Thinking Mathematically

Thinking Mathematically
Chapter 11: Counting Methods and Probability Theory

Section 1: The Fundamental Counting Principle

The Fundamental Counting Principle
If you can choose one item from a group of M items and a second item from a group of N items, then the total number of two-item choices is M  N. You the numbers! MULTIPLY

At breakfast, you can have eggs, pancakes or cereal. You get a free juice with your meal: either OJ or apple juice. How many different breakfasts are possible? eggs cereal pancakes apple OJ 1 2 OJ apple 3 4 apple OJ 5 6

Example: Applying the Fundamental Counting Principle
The Greasy Spoon Restaurant offers 6 appetizers and 14 main courses. How many different meals can be created by selecting one appetizer and one main course? Using the fundamental counting principle, there are 14  6 = 84 different ways a person can order a two-course meal.

Example: Applying the Fundamental Counting Principle
This is the semester that you decide to take your required psychology and social science courses. Because you decide to register early, there are 15 sections of psychology from which you can choose. Furthermore, there are 9 sections of social science that are available at times that do not conflict with those for psychology. In how many ways can you create two-course schedules that satisfy the psychology-social science requirement?

Solution The number of ways that you can satisfy the requirement is found by multiplying the number of choices for each course. You can choose your psychology course from 15 sections and your social science course from 9 sections. For both courses you have: 15  9, or 135 choices.

The number of ways a series of successive things can occur is found by multiplying the number of ways in which each thing can occur.

Example: Options in Planning a Course Schedule
Next semester you are planning to take three courses - math, English, and humanities. Based on time blocks and highly recommended professors, there are 8 sections of math, 5 of English, and 4 of humanities that you find suitable. Assuming no scheduling conflicts, there are: 8  5  4 = 160 different three course schedules.

Example Car manufacturers are now experimenting 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?

Solution This situation involves making choices with four groups of items. color - air-conditioning - removable roof - computer 9  2  2  2 = 72 Thus the car can be ordered in 72 different ways.

Example: A Multiple Choice Test
You are taking a multiple-choice test that 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?

Solution We DON’T blindly multiply the first two numbers we see. The answer is not 10  4 = 40. We use the Fundamental Counting Principle to determine the number of ways you can answer the test. Multiply the number of choices, 4, for each of the ten questions 4  4  4  4  4  4  4  4  4  4 =1,048,576

Example: Telephone Numbers in the United States
Telephone numbers in the United States begin with three-digit area codes followed by seven-digit local telephone numbers. Area codes and local telephone numbers cannot begin with 0 or 1. How many different telephone numbers are possible?

Solution We use the Fundamental Counting Principle to determine the number of different telephone numbers that are possible. 8  10  10  8  10  10  10  10  10  10 =6,400,000,000

Section 2: Permutations

Permutations A permutation is an arrangement of objects.
No item is used more than once. The order of arrangement makes a difference.

Example: Counting Permutations
Based on their long-standing contribution to rock music, you decide that the Rolling Stones should be the last group to perform at the four-group Offspring, Pink Floyd, Sublime, Rolling Stones concert. Given this decision, in how many ways can you put together the concert?

Solution We use the Fundamental Counting Principle to find the number of ways you can put together the concert. Multiply the choices: 3  2  1  1 = 6 Thus, there are six different ways to arrange the concert if the Rolling Stones are the final group to perform. 3 choices 2 choices 1 choice 1 choice whichever of the two remaining offspring pink floyd sublime only one remaining stones

Example: Counting Permutations
You need to arrange seven of your 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?

There are 5040 different possible permutations.
Solution You may choose any of the seven books for 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. 7  6  5  4  3  2  1 = 5040 There are 5040 different possible permutations.

