

But in this case, order does not matter, so we are going to use combinations.


If n is the number of teams we have to choose from, what do you think n is in this problem?

If you said n = 9 you are correct!!! There are 9 teams in the conference.

If r is the number of teams we are using at a time, what do you think r is?

If you said r = 2, pat yourself on the back!! 2 teams play per game. Let’s put those values into the combination formula and see what we get:

*n = 9, r = 2

*Eval. inside ( )

*Expand 9! until it gets to 7! which is the larger ! in the den.

*Cancel out 7!'s

If you have a factorial key, you can put it in as 9!, divided by 7!, divided by 2! and then press enter or =.


Wow, this means there are 36 different games in the conference.

Example 2: You are going to draw 4 cards from a standard deck of 52 cards. How many different 4 card hands are possible?

This would be a combination problem, because a hand would be a group of cards without regard to order.

Note that if we were putting these cards in any kind of order, then we would need to use permutations to solve the problem.



If n is the number of cards we have to choose from, what do you think n is in this problem? If you said n = 52 you are correct!!! There are 52 cards in a deck of cards. If r is the number of cards we are using at a time, what do you think r is?

If you said r = 4, pat yourself on the back!! We want 4 card hands. Let’s put those values into the combination formula and see what we get:

*n = 52, r = 4

*Eval. inside ( )


*Cancel out 48!'s



Wow, this means there are 270,725 different 4 card hands.

Example 3: 3 marbles are drawn at random from a bag containing 3 red and 5 white marbles. Answer the following questions (a - d):

3a. How many different draws are there?

This would be a combination problem, because a draw would be a group of marbles without regard to order. It is like grabbing a handful of marbles and looking at them.

Note that there are no special conditions placed on the marbles that we draw, so this is a straight forward combination problem.

Note that if we were putting these marbles in any kind of order, then we would need to use permutations to solve the problem.


First we need to find n and r:

If n is the number of marbles we have to choose from, what do you think n is in this problem?

If you said n = 8 you are correct!!! There are 3 red and 5 white marbles for a total of 8 marbles.

If r is the number of marbles we are drawing at a time, what do you think r is?

If you said r = 3, pat yourself on the back!! 3 marbles are drawn at a time. Let’s put those values into the combination formula and see what we get:

*n = 8, r = 3

*Eval. inside ( )


*Cancel out 5!'s



Wow, this means there are 56 different draws.

3b. How many different draws would contain only red marbles?


In part a above , we looked at all possible draws. From that list we only want the ones that contain only red. Let’s see what the draw looks like: we would have to have 3 red marbles to meet this condition:

3 RED


If n is the number of RED marbles we have to choose from, what do you think n is in this problem?

If you said n = 3 you are correct!!! There are a total of 3 red marbles.

If r is the number of RED marbles we are drawing at a time, what do you think r is?

If you said r = 3, pat yourself on the back!! 3 RED marbles are drawn at a time. Let’s put those values into the combination formula and see what we get:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut57_comb.htm#part3a

*n = 3, r = 3

*Eval. inside ( )

*Cancel out 3!'s



This means there is only 1 draw out of the 56 found in part a that would contain 3 RED marbles.

3c. How many different draws would contain 1 red and 2 white marbles?


In part a above , we looked at all possible draws. From that list we only want the ones that contain 1 RED and 2 WHITE marbles. Let’s see what the draw looks like: we would have to have 1 red and 2 white marbles to meet this condition:

1 RED 2 WHITE


Together that would make up 1 draw. We are going to have to use the counting principle to help us with this one. If you need a review on the Fundamental Counting Principle, feel free to got to Tutorial 55: The Fundamental Counting Principle.

Note how 1 draw is split into two parts - red and white. We can not combine them together because we need a particular number of each one. So we will figure out how many ways to get 1 RED and how many ways to get 2 WHITE, and using the counting principle, we will multiply these numbers together.

1 RED:


If you said n = 3 you are correct!!! There are a total of 3 RED marbles.


If you said r = 1, pat yourself on the back!! 1 RED marble is drawn at a time.

2 WHITE:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut55_count.htm



If n is the number of WHITE marbles we have to choose from, what do you think n is in this problem?

If you said n = 5 you are correct!!! There are a total of 5 WHITE marbles.

If r is the number of WHITE marbles we are drawing at a time, what do you think r is?

If you said r = 2, pat yourself on the back!! 2 WHITE marbles are drawn at a time. Let’s put those values into the combination formula and see what we get:

*RED: n = 3, r = 3 *WHITE: n = 5, r = 2

*Eval. inside ( )

*Expand 3! until it gets to 2! *Expand 5! until it gets to 3!

*Cancel out 2!'s and 3!'s

If you have a factorial key, you can put it in as 3!, times 5!, divided by 2!, divided by 1!, divided by 3!, divided by 2! and then press enter or =.


This means there are 30 draws that would contain 1 RED and 2 WHITE marbles.

3d. How many different draws would contain exactly 2 red marbles?


In part a above , we looked at all possible draws. From that list we only want the ones that contain 2 RED and 1 WHITE marbles. Remember that we need a total of 3 marbles in the draw. Since we have to have 2 red, that leaves us needing 1 white to complete the draw of 3. Let’s see what the draw looks like: we would have to have 2 red and 1 white marbles to meet this condition:

2 RED 1 WHITE






Note how 1 draw is split into two parts - red and white. We can not combine them together because we need a particular number of each one. So we will figure out how many ways to get 2 RED and how many ways to get 1 WHITE, and using the counting principle, we will multiply these numbers together.

