Chapter+14

Chapter 14: Counting and Probability ​ toc

In this chapter, we will learn about how to use sets, combinations, permutations, and probability.

=14.1: Sets and Counting​ =

**Set**—a well defined collection of distinct objects. Well defined means there is a rule that enables us to determine whether an object is in the set. A= {1, 2, 3, 4, 5}
 * Sets **

The objects in set are called **elements.** 1, 2, 3, 4, and 5 are elements of set A.

If a set has no elements it is called an **empty set** or a **null set**. The symbol refers to a null set. B= {} Don’t repeat elements of a set. Only write them once, even if they appear multiple times.

It also doesn’t matter what order you write the numbers in.

Example #1: Write the set with all the possible results of rolling a fair die. Solution: When you roll a die, you have an equal chance of it landing with any of the six numbers face up. Therefore, the set would look like this:  A= {1, 2, 3, 4, 5, 6}

If set B= {1, 2, 3, 4, 5, 6}, then you can say A=B, because A and B have exactly the same elements.

If set C= {1, 2, 3}, then C is a **subset** of A, because every single element of C is an element of A. This is written as CA. Since A has additional elements besides just the ones in C, C can also be called a **proper subset** of A, written as CA. REMEMBER: The empty set is a subset of all sets

Example #2: Find all the subsets of set {m, n, o} Solution: It is easier to find the subsets of you organize them based on number of elements.

0 elements: {} 1 element: {m}, {n}, {o} 2 elements: {m, n}, {m, o}, {n, o} 3 elements: {m, n, o}

Intersections and Unions **
 * Intersection**—the set that has elements belonging to both A and B. It is written A ∩ B.


 * Union**—the set that has elements belonging to either A, B, or both. It is written AUB.

Example #3: A= {1, 3, 5, 7, 9} B= {2, 4, 6, 8, 9} and C= {1, 2, 3} Find: a) AUB b) A ∩ B c)A ∩ (CUB) Solution: a) {1, 3, 5, 7, 9} U {2, 4, 6, 8, 9} = {1, 2, 3, 4, 5, 6, 7, 8, 9} b) {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8, 9} = {9} c) {1, 3, 5, 7, 9} ∩ ({1, 2, 3} U {2, 4, 6, 8, 9}) = {1, 3, 5, 7, 9} ∩ {1, 2, 3, 4, 6, 8, 9} = {1, 3, 9}

Complements When using sets, there is usually a **universal set (U)**, which contains all of the elements involved. If A is a set, A’ is the **complement** of A—the set containing all the elements in the universal set that are not in A.



This is a **Venn diagram**. It uses shaded circles and squares to represent sets, complements and the universal set.

Here are some more examples of Venn diagrams:



When two sets have no elements in common, they are said to be **disjoint:**



Counting We use the notation n(A) = 6 to say that there are 6 elements in set A. A set is **finite** if the number of elements is a nonnegative integer. Otherwise, the set is **infinite**. This section deals only with finite sets  Theorem: If A is a set with n elements, then A has 2ⁿ subsets Example #4: 100 Okemos High School students were surveyed. 27 were enrolled in Algebra 2 Honors and 52 were enrolled in American Literature. 9 students were enrolled in both classes.

a) How many students were enrolled in either Algebra 2 Honors or American Literature?

b) How many students were enrolled in both classes?

Solution:

a) Let A = the set of students in Algebra 2 Honors and B = the set of students in American Literature. The given information shows that: n(A) = 27, n(B) =52 and n(A ∩ B) = 9. Using the Venn diagram, we can see that to solve to problem we need to add up the number of students in ONLY Algebra 2 Honors or American Literature. That is (27 – 9) + (52 – 9) = 61. 61 students are enrolled in either Algebra 2 Honors or American Literature.

b) A total of 100 students were surveyed, and as we figured out in part a, 61 students were in either Algebra 2 Honors or American Literature class. If you subtract that from 100, you will find out how many students are not enrolled in either class. 100 – 61 = 39. 39 students surveyed were not enrolled in either class.