Factorial Notation If n is a positive integer, the notation n! is the product of all positive integers from n down through 1. 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 r items are taken from n items: n! nPr = = n(n – 1) (n – 2) (n – 3) (n – r + 1) (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)

Permutations of n Things Taken r at a Time
The number of permutations possible if r items are taken from n items: nPr: starting at n, write down r numbers going down by one: nPr = n(n – 1) (n – 2) (n – 3) (n – r + 1) 1 2 3 4 r

Problem A math club has eight members, and it must 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? 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. 8P5 = 8!/(8-5)! = 8!/3! = 8 · 7 · 6 · 5 · 4 = 6720

Permutations with duplicates.
In how many ways can you arrange the letters of the word minty? That's 5 letters that have to be arranged, so the answer is 5P5 = 5! = 120 But how many ways can you arrange the letters of the word messes? You would think 6!, but you'd be wrong!

messes m m e e s s s s e e s s m e s s e s m e s s e s m e s s e s m e
here are six permutations of messes m m e e s s s s e e s s 1 well, all 3! arrangements of the s's look the same to me!!!! This is true for any arrangement of the six letters in messes, so every six permutations should count only once. The same applies for the 2! arrangement of the e's m e s s e s 2 m e s s e s 3 m e s s e s 4 m e s s e s 5 6

How many ways can you arrange the letters of the word messes? The problem is that there are three s's and 2 e's. It doesn't matter in which order the s's are placed, because they all look the same! This is called permutations with duplicates.

Since there are 3! = 6 ways to arrange the s's, there are 6 permutations that should count as one. Same with the e's. There are 2! = 2 permutations of them that should count as 1. So we divide 6! by 3! and also by 2! There are 6!/3!2! = 720/12 = 60 ways to arrange the word messes.

In general if we want to arrange n items, of which m1, m2, .... are identical, the number of permutations is

Problem A signal can be formed by running 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 6!/3!2!1! = 720/(6)(2)(1) = 720/12 = 60

Section 3: Combinations

Combination: definition
A combination of items occurs when: The item are selected from the same group. No item is used more than once. The order of the items makes no difference.

How to know when the problem is a permutation problem or a combination problem
arrangement, arrange order matters Combination selection, select order does not matter.

Example: Distinguishing between Permutations and Combinations
For each of the following problems, explain if the problem is one involving permutations or combinations. Six students are running for student 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?

Solution Students are choosing three student 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.

Six people are on the volunteer board 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?

Solution A three-person committee is to be formed 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.

Baskin-Robbins offers 31 different 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?

Solution A three-scoop bowl of three different flavors is 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 of n Things Taken r at a Time
= nCr = r!(n – r)! r Note that the sum of the two numbers on the bottom (denominator) should add up to the number on the top (numerator).

Computing Combinations
Suppose we need to compute 9C3 r = 3, n – r = 6 The denominator is the factorial of smaller of the two: 3!

Suppose we need to compute 9C3 r = 3, n – r = 6 In the numerator write (the product of) all the numbers from 9 down to n - r + 1 = = 7: There should be the same number of terms in the numerator and denominator: 9  8  7

If called upon, there's a fairly easy way to compute combinations. Given nCr , decide which is bigger: r or n – r. Take the smaller of the two and write out the factorial (of the number you picked) as a product. Draw a line over the expression you just wrote.

If called upon, there's a fairly easy way to compute combinations. Now, put n directly above the line and directly above the leftmost number below. Eliminate common factors in the numerator and denominator. Do the remaining multiplications. You're done!

Suppose we need to compute 9C3 . n – r = 6, and the smaller of 3 and 6 is 3. 3 4 9  8  7 = 3  4  7 = 84 3  2  1 1 1

Problem A three-person committee is to be formed from the eight-person board of supervisors. We saw that this is a combination problem 2 = 8  7 = 56

A deck of cards suits ranks aces face cards

Problem How many poker hands (five cards) are possible using a standard deck of 52 cards? We need to pick any five cards from the deck. Therefore we are selecting 5 out of 52 cards.