2 RED:


If you said n = 3 you are correct!!! There are a total of 3 RED marbles.


If you said r = 2, pat yourself on the back!! 2 RED marble is drawn at a time.

1 WHITE:

If n is the number of WHITE marbles we have to choose from, what do you think nis in this problem?

If you said n = 5 you are correct!!! There are a total of 5 WHITE marbles.

If r is the number of WHITE marbles we are drawing at a time, what do you think r is?

If you said r = 1, pat yourself on the back!! 1 WHITE marble are drawn at a time. Let’s put those values into the combination formula and see what we get:

*RED: n = 3, r = 2 *WHITE: n = 5, r = 1

*Eval. inside ( )




This means there are 15 draws that would contain 2 RED and 1 WHITE marbles.

Practice Problems

Practice Problems 1a - 1b:

A teacher has 15 students and 5 are to be chosen to give demonstrations. How many different ways can the teacher choose the demonstrators given the following conditions.

1a. The order of the demonstrators is important?

1b. The order of the demonstrators is not important?

Practice Problems 2a - 2c:

8 students names will be drawn at random from a hat containing 14 freshmen names, 15 sophomore names, 8 junior names, and 10 senior names.

2a. How many different draws of 8 names are there overall?

2b. How many different draws of 8 names would contain only juniors?

2c. How many different draws of 8 names would contain exactly 4 juniors and 4 seniors?

Probability

In this tutorial we will be going over probability. This will include learning some of the terminology that goes with the different aspects of probability. Things like experiment, sample space and event to name a few. A lot of times people associate probability with gambling, like playing cards and lotto. It can be used to find out your chances of winning :-) or losing :-( a game of chance. But it can also be used in other areas like research and advertising. A doctor might want to weigh the risks of putting a patient on a new drug and may look at the probability of their chances of success on the drug. A company may take a survey of people on certain products to see what is the probability that they will pick their product. So probability can be fun and games, but it can also be useful in a professional setting. I think you are ready to venture into the world of probability.

Experiment

An experiment is an act for which the outcome is uncertain.

Examples of experiments are rolling a die, tossing a coin, surveying a group of people on their favorite soft drink, etc..

Sample Space

A sample space S for an experiment is the set of all possible outcomes of the experiment such that each outcome corresponds to exactly one element in S. The elements of S are called sample points. If there is a finite number of sample points, that number is denoted n(S), and S is said to be a finite sample space.

For example, if our experiment is rolling a single die, the sample space would be S = {1, 2, 3, 4, 5, 6}. If our experiment is tossing a single coin, our sample space would be S = {Heads, Tails}. If our experiment is surveying a group of people on their favorite soft drink, our sample space would be all of the soft drinks on the survey.

Event

Any subset E of a sample space for an experiment is called an event for that experiment.

For example, if our experiment is rolling a single die, an event E could be rolling an even number, thus E = {2, 4, 6}. If our experiment is tossing a single coin, an event E could be tossing a Tail, where E = {Tails}. If our experiment is surveying a group of people on their favorite soft drink, an event E could be picking a diet soft drink.

Empirical Probability

Finding the probability of an empirical event is specifically based on direct observations or experiences.

For example, a survey may have been taken by a group of people. If the data collected is used to find the probability of an event tied to the survey, it would be an empirical probability. Or if a scientist did research on a topic and recorded the outcome and the data from this is used to find the probability of an event tied to the research, it would also be an empirical probability.

Empirical Probability Formula

P(E) represents the probability that an event, E, will occur.

The numerator of this probability is the number of times or ways that specific event occurs.

The denominator of this probability is the overall number of ways that the experiment itself could occur.

Example 1: The table below lists the results of a student survey pertaining to favorite ethnic foods. Each student chose only one type of ethnic food for the survey.

Type Italian Chinese Japanese Thai Mexican Other

Number 15 20 3 4 30 10

Using the given table find the probability that a) a student’s favorite ethnic food is Chinese, and b) a student’s favorite ethnic food is Mexican. Round answers to three decimal places.

1a) a student’s favorite ethnic food is Chinese:

To find this probability, we need to know n(E) = n(students whose favorite ethnic food is Chinese), which is the number of students who preferred Chinese food and n(S) = n(students surveyed), which is the total number of students surveyed.

What do you think n(E) = n(students whose favorite ethnic food is Chinese) is?

If you said 20 you are correct. 20 students indicated that their favorite ethnic food is Chinese.

What do you think n(S) = n(students surveyed) is?

If you said 82 give yourself a pat on the back. If you total 15 + 20 + 3 + 4 + 30 + 10 you get 82. Putting this together in the Empirical Probability formula you get:

*putting in the numbers found above

*reducing the fraction *rounding to 3 decimal places

The probability that a student’s favorite ethnic food is Chinese is .244.

1b) a student’s favorite ethnic food is Mexican:

To find this probability, we need to know n(E) = n(students whose favorite ethnic food is Mexican), which is the number of students who preferred Mexican food and n(S) = n(students surveyed), which is the total number of students surveyed.

What do you think n(E) = n(students whose favorite ethnic food is Mexican) is?

If you said 30 you are correct. 30 students indicated that their favorite ethnic food is Mexican.


If you said 82 give yourself a pat on the back. As found in example 1a above, the total surveyed is 82. Putting this together in the Empirical Probability formula you get:

*putting in the numbers found above

*reducing the fraction *rounding to 3 decimal places

The probability that a student’s favorite ethnic food is Mexican is .366.