Example 5 is based on three major counting theorems:

Counting Formula: Theorem: n(AUB) = n(A) + n(B) – n(A ∩ B) when A and B are finite sets

Addition Principle of Counting: Theorem: if A and B have no elements in common, then n(AUB) = n(A) + n(B)

General Addition Principle of Counting: Theorem: If, for n sets A1, A2 no two have elements, in common, then n(A1 U A2 U…U An) = n(A1) + n(A2) +…+ n(An)

For videos and more explanations of sets, visit: [] = =

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=14.2: Permutations and Combinations =

**Permutations ** There are three types of Permutations:
 * Permutation: **  an ordered arrangement of r objects chosen from n objects.

Type #1 : n objects where repetition is allowed. Type #2 : n distinct objects where repetition is not allowed. Type #3 : Objects are not distinct. (In order to do a type 3 permutation, you need to know background knowledge in combinations.)

Question : A company is looking into making stocks available to the public and wants to know how many possibilities there are for a three letter stock code where repeated letters are allowed. Solution : There are 26 letters in the alphabet, so there are 26 possible letters for the first letter of the code, 26 possible letters for the second letter of the code and 26 possible letters for the third letter of the code. So you multiply 26 x 26 x 26 = 17,576 possible codes for the company's stock code.
 * Type #1 Examples:**

Question : How many possible 5 digit zip codes are there for California, where the first digit has to be a 9? Solution : There are 10 possible numbers that the zip code can be made of: (0,1,2,3,4,5,6,7,8,9). However, the first digit must be a 9, so there is only one possible choice for the first digit. So you would multiply 1 x 10 x 10 x 10 x 10 = 1,0000 possible zip codes for California. It helps to draw in slots for all the spaces you need to fill.

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 * Type #2 Example**:

Question : Dale wants to make up a password for his new e-mail account. He wants the password to be 4 letters and have no repeating letters. How many possibilities does Dale have for his password? Solution : For the first letter of the password we have 26 possible letters, for the second letter we have 25 possible letters, for the third letter we have 24 possible letters and for the fourth we have 23 possible letters. So then we multiply 26 x 25 x 24 x 23 = 358,800 possible passwords for Dale.



(Since we can't use the same letter, one letter will be removed from our total of 26 for every letter after our total of 26 i.e. 26, 25, 24, 23, 22, 21, 20......)

You can also solve this problem using: P(n,r) = n! / (n-r)! For this problem we would set up P(26, 4) since 26 is the total and we are choosing 4 letters from the total. So, P(26,4) = 26! / (26-4)! = 26! / 22! = 358,800 possible choices.

You could also compute P(26,4) using just a calculator. To do this (using a TI-83 Plus graphing calculator) you would first type in 26 to the main screen, then you would push the MATH button (which is under the ALPHA button), now arrow over to PRB, you are looking for nPr which should be number 2 on the list, now type 4 into the main screen. On the main screen you should see: 26 nPr 4 now push enter and you should get the answer of 358,800.


 * Practice problems:**

Find the value of the following: 1. P(5,2) 2. P(10,5) 3. List the ordered arrangements of 10 objects choosing 7 at a time without repetition. 4. In how many ways can 11 people be seated in a row of desks?

Answers (mouse over area next to number to see): 1. 20 2. 30,240 3. P(10,7) = 604,800 4. 11! = 39,916,800





**Combinations**

**Combination**: An arrangement without regard to order of // r // objects selected from // n // distinct objects without repetition, also // r // is less than or equal to // n. // **Multiplication Counting Principle**: P items for your first choice, Q items for your second choice, and R items for your third choice. The selections can be made in P x Q x R different ways.

Many people wonder why combinations and permutations are different, but don't let them fool you! From the definition above it may be hard to tell the difference, but trust me there is one. In any permutation it is very important to know the order of the arrangements. For example in a permutation ABCD, BCDA, CDAB, and DABC are all different arrangements. In a combination, however, those arrangements listed above would all be considered the same thing. Think of it as a game of poker. If you have the cards A, 2, 3, 4, and 5, you wouldn't consider that any different from having 2, 4, 3, A, 5...it'd still be a straight!

