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**Unformatted text preview: **Sets, Combinatorics and Probability Probability
Sets Counting Permutations and Combinations Probability Basic Set Concepts Basic
A set is a collection of objects. Each object set is called an element of the set. element Often the objects in a set are listed and are enclosed in “braces.” For example the set of integers that fall between 1 and 5 can be written {2 , 3 , 4}. Representing Sets Representing Word Description: Describe the set in Description your own words, but be specific so the elements are clearly defined elements
All the whole numbers from 1 to 20 Roster Method: List each element, Method separated by commas, in braces separated {1, 2, 3, 4, 5, 6, 7, 8,{1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} 9, 2, 3, . . ., 20} Set-Builder Notation: Notation {x | x is … word description} {x | x is a whole number and 1 ≤ x ≤ 20} The Empty Set The
The empty set, also called the null set, empty null set is the set that contains no elements. is elements. The empty set is represented by The {}
or by or Φ
We will use { } most of the time, because it's easier to understand it's Elements of a set Elements The symbol ∈ is used to indicate that an object is an element of a set. The symbol is used to replace the words: is is an element of is The symbol ∉ is used to indicate that an object is not an element of a set. The symbol is used to replace the words: symbol is not an element of not Standard Number Sets Standard e ˆ e ˆ e ˆ e ˆ e ˆ : integers integers : natural numbers (non-negative) natural : rational numbers (fractional) rational : real numbers real : complex numbers complex Definition of a Set’s Cardinal Number Definition The cardinal number of set A, The cardinal represented by |A|, is the number of |A| elements in set A. Cardinal Number Cardinal
If A ={a, b, c}, what is |A|? If }, |A
0 1 23 |A| = 3 If A = {}, what is |A|? If |A
0 |A| = 0 If A = { a, b, c, d, a , e}, what is |A|? |A| = 5 If d,
01 2 3 45 4 5 no element can be counted no twice, even if it's accidentally listed twice!!!!! twice!!!!! Definition of a Finite Set Definition
Set A is a finite set if |A| is a Set finite |A| natural number. A set that is not finite is called an infinite set. infinite The set of natural numbers, for example, is itself an infinite set. example, Definition of Equality of Sets Definition
Set A is equal to set B means that set Set equal A and set B contain exactly the same elements, regardless of order. We symbolize the equality of sets A and B using the statement A = B. Definition of Equivalent Sets Definition
Set A is equivalent to set B means that set A Set equivalent and set B contain the same number of elements. For equivalent sets, |A| = |B|. |A| The sets {George Washington, John Adams, Thomas Jefferson, James Madison} and {1789, 1797, 1801, 1809} are equivalent because they both have a cardinality of 4. Definition of a Universal Set Definition When we talk about a set, we ask what's When in the set and what's not in the set. in Well, pretty much anything you can think Well, of might not be in the set. of We limit ourselves to what makes sense. The universal set is one that contains all universal of the elements that are included in the discussion. discussion. Definition of a Universal Set Definition Examples: When talking about the set {a, b, c, e} our When universe most likely would be the set of English language lowercase letters. English When talking about students in this When classroom, my universe might be all FDU students enrolled for this course. It might also be all students at FDU. It might be all people in the U.S. all Definition of the Complement of a Set Definition
The complement of a set is the collection complement of all the objects in the universal set that are not in the given set. The complement of a set A is written A' . A' = {x | x ∈ U and x ∉ A}. Definition of a Subset Definition
Set A is a subset of set B, expressed as subset A⊆ B if every element in set A is also in set B.
Note that the set A could be equal to the set B. That's why there's a line at the bottom of the symbol. Think about how ≤ means less than or equal. Definition of a Proper Subset Definition
Set A is a proper subset of set B, Set proper expressed as A ⊂ B, if set A is a if subset of set B and sets A and B are not equal ( A ≠ B ).