Problem A poker hand of a full house consists of three cards of the same rank (e.g. three 8's or three kings or three aces.) and two cards of another rank. How many full houses are possible? First we need to select 1 out of the 13 possible ranks. 13 Then we need to select 3 out of the 4 cards of that rank. 4C3 = 4 By the fundamental counting principle, there are 13  4 = 52 ways to do this part of the problem.

Problem Similarly, we then need to select 1 out of the 12 remaining ranks. 12 Then select 2 out of the 4 cards of that rank. 4C2 = 4!/2!2! = 24/4 = 6 Once again by the fundamental counting principle, there are 12  6 = 72 ways to solve this part of the problem.

Problem Finally, in order to select a full house, we merely have to choose any of the 52 possible threes of a kind and choose any of the remaining 72 pairs. Therefore there are 52  72 = 3744 full houses!!!

Another Problem In poker, a straight is
5 cards in sequence, for example 3, 4, 5, 6 and a 7 8, 9, 10, jack, queen ace, 2, 3, 4, 5 (ace as low card) 10, jack, queen, king, ace. (ace as high card) and not all of the same suit (that's a straight flush!) How many possible straights can you be dealt? Think about it. I won't give the answer now.

Another Problem One part of the solution is straightforward. In how many ways can you select five cards in sequence? the low card of the sequence can not be greater than Getting a 10 as low card means that the high card is an ace. You can't go higher than ace! there are 4 possible suits for each card. So there are 10  1024 = 10,240 ways you can do this. 4  4  4  4  4 =1024

Another Problem But some of those include the ones where all the cards are of the same suit. Remember, there are 10 possible low cards There are 4 possible suits for that low card Once we have chosen one of those 40 possible cards, there is exactly one possibility for the next four cards. 10,240 – 40 = 10,200 That's the answer

Section 4: Fundamentals of Probability

Computing Theoretical Probability

Remember A probability can never be greater than 1 or less than 0.

Example: Computing Theoretical Probability
A die is rolled once. Find the probability of getting a number less than 5. Sample space: all possible outcomes: 1, 2, 3, 4, 5, 6 Event: the roll yields a number less than 5: 1, 2, 3, 4

Solution The event of getting a number less than 5 can occur in 4 ways: 1, 2, 3, 4. P(less than 5) = = 4/6 = 2/3

Example: Probability and a Deck of 52 Cards
You are dealt one card from a standard 52-card deck. Find the probability of being dealt a King.

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. P(King) = 4/52 = 1/13

Empirical Probability

Example: Computing Empirical Probability
There are approximately 3 million Arab Americans in 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.

Solution The probability of selecting a Catholic is the observed number of Arab Americans who are Catholic, 1.26 (million), divided by the total number of Arab Americans, 3 (million). P(selecting a Catholic from the Arab American Population) = 1.26/3 = 0.42

Section 5: Probability with the Fundamental Counting Principle, Permutations, and Combinations

Example: Probability and Permutations
Five groups in a tour, Offspring, Pink Floyd, Sublime, the Rolling Stones, and the Beatles, agree to determine the order of performance based on a random selection. Each band’s name is written on one of five cards. The cards are placed in a hat and then five cards are drawn out, one at a time. The order in which the cards are drawn determines the order in which the bands perform. What is the probability of the Rolling Stones performing fourth and the Beatles last?

Solution We begin by applying the definition of probability to this situation. P(Rolling Stones fourth, Beatles last) = We can use the Fundamental Counting Principle to find the total number of possible permutations. 5  4  3  2  1 = 120

Solution cont. We can also use the Fundamental Counting Principle to find the number of permutations with the Rolling Stones performing fourth and the Beatles performing last. You can choose any one of the three groups as the opening act. This leaves two choices for the second group to perform, and only one choice for the third group to perform. Then we have one choice for fourth and last. 3  2  1  1  1 = 6 There are six lineups with Rolling Stones fourth and Beatles last.