You can write a combination by writing **C(//n,r//)** (can also be written // **nCr** // ) in which case (from the definition) // r // is selected from // n. // You solve a combination with the formula n!/(n-r)! r!. (That equation can be also recalled as that of the binomial theorem from 13.5).

Listing Combinations : Sometimes when solving a combination problem it will ask you to list the combinations--let's look at an example: List all the combinations of the 5 objects 1, 2, 3, 4, 5 taken 2 at a time. What is C(5, 2)?

Solution <span style="color: #000000; font-family: 'Comic Sans MS',cursive; font-size: 16pt;">: One combination of 1, 2, 3, 4, 5 taken 2 at a time is: 1, 2 (you wouldn't include 2, 1 because in combinations the order does not matter!) The list of all the combinations is 1, 2 1, 3 1, 4 1, 5 2, 3 2, 4 2, 5 3, 4 3, 5 4, 5. Therefore: C(5,2) = 10. Now you are able to check your work using the formula n!/(n-r)! r! So plug the numbers in 5!/3! x 2! Cancel out the 3! and you get (5 x 4)/2 which is 20/2 which equals 10!

Now try one yourself: List all the combinations of the 10 objects 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 taken 2 at a time. What is C(10, 2)? Highlight below to see the answer: You should've gotten 45 as your answer. Because 10!/(8! x 2)= (10 x 9)/2 = 90/2 = 45.

Forming Committees : This is just a different kind of Combination problem in which you take a certain number of people and have to form different sized committees from the number of people. Let's try one: How many different committees of 5 people can be formed from a group of 8?

Solution <span style="color: #000000; font-family: 'Comic Sans MS',cursive; font-size: 16pt;">: The 8 people are distinct. The problem asks for the number of combinations of 8 objects taken 5 at a time. You would solve that using the C(n,r) formula. So essentially we need to figure out C(8, 5) and you set the problem up like this: 8!/(8-5)! x 5! = 8!/3! x 5! if you cancel out the 5! and 8! you get (8 x 7 x 6)/(3 x 2 x 1) now cancel out the 3 x 2 with the 6 and you get (8 x 7)/1 = 56/1 = 56 different committees can be formed. Now try one yourself: How many different committees of 7 people can be formed from from a group of 13? Highlight below to see the answer: Essentially in this problem you are trying to find C(13, 7). 13!/(13-7)! x 7! cancel out the 7! and the 13! and you get (13 x 12 x 11 x 10 x 9 x 8)/(6 x 5 x 4 x 3 x 2 x 1) cancel out a few things and you get (13 x 12 x 11)/1 which equals 1716/1 = 1716 committees can be formed.

Forming Different Words : This is the third type of permutation problem in which you must find how many words can be made (real or imaginary) by rearranging all the letters in a given word. Now let's look at an example: How many different words can be formed using all the letters in the word **COMMITTEE**?

Solution : Each word formed will have 9 letters (because that's how many that are in the original word) the letters are: 2 M's, 2 T's, 2 E's, 1 C, 1 O, and 1 I. To construct each word, we need to fill 9 positions with the 9 letters.

The process of forming a word consists of 6 tasks: Task 1: Choose the positions for the 2 M's. Task 2: Choose the positions for the 2 T's. Task 3: Choose the positions for the 2 E's. Task 4: Choose the positions for the 1 C. Task 5: Choose the positions for the 1 O. Task 6: Choose the positions for the 1 I.