Note that the set A can not be equal to the set B. That's why there isn't a line at the bottom of the symbol. The Empty Set as a Subset The
1. 2. For any set B, { } ⊆ B. For For any set B other than the empty For set, { } ⊂ B. Of course, { } might also be written as Of Φ 3. Sets of Sets Sets Number of subsets Number
How many subsets does {a, b, c} have? Let's count: 1. 2. 3. 4. 5. 6. 7. 8. choose a no {} yes {a} no {b} no {c} yes {a, b} yes {a, c} no {b, c} yes {a, b, c} choose b no no yes no yes no yes yes choose c no no no yes no yes yes yes Number of Subsets and Proper Subsets Number The number of subsets of any set is The Why is this correct? For each element of the set it is either contained or not contained in a subset. Two choices. n elements. Venn Diagrams Venn
Disjoint sets have no elements in common. The set B is a proper subset of A. The sets A and B have some common elements.
U A B U A U A B B Venn Diagrams Venn A general Venn Diagram looks like the general one below, with the understanding that the purple region might be empty or that one set might be inside the other. one A B Venn Diagrams Venn Consider the case the universe of {1, 2, ..., 8}, Consider with A = {1,2, 3, 4} and B = {2, 4, 6, 8}. Let's see how these values get placed in the Venn but 2 is but Diagram below:also in B! Diagram 4 is also in B
1 2 3 4 5 6 7 A
8 B Venn Diagrams Venn The area representing those elements of A The that don't belong to B is the region: is A B Venn Diagrams Venn The area representing those elements of B The that don't belong to A is the region: is A B Venn Diagrams Venn The area representing those elements that The are both in A and in B is: A B Venn Diagrams Venn The area representing those elements that The don't belong to either A or B is: A B Definition of Intersection of Sets Definition
The intersection of sets A and B, The intersection written A∩ B is the set of elements common to both set A and set B. This definition can be This expressed in set builder notation as follows: A ∩ B = { x | x ∈ A AND x ∈ B} Definition of Intersection of Sets Definition
The intersection of sets A and B The intersection A ∩ B = { x | x ∈ AAND x ∈ B}
what blinked both times? A B Definition of Union of Sets Definition
The union of sets A and B, written The union written A∪ B is the set of elements that are members of set A or of set B or of both sets. This definition can be expressed in set builder notation as follows: A ∪ B = { x | x ∈ A OR x ∈ B} Definition of Union of Sets Definition
The union of sets A and B The union A ∪ B = { x | x ∈ A OR x ∈ B} A B DeMorgan's Laws DeMorgan's Remember what happened when we Remember considered those don't belong to A or to B? That's the complement of A or B, namely (A ∪ B)' DeMorgan's Laws DeMorgan's
I Now, we blue red region below represents fThe combine these Venn Diagrams,' A' • Now, the region below represents B Therefore (region is = A' ∩ B' in )' The red + blue∪ B)' the same as purple A = purple, the purple since our previous slide: region represents A' ∩ B' DeMorgan's Laws DeMorgan's DeMorgan's Laws state that and (A ∪ B)' = A' ∩ B' )' (A ∩ B)' = A' ∪ B' )' Definition of the Cartesian Product of Sets Sets
The Cartesian product of sets A and B, The Cartesian written Ae B ˆ is the set of all pairs of elements taken from A and B. This definition can be expressed in This set builder notation as follows: A e B = { (x, y) | e x ∈ A and y ∈ B} ˆ ˆ Definition of the Cartesian Product of Sets Sets Countable and Uncountable Sets Countable
A denumerable set is called countable. denumerable countable The set of all integers is also denumerable. The (What is the mapping?) The set of rational numbers is also denumerable. (What is the mapping?) denumerable. Countable and Uncountable Sets Countable
Speaking informally, there are two types of Speaking "infinity." An infinite set is one whose infinite cardinality is boundless, (for example, the integers, the real numbers, the set of all strings over the alphabet {a,b}). A set is denumerable if we can place it into denumerable a one-to-one correspondence with the nonone-to-one negative integers. Clearly the non-negative negative integers is a denumerable set. integers Countable and Uncountable Sets Countable