Equiprobable Space

A sample space S is called an equiprobable space if and only if all the simple events are equally likely to occur.

Some quick examples of this are:

A toss of a fair coin. It is equally likely for a head to show up as it is for a tail.

Select a name at random from a hat. Since it is at random, each name is equally likely to be picked.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob.htm#ex1a%20n(S)

Throwing a well balanced die. Each number on the die has the same amount of chance of coming up.

Theoretical Probability

Theoretical probability is finding the probability of events that come from an equiprobable sample space or, in other words, a sample space of known equally likely outcomes.

For example, finding various probabilities dealing with the roll of a die, a toss of a coin, or a picking of a name from a hat.

Theoretical Probability Formula

If E is an event of sample space S, where n(E) is the number of equally likely outcomes of event E and n(S) is the number of equally outcomes of sample space S, then the probability of event E occurring can be found using the Theoretical Probability Formula above.

Mutually Exclusive

In general, events E and F are said to be mutually exclusive if and only if they have no elements in common.

For example, if the sample space is rolling a die, where S = {1, 2, 3, 4, 5, 6}, and E is the event of rolling an even number, E = {2, 4, 6} and F is the event of rolling an odd number, F = {1, 3, 5}, E and F are mutually exclusive, because they have NO elements in common. Now lets say that event G is rolling a number less than 4, G = {1, 2, 3}. Would event G and E be mutually exclusive? If you said no, you are correct, they have one element, the number 2, in common. G and F would not be mutually exclusive either.

Properties of Probability

Property 1 It stands to reason that if the bottom number of the probability is the total number - which is the highest number - than the probability will never exceed 1.

Property 2

Either an element is in E or it is not in E.

So if, P(E) = 1/4 then P(not E) = 3/4.

Property 3 This is just like Property 2 in reverse.

Property 4 "Or" probabilities with mutually exclusive events Since we are dealing with sets that are mutually exclusive, this means they have no elements in common. So we can just add the two probabilities together without running a risk of having something counted twice.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob.htm#mutual

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob.htm#prop2

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob.htm#equi

Property 5 "Or" probabilities with events that are NOT mutually exclusive

Since we are dealing with events that are NOT mutually exclusive, we run the risk of elements being counted twice if we just add them together as in Property 4 above.

You need to subtract the intersection to get rid of the elements that were counted twice. In other words, you may have some elements in common, so if we add the number of elements in E to the number of elements of F, we may be adding some elements twice, so to avoid this we need to subtract the number of elements in the intersection of the two events - which would be all the elements that are in both sets.

Example 2: A pair of fair dice is tossed. Determine the probability that a) at least one of the dice shows a 6 and b) the sum of the two numbers is 5. Round answers to three decimal places.

Here is a table of all the possible outcomes of having a pair of dice tossed:

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

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

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

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

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

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

2a) at least one of the dice shows a 6:To find this probability we need to know n(E) = n(at least one of the dice shows a 6), which is the number of rolls of two dice that have at least one 6 showing and n(S) = n(rolls of a pair of fair dice), which is the total number of ways a pair of fair dice can be rolled.

What do you think n(E) = n(at least one of the dice shows a 6) is? If you said 11 you are correct. Looking at the table above, there are 11 rolls of two dice that contain at least one 6.

What do you think n(S) = n(rolls of a pair of fair dice) is? If you said 36 give yourself a pat on the back. Looking at the table above, there are 36 possible rolls of two fair dice.

Putting this together in the Theoretical Probability formula you get:

*putting in the numbers found above *rounding to 3 decimal places

The probability that at least one of the dice shows a 6 is .306.

2b) the sum of the two numbers is 5:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob.htm#die%20table



To find this probability we need to know n(E) = n(the sum of the two numbers is 5), which is the number of rolls of two dice that have a sum of 5 and n(S) = n(rolls of a pair of fair dice), which is the total number of ways a pair of fair dice can be rolled.

What do you think n(E) = n(the sum of the two numbers is 5) is?

If you said 4 you are correct. Looking at the table above, 4 rolls of two die {(4, 1), (3, 2), (2, 3), (1, 4)} have a sum of 5.

What do you think n(S) = n(rolls of a pair of fair dice) is? If you said 36 give yourself a pat on the back. Looking at the table above, there are 36 possible rolls of two fair dice.


*putting in the numbers found above *reducing the fraction *rounding to 3 decimal places

The probability that the sum of the two numbers is 5 is .111.

Example 3: From a group of 10 women and 5 men, 2 people are selected at random to form a committee. Find the probability that a) only men are selected and b) exactly 1 man and 1 woman is selected. Round answers to three decimal places.

3a) only men are selected:To find this probability we need to know n(E) = n(only men are selected), which is the number of committees that contain 2 men and n(S) = n(2 person committees), which is the total number of 2 person committees.

What do you think n(E) = n(only men are selected) is?

Since we are counting committees that means we are counting groups of people, which means we need to use combinations to count them. If you need a review on combinations, feel free to go to Tutorial 57: Combinations.

If you said 10 you are correct.

Let's see how we get that number.

Here is what the committee looks like: we would have to have 2 men to meet this condition:

2 MEN

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut57_comb.htm



*Comb. of n = 5 men taken r = 2 at a time

*Eval. inside ( )


*Cancel out 3!'s

As mentioned above, n(E) = n(only men are selected) = 10.