Task 1 can be done in C(9, 2) ways. There then remain 7 positions to be filled, so Task 2 can be done in C(7, 2) ways. Now there are 5 positions remaining, C(5, 2). Now 3 spots remain, C(3, 1), then C(2, 1) and then finally, C(1, 1). Using the Multiplication Principle, the number of possible "words" that can be formed can be determined by: C(9, 2) x C(7, 2) x C(5, 2) x C(3, 1) x C(2, 1) x C(1,1) = (9! x 7! x 5! x 3! x 2! x 1!)/(7! x 2! x 5! x 2! x 3! x 1! x 2! x 1! 1! x 1!) As you can see many of these things cancel out with each other until all you are left with is: 9!/(2! x 2! x 1! x 1! x 1!). As we all know 1! = 1 so we can cancel all those out as well as one of the 2! and the 9! and you are left with: (9 x 8 x 7 x 6 x 5 x 4 x 3)/2 = 988848/2 = 494424 "words" can be made.

<span style="font-family: 'Comic Sans MS',cursive; font-size: 16pt;"> Now try one yourself : How many different "words" can be formed by using all the letters in the word **INDIVISIBILITY**? Highlight below to see the answer: <span style="color: #ffffff; font-family: 'Comic Sans MS',cursive; font-size: 16pt;">There are thirteen letters in this word; 6 I's, 1D, 1 V, 1 S, 1 B, 1 L, 1 T, and 1 Y. So you must do C(13, 6) x C(7, 1) x C(6, 1) x C(5, 1) x C(4, 1) x C(3, 1) x C(2, 1) x C(1, 1) after you figure out what that is you must cancel and you will end up with 13!/6! = 13 x 12 x 11 x 10 x 9 x 8 x 7 = 95135040 "words" can be made. If you need any more help with Combinations or Permutations visit this website: [] =<span style="background-color: #008080; color: #800080; display: block; font-family: 'Comic Sans MS',cursive; font-size: 22pt; text-align: center;">**14.3: Probability** = = = <span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">**Probability:** the chance of something happening.

When you flip a coin, the probability, or chance, that you get a heads or tails is ½. This is because there are only two possible outcomes, Heads or Tails and you can only get one.

The **Sample Space** is a set that represents the number of possible outcomes:



<span style="font-family: 'Comic Sans MS',cursive; font-size: 140%;">Th <span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">erefore, the Sample Space for flipping a coin would be S= {Heads, Tails}.

<span style="font-family: 'Comic Sans MS',cursive; font-size: 16pt;"> Example: Let’s say you wanted to find the chance of rolling a number 6 on a die. <span style="color: red; display: block; font-family: 'Comic Sans MS'; font-size: 16pt; line-height: 115%; text-align: left;">Solution: <span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">First you list out all the possible outcomes: S= {1,2,3,4,5,6} <span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;"> There are six different outcomes and since there is only one side on a die with the number six on it, the chances of rolling the number six would be P(6)= 1/6 To find the probability or chance of something, just divide the number of ways it could happen by the number of possible outcomes.

<span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;"> A **Probability Model** shows the sample space and the probability of the experiment in it.
 * Constructing a probability model**

Example : Lets say you have two spinners:



<span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">You have to construct a probability model for this experiment: Spin spinner 1 and 2. What is the probability of getting a 2 or 4 followed by a Blue?

Solution : First list the sample space: S= {1 Red, 1 Blue, 1 Green, 2 Red, 2 Blue, 2 Green, 3 Red, 3 Blue, 3 Green, 4 Red, 4 Blue, 4 Green} There are twelve different outcomes possible, and the probability of each is 1/12.

Therefore, to get the probability model for the experiment, 1/12 + 1/12 = 1/6, You add the probability for getting a 2 Blue and getting a 4 Blue.

The probabilty of getting a 2 or 4 and a Blue is P(2 or 4, Blue)= 1/6.

<span style="color: red; font-family: 'Comic Sans MS',cursive; font-size: 16pt; line-height: 115%;">Here is another example (using the same spinners) <span style="color: #000000; font-family: 'Comic Sans MS',cursive; font-size: 16pt; line-height: 115%;">: <span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">What is the probability of getting a 2, followed by a 2 or a 4, followed by a red or green?

Solution : First, you need to find the total number of possible outcomes. Since there could be many outcomes, we won't write them all out to find out how many there are. Instead, we just multiply the number of possibilities for each number and color.