The set of rational numbers is also denumerable. The (What is the mapping?) (What 1/1 1/2 1/3 1/4 1/5 … 1/1 2/1 2/2 2/3 2/4 2/5 … 2/1 3/1 3/2 3/3 3/4 3/5 … 3/1 4/1 4/2 4/3 4/4 4/5 … 4/1 . .. .. . .. .. . .. .. Venn Diagrams - Two Sets Venn
RegionIV:In Inautb B B in Region III: A nd o I: Not The RegionregionsinofnotnotVenn Diagram four II: In A bB Autr in B A the A I II B III IV Venn Diagrams - Three Sets Venn
The eight regionsCofutCnotBinCorC C Region III: InNotbnd Andutrnot or Diagram RegionIV: In AInin a , b in and in B Region I: V:B a ut the o BA CA Region In b not Venn VIII: VI: A A B B A VII: II: A I II B IV V III VI VII
C VIII Example: Blood Typing Example: Blood is characterized by examining Blood components called antigens. antigens Two of these antigens are called type A Two type and type B. type We name a person's blood on whether We or not they have these antigens: or A: has antigen type A has B: has antigen type B has AB: has both O: has neither. has Example: Blood Typing Example:
Look at the diagram below:
has type A has type B has both A and B has neither A nor B A AB O B Example: Blood Typing Example: But there's a third antigen as well: the But Rh antigen. Rh antigen. Blood with Rh is said to be positive: + Blood Rh Blood without Rh is said to be negative: Blood Rh - Example: Blood Typing Example:
Look at the diagram below: A A ABB BAB AAB+ B+ A+ O O+ ORh B + Example: Blood Typing Example:
When you receive blood, your blood must have all the antigens found in the donor's blood. Who can receive ANY type of blood? A AA+ O- ABAB+ B BB+ O+ Rh Universal acceptor: AB+ Example: Blood Typing Example:
When you donate blood, the acceptor's blood must have all the antigens found in your blood. Who can donate to everyone? A AA+ O- ABAB+ B BB+ O+ Rh Universal donor: O- Cardinal Number of the Union of Two Sets Two Suppose a class has 16 students with brown Suppose hair and that it has 12 students who wear glasses. glasses. How many students in the class either have How brown hair or wear glasses? brown Since a student in this group can either have Since brown hair (16 students) or wear glasses (12 students) a good guess is that there are 16 + 12 = 28 students in this group. 12 Let's count! Everyone with brown hair, blink for the Everyone haven't glasses, blink for me. Everyone I wearingmentioned, AND have Blink for me if you wear glasses leaveme. room. brown hair. 11 1 4 3 6 4 94 2 7 13 7 19 12 5 10 10 14 11 4 13 20 11 36 8 15 8 21 14 16 9 22 15 2 2 3 5 5 7 12 9 17 11 18 10 + 23 5 16 28 12 23 8 6 2 3 Cardinal Number of the Union of Two Sets Two What happened there? When counting heads, we counted the When intersection twice, first as having brown hair and then again as wearing glasses. hair We have to take that into account. We We subtract that number in the We intersection from the total. intersection Cardinal Number of the Union of Two Sets Two The number of students who have brown The hair or who wear glasses is the UNION of or two sets. (Remember, or means union.) or union The number of students who have brown The hair and who wear glasses is the and INTERSECTION of two sets. INTERSECTION (Remember, and means intersection.) (Remember, and intersection Cardinal Number of the Union of Two Sets Two
brown hair 5 glasses 11 16 12 7 n(brown hair) = 16, n(glasses) = 12, Region 1: 11, Region II: 5 and n(brown hair e glasses) = 5 Region III: 7. 11 + 5 + 7 = 23 students 16 + 12 – 5 = 23 students. Solving Survey Problems Solving
A class has 28 students. Of the 15 class female students, 8 wear glasses. Half the class (14 students) wear glasses. How many students are either male or don’t wear glasses? don’t Venn Diagrams Venn Female15 14 Glasses 8 6 7 7 28 students 15 female students 14 wear glasses 8 female students wear glasses Venn Diagrams Venn Female 8 7 7 Glasses 6 students who are male or don't wear glasses include all the male students (7+6, or 28 - 15) and all the female students who don't wear glasses (7) = Or Use DeMorgan's Law Or Male = not Female = F ' don’t wear glasses = G ' don’t F ' ∪ G' = (F ∩ G)' |F ∩ G) = 8 |(F ∩ G)') = 28 – 8 = 20 Venn Diagrams - Three Sets Venn