If you don't have a factorial key, you can simplify it as shown above and then enter it in.

What do you think n(S) = n(2 person committees) is?

If you said 105 give yourself a pat on the back.

Again you would have to use combinations to figure this one out. Overall, there are 10 women and 5 men for a total of 15 people, taken 2 at a time.

*Comb. of n = 15 people taken r = 2 at a time

*Eval. inside ( )


*Cancel out 13!'s

As mentioned above, n(S) = n(2 person committees) = 105.





The probability that only men are selected is .095.

3b) exactly 1 man and 1 woman:To find this probability we need to know n(E) = n(1 man and 1 woman are selected), which is the number of committees that contain 1 man and 1 woman and n(S) = n(2 person committees), which is the total number of 2 person committees.

What do you think n(E) = n(1 man and 1 woman are selected) is?

Since we are counting committees that means we are counting groups of people, which means we need to use combinations to count them. If you need a review on combinations, feel free to go to Tutorial 57: Combinations.



Here is what the committee looks like: we would have to have 1 man and 1 woman to meet this condition:

1 MAN 1 WOMAN

*Comb. of n = 5 men taken r = 1 at a time *Comb. of n = 10 women taken r = 1 at a time

*Eval. inside ( )

*Expand 5! until it gets to 4! and 10! until it gets to 9! which are the larger !'s in the den.


As mentioned above, n(E) = n(1 man and 1 woman selected) = 50.

If you have a factorial key, you can put it in as 5!, divided by 4!, divided by 1! times 10!, divided by 9!, divided by 1! and then press enter or =.


What do you think n(S) = n(2 person committees) is?


If you said 105 give yourself a pat on the back. As found in example 3a above, the total number of 2 person committees is 105.



The probability that exactly one man and one woman are selected is .476.

Example 4: One student’s name will be picked at random to win a CD player. There are 12 male seniors, 15 female seniors, 10 male juniors, 5 female juniors, 2 male sophomores, 4 female sophomores, 11 male freshmen and 12 female freshman. Find the probability that a) a senior or a junior is picked, b) a freshman or a female is picked, and c) a freshman is NOT picked. Round answers to three decimal places.

4a) a senior or a junior is picked:Note that these two events are mutually exclusive. You CANNOT be a senior and a junior at the same time. So

we can use the formula (property 4):

Let’s break it down first by finding each separate probability.

P(a senior is picked)

To find this probability we need to know n(E) = n(seniors), which is the number of seniors and n(S) = n(students), which is the total number of students.

What do you think n(E) = n(seniors) is?

If you said 27 you are correct. There are 12 male seniors and 15 female seniors for a total of 27 seniors.

What do you think n(S) = n(students) is?

If you said 71 give yourself a pat on the back. If we add all the students together we get 71.



P(a junior is picked)



http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob.htm#ex3a

To find this probability we need to know n(E) = n(juniors), which is the number of juniors and n(S) = n(students), which is the total number of students.

What do you think n(E) = n(juniors) is? If you said 15 you are correct. There are 10 male juniors and 5 female juniors for a total of 15 juniors.

What do you think n(S) = n(students) is?If you said 71 give yourself a pat on the back. If we add all the students together we get 71.



Putting the two probabilities of mutually exclusive events together we get:

*OR for mutually exclusive


The probability that a senior or a junior is picked is .592.

4b) a freshman or a female is picked:Note that these two events are NOT mutually exclusive. You CAN be a freshman and a female at the same time.

So we can use the formula (property 5):


P(a freshman is picked)To find this probability we need to know n(E) = n(freshmen), which is the number of freshmen and n(S) = n(students), which is the total number of students.

What do you think n(E) = n(freshmen) is?

If you said 23 you are correct. There are 11 male freshmen and 12 female freshmen for a total of 23 freshmen.







P(a female is picked)To find this probability we need to know n(E) = n(females), which is the number of females and n(S) = n(students), which is the total number of students.

What do you think n(E) = n(females) is?

If you said 36 you are correct. There are 15 female seniors, 5 female juniors, 4 female sophomores and 12 female freshman for a total of 36 females.





P(a freshman AND female is picked)To find this probability we need to know n(E) = n(freshman AND female), which is the number of female freshmen and n(S) = n(students), which is the total number of students.

What do you think n(E) = n(freshmen AND female) is?

If you said 12 you are correct. There are 12 female freshmen.





Putting the probabilities of non mutually exclusive events together we get:

*OR for not mutually exclusive *putting in the numbers found above *rounding to 3 decimal places

The probability that a freshman or a female is picked is .662.

4c) a freshman is NOT picked:

One way we can approach this is to use Property 2: To find this probability we need to know n(E) = n(freshmen), which is the number of freshman and n(S) = n(students), which is the total number of students.

What do you think n(E) = n(freshmen) is?

If you said 23 you are correct. There are 11 male freshmen and 12 female freshmen for a total of 23 freshman.





So, the probability a student picked is NOT a freshman is



*Property 2


The probability that a freshman is NOT picked is .676.

Independent Events

E and F are Independent Events if an only if

Two events are independent of each other if the outcome of one event does not affect the outcome of the other event. You can use this formula to show if two events are independent or not.Example 5: If two cards are drawn from a deck of cards, WITH REPLACEMENT, find the probability that the 1st card is a heart AND the 2nd card is an ace. Round answer to three decimal places.

Since it said WITH REPLACEMENT, that means the outcome of the 1st card does NOT affect the outcome of the 2nd card, which means these two events are independent. So we can use the

formula to find our probability.