Solution cont. Now we can return to our probability fraction.
P(Rolling Stones fourth, Beatles last) = = 6/120 = 1/20 The probability of the Rolling Stones performing fourth and the Beatles last is 1/20.

Example: Probability and Combinations
A club consists of five men and seven women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of: three men. one man and two women.

Solution We begin with the probability of selecting three men.
P(3 men)= 12C3 = 12!/((12-3)!3!) = 220 5C3 = 5!/((5-3)!(3!)) = 10 P(3 men) = 10/220 = 1/22

Solution part b cont. 5C1  7C2 = 5  21 = 105 = 105/220 = 21/44
There are 5 men. We can select 1 man in 5C1 ways. There are 7 women. We can select 2 women in 7C2 ways. By the Fundamental Counting Principle, the number of ways of selecting 1 man and 2 women is 5C1  7C2 = 5  21 = 105 Now we can fill in the numbers in our probability fraction. P(1 man and 2 women) = = 105/220 = 21/44

Section 6: Events Involving Not and Or; Odds

The Probability of an Event Not Occurring
The probability that an event E will not occur is equal to one minus the probability that it will occur. P(not E) = 1 - P(E)

Mutually Exclusive Events
If it is impossible for events A and B to occur simultaneously, the events are said to be mutually exclusive. 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
If A and B are not mutually exclusive events, then P(A or B) = P(A) + P(B) - P(A and B)

Examples In picking a card from a standard deck,
1. What is the probability of picking a red King? 2. What is the probability of not picking a red King? 3. What is the probability of picking a Heart or a face card (a face card is either a Jack, Queen or King)?

Example In picking a card from a standard deck,
1. What is the probability of picking a red King? A red King is either a King of Hearts or a King of Diamonds. Since a card can't be both a Heart and a Diamond, the events are mutually exclusive: the probability is 1/52 + 1/52 = 1/26.

2. What is the probability of not picking a red King? Since P(not A) = 1 – P(A) and we saw that P(A) = 1/26, the probability is 1 – 1/26 = 25/26.

3. What is the probability of picking a Heart or a face card (a face card is either a Jack, Queen or King)? Since some Hearts are face cards, the events are not mutually exclusive. There are 13 Hearts, so P(Heart) = 13/52. There are 12 face cards, so P(Face Card) = 12/52. There are 3 cards that are both a Heart and a Face Card. (Namely the Jack, Queen and King of Hearts). P(Heart or Face Card) = 13/ /52 – 3/52 = 22/52

Odds For some given event, we sometimes express the chances for an outcome by odds. The odds in favor of something happening is the ratio of the probability that it will happen to the probability that it won't. Odds in favor of E = When the Odds in favor of E are a to b, we write it as a:b.

Odds Just as we can talk about the odds in favor of something happening, we can also talk about the odds against something happening: Odds against E = When the Odds against E are b to a, we write it as b:a.

Odds Suppose you roll a pair of dice.
There are 6  6 = 36 possible outcomes. There are three ways of rolling an 11 or higher: a 5 and a 6, a 6 and a 5, and two 6's. The probability of rolling an 11 or higher is , or . The probability of not rolling an 11 or higher is 1 – , or .

Odds Suppose you roll a pair of dice.
The odds in favor of rolling an 11 or better are The odds against rolling an 11 or better are

Finding Probabilities from Odds
If the odds in favor of an event E are a to b, then the probability of the event is given by

If the odds against an event E are a to b, then the probability of the event is given by

Example: Suppose Bluebell is listed as 7:1 in the third race at the Meadowlands. The odds listed on a horse are odds against that horse winning, that is, losing. The probability of him losing is 7 / (7+1) = 7/8. The probability of him winning is 1/8.

Example: Suppose Bluebell is listed as 7:1 in the third race at the Meadowlands. (a:b against) The odds listed on a horse are odds against that horse winning, that is, losing. The probability of him losing is 7 / (7+1) = 7/8. The probability of him winning is 1/8.