Region III: In AInnd anotutinA, rCor C RegionIV: In BC ut not in andBr C Region I: V:A a butnd B BA in A Region In b A C b A o VI: VII: II: The set of elements Bnotinnot o CB A I
VIII II B IV V III VI VII
C Venn Diagrams - Three Sets Venn
Region I: In A but not in B or C Region II: In A and B but not in C Region III: In B but not in A or C Region IV: In A and C but not in B Region V: In A and B and C Region VI: In B and C but not in A Region VII: In C but not in A or B Region VIII: Not in A, B or C U A
I IV II V VII VI B
III VIII C Solving Survey Problems Solving
1. Sixty people were contacted and responded Sixty to a movie survey. The following results were obtained. were
1. 2. 3. 4. 5. 6. 7. 6 people liked comedies, dramas AND sci-fi. 13 people liked comedies and dramas. 10 people liked comedies and sci-fi. 11 people liked dramas and sci-fi. 26 people liked comedies. 21 people liked dramas. 25 people liked sci-fi. Solving Survey Problems Solving
60 people took the survey
1. 6 people liked comedies, dramas AND scifi. 2. 13 people liked comedies and dramas. 3. 10 people liked comedies and sci-fi. 4. 11 people liked dramas and sci-fi. 5. 26C people likedD comedies. start in the work towards U 16 60 = 9 ctenter. he outside 9 7 liked dramas. + 7 6. 21 people 3 +3+4+6 6 liked sci-fi. +15= 61+ + 21 = 7.+ 25 people 5 26 =7 +5 7 4 1 + 05
25 = 4 + 5 +6+30 4+6+9 +6+1 10 SF 10 = 6 + 4 13 = 6 + 7 16 Solving Survey Problems Solving
1. 6 people liked comedies, dramas AND scifi. 2. 13 people liked comedies and dramas. 3. 10 people liked comedies and sci-fi. How many How many 4. 11 people liked dramas and sci-fi. people liked people don't 5. 26 people liked comedies. only movies atof like one type U 16 21C C 6. people likedD dramas. mll? 7 a ovie? 9 3 7. 25 people liked sci-fi. 6 9 + 3 + 10 =
4 5 10 SF 16 22 Formula for the Cardinal Number of the Union of Two Sets of
Principle of Inclusion and Exclusion Principle (Two Sets) |A e B|| = |A| + |B| - |A ˆ B|| ˆB |B |A B
To find the cardinal number in the union of To sets A and B, add the number of elements add in sets A and B and then subtract the number of elements common to both sets. Formula for the Cardinal Number of the Union of Three Sets of
Principle of Inclusion and Exclusion Principle (Three Sets) ||A e B e C|| = |A| + |B| + |C| Aˆ ˆ C |A| |B |C - |A e B| - |A e C| - |B e C| ˆ ˆ ˆ + |A e B e C| ˆ ˆ Counting Counting
Addition Principle If two events are mutually exclusive, that If mutually that is the events do not overlap, then there are n + m ways of performing one or the other event. other Counting Counting
Multiplication Principle Whenever two independent events are to be Whenever performed in sequence, if there are n ways of performing the first and m ways of performing the second, there are ne m ways ˆ of performing the sequence. of Example Example
Car manufacturers are now experimenting Car with lightweight three-wheeled cars, designed for a driver and one passenger, and considered ideal for city driving. Suppose you could order such a car with a choice of 9 possible colors, with or without air-conditioning, with or without a removable roof, and with or without an onboard computer. In how many ways can this car be ordered in terms of options? options? Solution Solution
This situation involves making choices This with four groups of items. with
color - air-conditioning - removable roof color computer computer 9 × 2 × 2 × 2 = 72 Thus the car can be ordered in 72 different Thus ways. ways. Example A Multiple Choice Test Example
You are taking a multiple-choice test that You has ten questions. Each of the questions has four choices, with one correct choice per question. If you select one of these options per question and leave nothing blank, in how many ways can you answer the questions? the Solution Solution
We use the Fundamental Counting We Principle to determine the number of ways you can answer the test. Multiply the number of choices, 4, for each of the ten questions ten 4× 4× 4× 4× 4× 4× 4× 4× 4× 4 =1,048,576 =1,048,576 Permutations Permutations A permutation is an arrangement of permutation arrangement objects. objects.