Let’s break it down.

P(1st card is a heart)To find this probability we need to know n(E) = n(hearts), which is the number of hearts in a deck of cards and n(S) = n(cards), which is the total number of cards in a deck.

What do you think n(E) = n(hearts) is?

If you said 13 you are correct. There are 13 cards in a deck that have heart on them.

What do you think n(S) = n(cards) is?

If you said 52 give yourself a pat on the back. There are 52 cards in a deck.




P(2nd card is an ace)To find this probability we need to know n(E) = n(aces), which is the number of aces in a deck of cards and n(S) = n(cards), which is the total number of cards in a deck.

What do you think n(E) = n(aces) is?

If you said 4 you are correct. There are 4 cards in a deck that have an ace on them.

What do you think n(S) = n(cards) is?




Putting the probabilities of two events that are independent together we get:


Practice Problems


The table below lists the results of a student survey pertaining to favorite types of music.

Music Rock Jazz Country Classical Rap

Number 51 24 33 5 15

Using the given table find the probability that

1a. a student’s favorite type of music is jazz. Round answer to three decimal places.

1b. a student’s favorite type of music is classical. Round answer to three decimal places.

Practice Problems 2a - 2e: You are dealt one card. Find the probability that you are dealt

2a. a jack. Round answer to three decimal places.

2b. a heart or club. Round answer to three decimal places.

2c. a heart and a club. Round answer to three decimal places.

2d. an ace and a spade. Round answer to three decimal places.

2e. an ace or a spade. Round answer to three decimal places.


From a group of 7 freshmen, 9 sophomores, 8 juniors, and 10 seniors, 6 names will be drawn. Find the probability that

3a. 2 freshmen, 2 sophomores and 2 juniors are selected. Round answer to three decimal places.

3b. only seniors are selected. Round answer to three decimal places.


A single die is rolled twice. Find the probability that

4a. a six is rolled the 1st time and an odd number is rolled a 2nd time. Round answer to three decimal places.

4b. a number less than or equal to 2 is rolled the 1st time and an even number is rolled the 2nd time. Round answer to three decimal places.

Answers: The Fundamental Counting Principle

Answer/Discussion to 1a

One quarter, one dime and one six-sided die are tossed. How many results are possible?


There are 3 stages or events: one quarter, one dime, and one six-sided die.

Each coin has 2 possible outcomes, either a tail or a head.

The die has 6 possible outcomes.


quarter dime die # of possible outcomes2 x 2 x 6 = 24

So there are 24 different possible outcomes.

Answer/Discussion to 1b

Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?


There are 4 stages or events: a science class, a math class, a history class and an english class.

There are 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to pick from.


Science Math History English # of schedules4 x 3 x 2 x 3 = 72

So there are 72 different schedules possible.

Answer/Discussion to 1c

Six students in a speech class all have to give there speech on the same day. One of the students insists on being first. If this student’s request is granted, how many different ways are there to schedule the speeches?


There are 6 stages or events: speaker 1, speaker 2, speaker 3, speaker 4, speaker 5, and speaker 6.

There is only one possibility for speaker 1.

That leaves 5 possibilities for speaker 2, which leaves 4 for speaker 3, which leaves 3 for speaker 4, which leaves 2 for speaker 5 which leaves 1 for speaker 6.


speaker 1 speaker 2 speaker 3 speaker 4 speaker 5 speaker 6# of ways

1 x 5 x 4 x 3 x 2 x 1 = 120

So there are 120 different ways they can be scheduled to speak.

Answers:Permutations


A company issues a questionnaire whereby each employee must rank the 5 items with which he or she is most satisfied. The items are wages, work environment, vacation time, job security, supervisors, health insurance, break time, and retirement plan.

The ranking is to be indicated by the numbers 1, 2, 3, 4 and 5, where 1 indicates the item involving the greatest satisfaction and 5 the least. In how many ways can an employee answer this questionnaire?

Since we are ranking these items, this means order is important. So we can use permutations to help us out here.


n is the number of items we have to choose from. What do you think n is in this problem?

If you said n is 8 you are correct!!! There are 8 items listed.

r is the number of items we are ranking at a time. What do you think r is?

If you said r is 5, pat yourself on the back!! We are ranking the top 5.

Putting this into the permutation formula we get:

*n = 8, r = 5


*Cancel out 3!'s



Wow, this means there are 6720 different ways to rank these items.


A key pad lock has 10 different digits, and a sequence of 5 different digits must be selected for the lock to open. How many key pad combinations are possible?

Since we are counting the number of 5 digit sequences, this means order is important. So we can use permutations to help us out here.


n is the number of digits we have to choose from. What do you think n is in this problem?

If you said n is 10 you are correct!!! There are 10 digits.

r is the number of digits being used at a time. What do you think r is?

If you said r is 5, pat yourself on the back!! There are 5 digits in the sequence.


*n = 10, r = 5


*Cancel out 5!'s



Wow, this means there are 30240 different combinations.


In how many ways can 7 books be arranged on a shelf?

Since we are arranging these books that means order is important. So we can use permutations to help us out here.


n is the number of books we have to choose from. What do you think n is in this problem?

If you said n is 7 you are correct!!! There are 7 books in this problem.

r is the number of books we are using at a time. What do you think r is?

If you said r is 7, pat yourself on the back!! We are arranging all 7 books on the shelf.