So multiply the probability of getting a number (1/4) by the probability of getting another number (1/4) and by getting a color (1/3)... 1/4 x 1/4 x 1/3 = 1/48.

Now we know that the probability of getting any random choice would be 1/48. However, the choices we would want would be 2 followed by a 2 or 4 followed by a Red or Green.

Therefore, to find the probability of getting that, you would add the probability of getting a (2 2 red) (2 4red) (2 2 green) (2 4 green). We know that each thing in the sample space has a probability of 1/48, so you would add 1/48 four times.

1/48 + 1/48 + 1/48 + 1/48 = 1/12

The probability of getting a 2 followed by a 2 or 4 followed by a red or green would be 1/12 Compound Probabilities **

**Compound Probabilities** are basically more than one event occuring at a time.

Here is an example of a compound probability event: Lets say you were rolling a die twice. A= rolling an odd number. B= rolling a 1 or 2. We have two different events.



<span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">a) Find the event AUB. b) Find the event A ∩B c) Compute P(A) and P(B) d) Compute P(A ∩B) e) Compute P(AUB)

<span style="color: #ff0000; font-family: 'Comic Sans MS',cursive; font-size: 16pt;">Solution: a) AUB = {1,2,3,5} b) A <span style="font-family: 'Arial','sans-serif'; font-size: 16pt; line-height: 115%;">∩ B = {1} c) P(A) = 3/6 or 1/2. P(B) = 2/6 or 1/3. d) P(A <span style="font-family: 'Arial','sans-serif'; font-size: 16pt; line-height: 115%;">∩ B) = 1/6. e) To find P(AUB) you could use the formula...

P(AUB) = P(A) + P(B) - P(A ∩ B) <span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;"> With this formula, P(AUB) = 3/6 + 2/6 - 1/6 = 4/6 or 2/3.

**Finding the Probabilities Using Complements** Use this formula to find the probability using complements:



Here is an example: On the news, a weather reporter said that the probability of rain is 40%. What is the probability that it wil // not // rain?

Solution: To find out the probability of it not raining, we find the complement of the probability that it would rain.

We use the formula: P(no rain) = 1 - P(rain) This turns out to be: P(no rain) = 1 - .40 = .60

Therefore, the chance of no rain would be 60%.

Here is another example: <span style="color: #000000; font-family: 'Comic Sans MS',cursive; font-size: 160%;">Out of 10 people, what is the probability that at least two people have the same birthday? (365 days in a year)

Solution: <span style="color: #000000; font-family: 'Comic Sans MS',cursive; font-size: 168%;">1 -- P(no one has the same birthday) 1 -- P(365 x 364 x 363 x 362 x 361 x ...... x 356 divided by 365^10) 1 -- .883 = .117 There is a 12% chance that two people have the same birthday.

<span style="font-family: 'Comic Sans MS',cursive; font-size: 160%;">** Probabilities involving combinations and permutations ** In most probability experiments, you will use permutations and combinations to solve the problem.

Here is an example: There are 5 broken graphing calculators that were packaged with 15 good ones. You go to buy three of them, though you don't know that some are defective.

a) What is the probability that all three will be defective? b) What is the probability that exactly two will be defective? c) What is the probability that at least two will be defective?

Solution: a) <span style="color: #000000; font-family: 'Comic Sans MS',cursive; font-size: 130%;"> Therefore, the probability of choosing three defective phones is 9%

b) <span style="font-family: 'Comic Sans MS',cursive; font-size: 130%;"> Therefore, the probability of getting exactly 2 broken calculators is 13%. c) <span style="font-family: 'Comic Sans MS',cursive; font-size: 168%;">Therefore, the probability of getting at least two broken calculators is 14%.


 * <span style="color: red; font-family: 'Comic Sans MS'; font-size: 16pt; font-weight: normal; line-height: 115%;">Here is another example (video): **

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If you’d like to see more examples and solutions, go to: []