No item is used more than once. The order of arrangement makes a The difference. difference. Example Counting Permutations Example
You need to arrange seven of your You favorite books along a small shelf. How many different ways can you arrange the books, assuming that the order of the books makes a difference to you? books Solution Solution
You may choose any of the seven books for You the first position on the shelf. This leaves six choices for second position. After the first two positions are filled, there are five books to choose from for the third position, four choices left for the fourth position, three choices left for the fifth position, then two choices for the sixth position, and only one choice left for the last position. last 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 There are 5040 different possible permutations. Factorial Notation Factorial
If n is a positive integer, the notation n! is If is the product of all positive integers from n down through 1. down n! = n(n-1)(n-2)…(3)(2)(1) note that 0!, by definition, is 1. 0!=1 Permutations of n Things Taken r at a Time The number of permutations possible if The r items are taken from n items:
n! (n – r)!
n! = n(n – 1) (n – 2) (n – 3) . . . (n – r + 1) (n - r) (n - r - 1) . . . (2)(1) (n – r)! = (n - r) (n - r - 1) . . . (2)(1) Problem Problem
A math club has eight members, and it must math choose 5 officers --- president, vice-president, secretary, treasurer and student government representative. Assuming that each office is to be held by one person and no person can hold more than one office, in how many ways can those five positions be filled? positions
We are arranging 5 out of 8 people into the five distinct offices. Any of the eight can be president. Once selected, any of the remaining seven can be vice-president. Clearly this is an arrangement, or permutation, problem.
85 P = 8!/(8-5)! = 8!/3! = 8 · 7 · 6 · 5 · 4 = 6720 Permutations with duplicates. Permutations In how many ways can you arrange the In letters of the word minty? minty That's 5 letters that have to be arranged, That's so the answer is 5P5 = 5! = 120 But how many ways can you arrange the But letters of the word messes? messes You would think 6!, but you'd be wrong! Permutations with duplicates. Permutations How many ways can you arrange the How letters of the word messes? messes The problem is that there are three s''s and s 2 e''s. It doesn't matter in which order the s. s''s are placed, because they all look the s same! same! This is called permutations with This duplicates. duplicates. Permutations with duplicates. Permutations Well, there are 3! = 6 ways to arrange the Well, s''s. So there will be 6 permutations that s. should count as one. Same with the e''s. s. There are 2! = 2 permutations of them that should count as 1. that So we divide 6! by 3! and also by 2! There are 6!/3!2! = 720/12 = 60 ways to There arrange the word messes. messes Permutations with duplicates. Permutations In general if we want to arrange n items, of In n! m1! m2 ! m3 ! Problem Problem
A signal can be formed by running signal different colored flags up a pole, one above the other. Find the number of different signals consisting of 6 flags that can be made if 3 of the flags are white, 2 are red, and 1 is blue white,
6!/3!2!1! = 720/(6)(2)(1) = 720/12 = 60 A combination of items occurs when: combination of The item are selected from the same The group. group. No item is used more than once. The order of the items makes no The difference. difference. How to know when the problem is a permutation problem or a combination problem problem Permutation: Permutation:
arrangement, arrange arrangement, order matters Combination Combination selection, select order does not matter. Example Distinguishing between Permutations and Combinations Permutations
For each of the following problems, explain For if the problem is one involving permutations or combinations. permutations Six students are running for student Six government president, vice-president, and treasurer. The student with the greatest number of votes becomes the president, the second biggest vote-getter becomes vice-president, and the student who gets the third largest number of votes will be student government treasurer. How many different outcomes are possible for these three positions? these Solution Solution Students are choosing three student Students government officers from six candidates. The order in which the officers are chosen makes a difference because each of the offices (president, vice-president, treasurer) is different. Order matters. This is a problem involving permutations. This Example Distinguishing between Permutations and Combinations Permutations Six people are on the volunteer board Six of supervisors for your neighborhood park. A three-person committee is needed to study the possibility of expanding the park. How many different committees could be formed from the six people on the board of supervisors? supervisors? Solution Solution A three-person committee is to be formed three-person from the six-person board of supervisors. The order in which the three people are selected does not matter because they are not filling different roles on the committee. Because order makes no difference, this is a problem involving combinations. combinations. Example Distinguishing between Permutations and Combinations Permutations Baskin-Robbins offers 31 different Baskin-Robbins flavors of ice cream. One of their items is a bowl consisting of three scoops of ice cream, each a different flavor. How many such bowls are possible? possible? Solution Solution A three-scoop bowl of three different flavors is three-scoop to be formed from Baskin-Robbin’s 31 flavors. The order in which the three scoops of ice cream are put into the bowl is irrelevant. A bowl with chocolate, vanilla, and strawberry is exactly the same as a bowl with vanilla, strawberry, and chocolate. Different orderings do not change things, and so this problem is combinations. combinations. Combinations of n Things Taken r at a Time Time
n r
n! r!(n – r)! =C= Problem Problem A three-person committee is to be formed three-person from the eight-person board of supervisors.