Section 7: Events Involving And; Conditional Probability

Independent Events Two events are independent events if the occurrence of either of them has no effect on the probability of the other. For example, if you roll a pair of dice two times, then the two events are independent. What gets rolled on the second throw is not affected by what happened on the first throw.

And Probabilities with Independent Events
If A and B are independent events, then P(A and B) = P(A)  P(B) The example of choosing from four pairs of socks and then choosing from three pairs of shoes (= 12 possible combinations) is an example of two independent events.

Dependent Events Two events are dependent events if the occurrence of one of them does have an effect on the probability of the other. Selecting two Kings from a deck of cards by selecting one card, putting it aside, and then selecting a second card, is an example of two dependent events. The probability of picking a King on the second selection changes because the deck now contains only 51, not 52, cards.

And Probabilities with Dependent Events
If A and B are dependent events, then P(A and B) = P(A)  P(B given that A has occurred) written as P(A)  P(B|A)

Conditional Probability
The conditional probability of B, given A, written P(B|A), is the probability that 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
Example: Suppose you are picking two cards from a deck of cards. What is the probability you will pick a King and then another face card? The probability of an King is = Once the King is selected, there are 11 face cards left in a deck holding 51 cards. P(A) = P(B|A) = The probability in question is 

Applying Conditional Probability to Real-World Data
P(B|A) = observed number of times B and A occur together observed number of times A occurs

Review P(not E) 1 – P(E) P(A or B) P(A) + P(B) – P(A and B)
mutually exclusive: P(A) + P(B) P(A and B) P(A)  P(B|A) independent:P(A)  P(B) Odds in favor - a:b P(E) / P(not E) probability is a/(a+b) Odds against - a:b P(not E) / P(E) probability is b/(a+b)

Section 8: Expected Value

Expected Value E = 1 + 2 + 3 + 4 + 5 + 6
Expected value is a mathematical way to use probabilities to determine what to expect in various situations over the long run. For example, we can use expected value to find the outcomes of the roll of a fair dice. The outcomes are 1, 2, 3, 4, 5, and 6, each with a probability of . The expected value, E, is computed by multiplying each outcome by its probability and then adding these products. E = 1 + 2  + 4   = ( )/6 = = 3.5 1 6 1 6 1 6 1 6 1 6 1 6 1 6 21 6

Expected Value E = 1 + 2 + 3 + 4 + 5 + 6
= ( )/6 = = 3.5 Of course, you can't roll a 3½ . But the average value of a roll of a die over a long period of time will be around 3½. 1 6 21 6

Example Expected Value and Roulette
A roulette wheel has 38 different "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
38 different numbers. If the ball lands on your number, you win awarded $35 and you keep the $1 you paid to play the game. If the ball lands on any of the other 37 slots, you are awarded nothing and you lose the $1 you bet. Find the expected value for playing roulette if you bet $1 on number 11 every time. Describe what this means.

Solution E = $35( ) + (-$1)( ) = $ - $ = -$ ≈ -$0.05
Outcome Gain/Loss Probability 11 Not 11 E = $35( ) + (-$1)( ) = $ - $ = -$ ≈ -$0.05 This means that in the long run, a player can expect to lose about 5 cents for each game played.

Expected Value A real estate agent is selling a house. She gets a 4-month listing. There are 3 possibilities: she sells the house: (30% chance) earns $25,000 another agent sells the house: (20% chance) earns $10,000 house not sold: (50% chance) loses $5,000 What is the expected profit (or loss)? If the expected profit is at least $6000 she would consider it a good deal.

Expected Value Outcome Probability Profit or loss product she sells
0.3 +$25,000 other sells 0.2 +$10,000 doesn't sell 0.5 -$5,000 +$7,500 +$2,000 -$2,500 +$7,000 The realtor can expect to make $7,000. Make the deal!!!!

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