*n = 7, r = 7

*0! = 1

*Expand out 7!



Wow, this means there are 5040 different ways to arrange these books.

Answers: Combinations


The order of the demonstrators is important?

Keeping in mind that order is important, would this be a permutation or a combination?

If you said a permutation problem, give yourself a pat on the back..


If n is the number of students we have to choose from, what do you think n is in this problem?

If you said n = 15 you are correct!!! There are 15 students.

If r is the number of students chosen at a time, what do you think r is?

If you said r = 5, pat yourself on the back!! 5 students are chosen to give demonstrations.

Let’s put those values into the permutation formula and see what we get:

*n = 15, r = 5

*Eval. inside ( )

*Expand 15! until it gets to 10!

*Cancel out 10!'s

If you have a factorial key, you can put it in as 15!, divided by 10! and then press enter or =.


Wow, this means there are 360360 different ways the teacher can set up the demonstrators with regard to order.


The order of the demonstrators is not important?

Keeping in mind that order is NOT important, would this be a permutation or a combination?

If you said a combination problem, give yourself a pat on the back.


If n is the number of students we have to choose from, what do you think n is in this problem?

If you said n = 15 you are correct!!! There are 15 students to chose from.

If r is the number of students chosen at a time, what do you think r is?

If you said r = 5, pat yourself on the back!! 5 students are chosen to give demonstrations.

Let’s put those values into the combination formula and see what we get:

*n = 15, r = 5

*Eval. inside ( )


*Cancel out 10!'s



Wow, this means there are 3003 different ways the teacher can set up the demonstrators without regard to order.


How many different draws of 8 names are there overall?

This would be a combination problem, because a draw would be a group of names without regard to order.

Note that there are no special conditions placed on the names that we draw, so this is a straight forward combination problem.


If n is the number of names we have to choose from, what do you think n is in this problem?

If you said n = 47 you are correct!!! There are 14 freshmen names, 15 sophomore names, 8 junior names, and 10 senior names for a total of 47 names.

If r is the number of names we are drawing at a time, what do you think r is?

If you said r = 8, pat yourself on the back!! 8 names are drawn at a time.


*n = 47, r = 8

*Eval. inside ( )


*Cancel out 39!'s



Wow, this means there are 314,457,495 different draws.


How many different draws of 8 names would contain only juniors?

This is a combination problem with a twist. In part a we looked at all possible draws. From that list we only want the ones that contain only juniors.

Let’s see what the draw looks like: we would have to have 8 juniors to meet this condition:

8 JUNIORS


If n is the number of JUNIORS we have to choose from, what do you think n is in this problem?

If you said n = 8 you are correct!!! There are a total of 8 JUNIORS.

If r is the number of JUNIORS we are drawing at a time, what do you think r is?

If you said r = 8, pat yourself on the back!! 8 JUNIORS are drawn at a time.


*n = 8, r = 8

*Eval. inside ( )

*Cancel out 8!'s



This means there is only 1 draw that would contain only juniors.


How many different draws of 8 names would contain exactly 4 juniors and 4 seniors?

Again we have a combination with a condition attached. Whenever you have a condition, you want to illustrate to yourself what would be in the group or draw. In this case, the only possible way this could come out is if you have 4 juniors and 4 seniors :

Let’s see what the draw looks like: we would have to have 4 juniors and 4 seniors to meet this condition:

4 JUNIORS 4 SENIORS



Note how 1 draw is split into two parts - juniors and seniors. We can not combine them together because we need a particular number of each one. So we will figure out how many ways to get 4 JUNIORS and how many ways to get 4 SENIORS, and using the counting principle, we will multiply these numbers together.

4 JUNIORS:

If n is the number of JUNIORS we have to choose from, what do you think n is in this problem?

If you said n = 8 you are correct!!! There are a total of 8 JUNIORS.

If r is the number of JUNIORS we are drawing at a time, what do you think r is?

If you said r = 4, pat yourself on the back!! 4 JUNIORS ARE drawn at a time.


4 SENIORS:

If n is the number of SENIORS we have to choose from, what do you think n is in this problem?

If you said n = 10 you are correct!!! There are a total of 10 SENIORS.

If r is the number of SENIORS we are drawing at a time, what do you think r is?

If you said r = 4, pat yourself on the back!! 4 SENIORS are drawn at a time.


*JUNIORS: n = 8, r = 4 *SENIORS: n = 10, r = 4

*Eval. inside ( )





This means there are 14,700 different draws that would contain 4 juniors and 4 seniors.

Answers: Probability


Find the probability that a student’s favorite type of music is jazz. Round answer to three decimal places.

To find this probability, we need to know n(E) = n(students whose favorite type of music is jazz), which is the number of students who preferred jazz and n(S) = n(students surveyed), which is the total number of students surveyed.

What do you think n(E) = n(students whose favorite type of music is jazz) is?

If you said 24 you are correct. 24 students indicated that their favorite type of music is jazz.


If you said 128 give yourself a pat on the back. If you total 51 + 24 + 33 + 5 + 15 you get 128.

Putting this together in the Empirical Probability formula you get:


The probability that a student’s favorite type of music is jazz is .188.


Find the probability that a student’s favorite type of music is classical. Round answer to three decimal places.

To find this probability, we need to know n(E) = n(students whose favorite type of music is classical), which is the number of students who preferred classical and n(S) = n(students surveyed), which is the total number of students surveyed.

What do you think n(E) = n(students whose favorite type of music is classical) is?