We saw that this is a combination problem 8! 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 8 × 7 × 6 = = = 56 8 C3 = 3!5! 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1 3 × 2 × 1 Problem Problem A jeweller is creating a ring that has a jeweller cluster of five gems on it. He's not really concerned about exactly where each jewel will go. The jewels will be rubies, emeralds and diamonds, taken from an unlimited supply of each unlimited Solution Solution This is a difficult problem and requires a This bit of serendipity. Remember that from the previous chapter? the Think of the five gems to be chosen being laid out in a line on a table, with all diamonds on the left, rubies in the middle and emeralds on the right: and Solution Solution Lay out any five gems on the table, with Lay diamonds, rubies and emeralds going from left to right. from Now lay down partitions to separate the three types of gems. three Solution Solution There are two partitions needed and any There different partition placements represents different numbers of the three types of jewels. The above example represents one diamond, three rubies and one emerald. diamond, Solution Solution Therefore this is a permutation Therefore problem with five indistinguishable circles and two indistinguishable lines. circles The formula is therefore: 7! 5!2! There are 21 different possibilities Solution Solution In general, creating such a ring with n gems In choosing from r different kinds of gems (with an unlimited supply) is an ( n + r − 1)! n! ( r − 1)! Computing Theoretical Probability Probability
number of outcomes in event E P( E ) = number of outcomes in sample space S Example Computing Theoretical Probability Probability A die is rolled once. Find the Probability die of getting a number less than 5. of Solution Solution
The event of getting a number less than 5 The can occur in 4 ways: 1, 2, 3, 4. can P(less than 5) =
(number of ways a number less than 5 can occur) (total number of possible outcomes) = 4/6 = 2/3 4/6 Example Probability and a Deck of 52 Cards Cards
You are dealt one card from a standard 52card deck. Find the probability of being card dealt a King. dealt Solution Solution
Because there are 52 cards, the total number of possible ways of being dealt a single card is 52. We use 52, the total number of possible outcomes, as the number in the denominator. Because there are 4 Kings in the deck, the event of being dealt a King can occur 4 ways. being P(King) = 4/52 = 1/13 Empirical Probability Empirical
observed number of times E occurs P( E ) = total number of observed occurrences Example Computing Empirical Probability Probability There are approximately 3 million Arab Americans in There America. The circle graph shows that the majority of Arab Americans are Christians. If an Arab American is selected at random, find the empirical probability of selecting a Catholic. selecting Solution Solution
The probability of selecting a Catholic is the The observed number of Arab Americans who are Catholic, 1.26 (million), divided by the total number of Arab Americans, 3 (million). number P(selecting a Catholic from the Arab P(selecting American Population) = 1.26/3 = 0.42 American Example Probability and Combinations Example A club consists of five men and seven club women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of: consists three men. one man and two women. a. b. Solution Solution
We begin with the probability of selecting We three men.
P( 3 men)=number of ways of selecting 3 men total number of possible combinations
12 5 C3 = 12!/((12-3)!3!) = 220 C3 = 5!/((5-3)!(3!)) = 10 P(3 men) = 10/220 = 1/22 Solution part b cont. Solution
There are 5 men. We can select 1 man in 5C1 ways. There ways There are 7 women. We can select 2 women in 7C2 There ways. ways. By the Fundamental Counting Principle, the number By of ways of selecting 1 man and 2 women is of
5 C1 × 7C2 = 5 × 21 = 105 Now we can fill in the numbers in our probability Now fraction. P(1 man and 2 women) = number of ways of selecting 1 man and 2 women number total number of possible combinations = 105/220 = 21/44 Thinking Mathematically Mathematically
Events Involving Not and Or; Events Odds Odds The Probability of an Event Not Occurring The The probability that an event E will not The occur is equal to one minus the probability that it will occur. probability P(not E) = 1 - P(E) (not Mutually Exclusive Events Mutually
If it is impossible for events A and B to If occur simultaneously, the events are said to be mutually exclusive. mutually
If A and B are mutually exclusive events, then P(A or B) = P(A) + P(B). Or Probabilities with Events That Are Not Mutually Exclusive Mutually If A and B are not mutually exclusive If events, then events, P(A or B) = P(A) + P(B) - P(A and B) Example Example
In picking a card from a standard deck, 1. What is the probability of picking a red King? 1. red A red King is either a King of Hearts or a King of Diamonds. red King King of Hearts King of Diamonds Since a card can't be both a Heart and a Diamond, the Heart Diamond the probability is 1/52 + 1/52 = 1/26. probability 2. What is the probability of not picking a red King? 2. not red King Just 1 - the answer to part 1, 1 - 1/26 = 25/26 Just 3. What is the probability of picking a Heart or a face card (a Heart face card face card is either a Jack, Queen or King)? Jack Queen King There are 13 Hearts, so P(Heart) = 13/52. There are 12 face There Hearts so P(Heart cards, so P(Face Card) = 12/52. There are three cards that are P(Face Card both a Heart and a Face Card. P(Heart or Face Card) = Heart Face Card P(Heart or Face Card 13/52 + 12/25 – 3/52 = 22/52 13/52 Playing Poker Playing In poker, played with a regular 52 card deck, the hands arranged from best to worst are: the straight flush: consecutive cards of the same suit. four of a kind full house: two of one kind, three of another. flush: cards of the same suit, but not consecutive. flush: cards straight: consecutive cards, not all of the same suit. straight: consecutive three of a kind (and two other different cards) three (and two of a kind (and three other different cards) two (and What is the probability of getting each possible What hand? hand? Playing Poker Playing First of all there are 52C5 First 52 Straight flush: = 2,598,960 possible individual hands.
Choose one suit out of the four. 4 possibilities Choose the high card in the hand, from Ace down to Choose five. 10 possibilities. five. By the principle of multiplication are are 40 By possible straight flushes possible P(straight flush) = 40/ 52C5 = 40/2598960 = .0015% P(straight 40/2598960 52 Playing Poker Playing First of all there are 52C5 possible First 52 individual hands. Four of a kind:
Choose one of thirteen possible cards (from 2 up to Ace) Choose one other card from the remaining 48. By the principle of multiplication are are 13 ˆ 48 = 624 possible four of a kinds. 624 P(straight flush) = 624/ 52C5 P(straight 52 It is about 15 times more likely to get four of a It kind than a straight flush. kind Playing Poker Playing First of all there are 52C5 possible individual First 52 hands. Full house: Choose one of 13 possible cards (from 2 up to Ace) to get Choose your three of a kind. your Choose one of 12 possible cards (from 2 up to Ace) to get Choose your two of a kind. your There are 4C3 possibilities within the suits for your three There of a kind and 4C2 possibilities of your two of a kind. There are 13 ˆ 12 ˆ 12
4 C3 ˆ 4C2 = 3744 possible full houses. 3744 Playing Poker Playing First of all there are 52C5 possible individual First 52 hands. Flush: Choose one of the four suits. Choose 5 of the 13 cards of Choose that suit. However, we have seen that 40 of these choices are the better hand of a straight flush are 4 ˆ 13C5 – 40 = 4 ˆ 1287 – 40 = 5108 Playing Poker Playing First of all there are 52C5 possible individual First 52 hands. Straight: Choose the high card (Ace down to 5 of any suit). 40 Choose possibilities. possibilities. 4 possibilities for the next lower card, 4 for the next, 4 for possibilities the next and 4 for the last. 40 ˆ 4 ˆ 4 ˆ 4 ˆ 4. However, 40 of these are straight flushes. of 40 ˆ 4 ˆ 4 ˆ 4 ˆ 4 - 40 =10200 possible straights. 40 Independent Events Independent