If you said 5 you are correct. 5 students indicated that their favorite type of music is classical.


If you said 128 give yourself a pat on the back. If you total 51 + 24 + 33 + 5 + 15 you get 128.

Putting this together in the Empirical Probability formula you get:


The probability that a student’s favorite type of music is classical is .039.


Find the probability that you are dealt a jack. Round answer to three decimal places.

To find this probability we need to know n(E) = n(jacks), which is the number of jacks in a deck of cards and n(S) = n(cards in a deck), which is the total number of cards in a deck.

What do you think n(E) = n(jacks) is?

If you said 4 you are correct. There are 4 jacks in a deck of cards.

What do you think n(S) = n(cards in a deck) is?




The probability that a jack is dealt is .077.


Find the probability that you are dealt a heart or club. Round answer to three decimal places.

Note that these two events are mutually exclusive. A card CANNOT have a heart and a club on it at

the same time. So we can use the formula (property 4):


P(a heart is dealt)

To find this probability we need to know n(E) = n(hearts), which is the number of hearts in a deck of cards and n(S) = n(cards in a deck), which is the total number of cards in a deck.

What do you think n(E) = n(hearts) is?

If you said 13 you are correct. There are 13 hearts in a deck of cards.





P(a club is dealt)

To find this probability we need to know n(E) = n(clubs), which is the number of clubs in a deck of cards and n(S) = n(cards in a deck), which is the total number of cards in a deck.



What do you think n(E) = n(clubs) is?

If you said 13 you are correct. There are 13 clubs in a deck of cards.





Putting the two probabilities of mutually exclusive events together we get:

*OR for mutually exclusive


The probability that a heart or club is dealt is .5.


Find the probability that you are dealt a heart and a club. Round answer to three decimal places.

As mentioned in the Answer/Discussion to problem 2b, these two sets are mutually exclusive. This time we are looking for the probability that is a heart AND a club is dealt. This means the card must have both attributes. Since they are mutually exclusive, this means they have nothing in common. In other words, a card cannot be both a heart and a club at the same time.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob_ans.htm#ad2b


So what would the probability of this event be????

If you said 0, you are right on. Since n(a card with a heart and a club) is 0, the probability will be 0.

Answer/Discussion to 2d

Find the probability that you are dealt an ace and a spade. Round answer to three decimal places.

To find this probability we need to know n(E) = n(ace AND spade), which is the number of cards with an ace and a spade on it in a deck of cards and n(S) = n(cards in a deck), which is the total number of cards in a deck.

What do you think n(E) = n(ace AND spade) is?

If you said 1 you are correct. There is 1 card in a deck of cards that has both an ace and a spade on it.





The probability that a card with an ace and spade on it is dealt is .019.

Answer/Discussion to 2e

Find the probability that you are dealt an ace or a spade. Round answer to three decimal places.

Note that these two events are NOT mutually exclusive. A card CAN have an ace or a spade at the

same time. So we can use the formula (property 5):


P(an ace is dealt)

To find this probability we need to know n(E) = n(aces), which is the number of aces in a deck of cards and n(S) = n(cards in a deck), which is the total number of cards in a deck.

What do you think n(E) = n(aces) is?

If you said 4 you are correct. There are 4 aces in a deck of cards.





P(a spade is dealt)

To find this probability we need to know n(E) = n(spades), which is the number of spades in a deck of cards and n(S) = n(cards in a deck), which is the total number of cards in a deck.

What do you think n(E) = n(spades) is?

If you said 13 you are correct. There are 13 spades in a deck of cards.






P(an ace AND a spade is dealt)

We found this probability to be 1/52 in the Answer/Discussion to problem 2d.

Putting the probabilities of non mutually exclusive events together we get:

*OR for not mutually exclusive


The probability that a card that is dealt has an ace or a spade on it is .308.


Find the probability that 2 freshmen, 2 sophomores and 2 juniors are selected. Round answer to three decimal places.

To find this probability we need to know n(E) = n(2 freshmen, 2 sophomores and 2 juniors are selected), which is the number of ways 6 names drawn can contain 2 freshmen, 2 sophomores and 2 juniors and n(S) = n(6 name drawings), which is the total number of ways 6 names can be drawn.

What do you think n(E) = n(2 freshmen, 2 sophomores and 2 juniors are selected) is?



http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob_ans.htm#ad2d

Since we are counting draws that means we are counting groups of names, which means we need to use combinations to count them. If you need a review on combinations, feel free to go to Tutorial 57: Combinations.



Here is what the committee looks like: we would have to have 2 freshmen, 2 sophomores and 2 juniors to meet this condition:

2 FRESHMEN 2 SOPHOMORES 2 JUNIORS

*Comb. of n = 7 freshmen taken r = 2 at a time *Comb. of n = 9 sophomores taken r = 2 at a time

*Comb. of n = 8 juniors taken r = 2 at a time

*Expand 7! until it gets to 5!, 9! until it gets to 7!, and 8! until it gets to 6! which are the larger !'s in the den.

*Cancel out 5!'s, 7!'s, and 6!'s

As mentioned above, n(E) = n(2 freshmen, 2 sophomores and 2 juniors are selected) = 21168.

If you have a factorial key, you can put it in as 7!, divided by 5!, divided by 2! times 9!, divided by 7!, divided by 2!, times 8!, divided by 6!, divided by 2! and then press enter or =.


What do you think n(S) = n(6 name drawings) is?

If you said 1344904 give yourself a pat on the back.

Again you would have to use combinations to figure this one out. Overall, there are 7 freshmen, 9 sophomores, 8 juniors, and 10 seniors for a total of 34 students, taken 6 at a time.



*Comb. of n = 34 students taken r = 6 at a time

*Eval. inside ( )


*Cancel out 26!'s

As mentioned above, n(S) = n(6 name drawings) = 1344904.





The probability that 2 freshmen, 2 sophomores and 2 juniors are drawn is .016.


Find the probability that only seniors are selected. Round answer to three decimal places.

To find this probability we need to know n(E) = n(only seniors are drawn), which is the number of ways 6 names drawn can contain only seniors and n(S) = n(6 name drawings), which is the total number of ways 6 names can be drawn.

What do you think n(E) = n(seniors) is?

Since we are counting draws that means we are counting groups of names, which means we need to use combinations to count them. If you need a review on combinations, feel free to go to Tutorial 57: Combinations.



Here is what the committee looks like: we would have to have only seniors to meet this condition:

6 SENIORS

*Comb. of n = 10 seniors taken r = 6 at a time *Eval. inside ( )


*Cancel out 6!'s

As mentioned above, n(E) = n(only seniors are drawn) = 210.



What do you think n(S) = n(6 name drawings) is?

As shown in Answer/Discussion to problem 3a, n(S) = n(6 name drawings) = 1344904.



http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut58_prob_ans.htm#ad3a



The probability that only seniors are drawn is .0002.

Note I carried this out 4 decimal places since that was the first non zero digit.


Find the probability that a six is rolled the 1st time and an odd number is rolled a 2nd time. Round answer to three decimal places.

Since the outcome of the 1st roll of the die does NOT affect the outcome of the 2nd roll of the die, this

means these two events are independent. So we can use the formula to find our probability.


P(1st roll is a 6)

To find this probability we need to know n(E) = n(6's on the die), which is the number of 6's on a die and n(S) = n(numbers on a die), which is the total number of numbers on a die.

What do you think n(E) = n(6's on the die) is?

If you said 1 you are correct. There is 1 number 6 on a die.

What do you think n(S) = n(numbers on a die) is?

If you said 6 give yourself a pat on the back. There are 6 numbers on a die.



P(2nd roll is odd)

To find this probability we need to know n(E) = n(odd numbers on a die), which is the number of odd numbers on a die and n(S) = n(numbers on a die), which is the total number of numbers on a die.

What do you think n(E) = n(odd numbers on a die) is?

If you said 3 you are correct. There is 3 odd numbers on a die.








Find the probability that a number less than or equal to 2 is rolled the 1st time and an even number is rolled the 2nd time. Round answer to three decimal places.

Since the outcome of the 1st roll of the die does NOT affect the outcome of the 2nd roll of the die, this

means these two events are independent. So we can use the formula to find our probability.


P(1st roll is less than or equal to 2)

To find this probability we need to know n(E) = n(numbers less than or equal to 2), which is the number of numbers that are less than or equal to 2 on a die and n(S) = n(numbers on a die), which is the total number of numbers on a die.

What do you think n(E) = n(numbers less than or equal to 2) is?

If you said 2 you are correct. There are 2 numbers that are less than or equal to 2 on a die.





P(2nd roll is even)

To find this probability we need to know n(E) = n(even numbers on a die), which is the number of even numbers on a die and n(S) = n(numbers on a die), which is the total number of numbers on a die.

What do you think n(E) = n(even numbers on a die) is?

If you said 3 you are correct. There is 3 even numbers on a die.







Miscellaneous 1

One Page Tutorial: Combination

In many problems we are interested in the number of ways of selecting objects from a total number of n without regard to the order. For example, consider the group of the club with 15 members and the selection of four officers on the committee are to be assigned. How many ways can a committee of four be chosen (and not arranged in order) from the 15 members?

This problem can be solved by combinations. The number of combinations of n distinct objects taken r at a time is:

Example:

How many ways can we select three letters from the letters of RSTUV?

n = 5 r = 3

These are: RST, RSU, RSV, RTU, RTV, RUV, STU, STV, SUV and TUV.

Poker Example:

Poker. Find the probability of a full house (three of a kind and two of another kind), say three kings and two aces.

Let A be the event 'full house'.

Miscellaneous 2

Imagine you order a pizza, and you want a three topping pizza. You will have pepperoni, mushrooms, and onions. The waiter asks you, "which topping do you want on your pizza first?"

What would you answer? Would you ever expect this question? No. Because the order of the toppings wouldn't matter. Your pizza will have three toppings, end of story.

Now, if you did want to know how many ways you could put those three toppings on your pizza, you could figure it out with permutations: 3 x 2 x 1 = 6. But at the end of the day, they're all the same, so you don't count them.

What would be different? Well, a different pizza, that's what: sausage, pinapple, and onion would be a different pizza. Again, you wouldn't care about the order, but it would be different.

That's the difference between permutations and combinations. In permutations, we count all the orderings of every different group. In that case, the two pizza examples would be 12 different permutations. But in combos, we don't care about the order, so we neutralize the copies. Take that 12 permutations from above, divide by the number of copies of each group, and that's combos, in this case: 2.

So combinations is a permutations with a tax on it. You manipulate a permutation question to get a combinations answer. And you do it by dividing by the number of spaces in the permutation, factorial.

And that's how it goes. Look for examples of each on the web - the common divisor between all combos is that we won't care about the order.

Documents

Permutation Combination Probability