Two events are independent events if the Two independent occurrence of either of them has no effect on the probability of the other. on And Probabilities with Independent And Events Events
• If A and B are independent events, then If P(A and B) = P(A) ˆ P(B) and The example of four pairs of socks and The three pairs of shoes (= 12 possible combinations) is an example of independent events. independent • Dependent Events Dependent
Two events are dependent events if the Two dependent occurrence of one of them does have an effect on the probability of the other. effect And Probabilities with Dependent And Events Events If A and B are dependent events, then If are P(A and B) = P(A) e P(B given that A has occurred) ˆ given has written as written P(A) e P(B|A) ˆ Conditional Probability Conditional
• The conditional probability of B, given The conditional given A, written P(B|A), is the probability that written event B will occur computed on the assumption that event A has occurred. Notice that when the two events are independent, P(B|A) = P(B). • Conditional Probability Conditional Example: Suppose you are picking two cards from a deck Suppose of cards. What is the probability you will pick a King and then another face card? 1 King face card 4 The probability of an King is 52 = 13 . The King Once the King is selected, there are 11 face cards Once King left in a deck holding 51 cards. left 1 11 P(A) = P(A) 13 51 . P(B|A) = P(B|A) 1 11 13 51 Applying Conditional Probability to RealApplying World Data P(B|A) =
observed number of times B and A occur together observed and occur observed number of times A occurs Review Review
P(not E) P(A or B) P(A and B) 1 – P(E) P(A) + P(B) – P(A and B) P(A) + P(B) P(A) e P(B|A) P(A) e P(B) Expected Value Expected
Expected value is a mathematical way to use probabilities to determine what to expect in various situations over the long run. over For example, we can use expected value to find the outcomes of the roll of a fair dice. outcomes The outcomes are 1, 2, 3, 4, 5, and 6, each with a The probability of 1 . The expected value, E, is computed 6 by multiplying each outcome by its probability and then adding these products. then E = 1e 1 + 2e 1 + 3e 1+ 4e 1+ 5e 1 6e 1 + ˆ6 ˆ6 ˆ6ˆ6ˆ 6ˆ 6 = (1+2+3+4+5+6)/6 = 21 = 3.5 (1+2+3+4+5+6)/6 6 Expected Value Expected
E = 1e 1 + 2e 1+ 3e 1+ 4e 1 5e + 6e 1 + ˆ6 ˆ6 ˆ 6 ˆ 6 ˆ 1 ˆ 6 6 21 = (1 + 2 + 3 + 4 + 5 + 6)/6 = 6 = 3.5 Of course, you can't roll a 3½ . But the Of average value of a roll of a die over a long period of time will be around 3½. Example Expected Value and Roulette Example
A roulette wheel has 38 different roulette "numbers." "numbers." One way to bet in roulette is to place $1 on a single number. If the ball lands on that number, you are awarded $35 and get to keep the $1 that you paid to play the game. If the ball lands on any one of the other 37 slots, you are awarded nothing and the $1 you bet is collected. Example Expected Value and Roulette Example 38 different numbers. If the ball lands on your number, If you win awarded $35 and you keep the $1 you paid to play the game. If the ball lands on any of the other If 37 slots, you are awarded nothing and you lose the $1 you bet. and Find the expected value for playing roulette if you bet $1 on number 11 every time. Describe what this means. Solution Solution
Outcome Gain or Loss Probability
1 38 37 38 11 Not 11 $35
− $1 1 ) + (-$1)( 37) E = $35( 38 38 35 37 2 = $38 - $ 38 = -$ 38 ≈ -$0.05 This means that in the long run, a player can This expect to lose about 5 cents for each game played. Expected Value Expected A real estate agent is selling a house. real gets a 4-month listing. There are 3 possibilities: possibilities: She she sells the house: she another agent sells the house: house not sold: (30% chance) earn $25,000 (30% (20% chance) earn $10,000 (50% chance) lose $5,000 What is the expected profit (or loss)? If the expected profit is at least $6000 she If would consider it a good deal. would Expected Value Expected
Outcome she sells other sells doesn't sell Probability 0.3 0.2 0.5 Profit or loss +$25,000 +$10,000 -$5,000 product +$7,500 +$2,000 -$2,500 +$7,000
The realtor can expect to make $7,000 . Make the deal!!!! ...

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