advanced functions and modeling -...

Advanced Functions and Modeling Workshop Summer 2004

Module 3 Probability

prepared by Jennifer Borchert Athena Kellogg Chuckie Hairston Courtney Overby

Kassidy McGee

Module 3: Probability 43

Chapter 3 3.1. North Carolina Standard

Course of Study....................................44 3.2. vocabulary .......................................................44 3.3. timeline & assignments..................44 3.4 introduction to mathematical

probability..........................................................46 3.4.1 permutations ......................................................46

3.4.2 permutations of objects not all different ...................................................................48

3.4.3 combinations.........................................................48

3.4.4 sample space........................................................49

3.4.5 probability..................................................................50

3.4.6 mutually exclusive.........................................50

3.4.7 independent events ................................51

3.4.8 dependent events.....................................52

3.4.9 expected value/fair games ........52

3.5 coin toss............................................................59 3.6. introductory activity..................62 3.7. genetics..............................................................64 3.8 hazardous waste in who’s

backyard?........................................................65 3.9. geometric probability......................66 3.10. mutually exclusive /

independent / dependent.......70 3.11. permutations................................................70 3.12 combinations..................................................70 3.13 probability paradox...........................72 3.14 a guaranteed millionaire..............74 3.15 group project .............................................76 3.16 conditional probabilities & false

positives ..............................................................77 3.17 do you feel lucky? ................................82 3.18 exercises............................................................84

3.19 references.....................................................88


3.1. North Carolina Standard Course of Study

GOAL 1: The learner will analyze data and apply probability concepts to solve problems.

1.03 Use theoretical and experimental probability to model and solve problems.

a) Use addition and multiplication principles. b) Calculate and apply permutations and combinations. c) Create and use simulations for probability models. d) Find expected values and determine fairness. e) Identify and use discrete random variables to solve problems. f) Apply the Binomial Theorem.

3.2. Vocabulary

Counting

Random Event

Success

Trial

Sample Spaces

Dependent

Independent

Compound

Mutually Exclusive

Conditional

Binomial Probability

Expected Value

Random Variable

Fairness

Game Theory

Markov Chains

Simulation

Combination

Permutation

3.3 Timeline and Assignments

Day 1 – Law of Large Numbers – Coin Toss Empirical, Theoretical Probability Vocabulary

Day 2 – Introductory Activity – Dice Game Basic Probability – worksheet Genetics Activity Hazardous Waste in Who’s Backyard?

Day 3 – Geometric Probability Fairness



Day 4 – Binomial Probability Theorem Mutually Exclusive Independent/Dependent Events Expected Value

Day 5 – Permutations

Day 6 – Combinations Odds

Day 7 – Test Group Presentation preparation

Day 8 – Group Presentations



3.4 Introduction to Mathematical Probability

The theory of mathematical probability is a little over three hundred years old. Since its discovery by Pascal and Fermat in the seventeenth century it has become an integral part of our everyday lives. At one time or another all of us have made decisions by taking chances, throwing dice, or drawing cards. Frequently we have made and will make judgments based on the likelihood or probability that a certain event will happen. These may be as simple as deciding whether or not to take an umbrella on a cloudy day to as complicated as deciding on the best time to sell an investment. Developments in the theory and applications of probability range from simple informal activities to important fields as the physical sciences, genetics, the social sciences, economics, industry, engineering and insurance.

In this module we will work on improving students’ understanding of the elementary ideas included in probability theory and removing some of the simple misconceptions with probability. We will look at several applications of the theory of probability to practical problems. We have activities, and sometimes several activities, that correspond to each topic given. There are several basic objectives for this unit of study. Upon completion of the unit, the student will be able to:

• understand and appreciate the use of probability in everyday life; • solve problems involving permutations and combinations; • define basic terms used in mathematical probability: • compute the probability that a certain event will happen.

3.4.1 Permutations

A permutation is an arrangement of a group of objects in a particular order. The four numbers 5, 6, 7, and 8 can be arranged in twenty-four different ways:

5,6,7,8 5,6,8,7 5,7,6,8 5,7,8,6 5,8,6,7 5,8,7,6 6,7,8,5 6,7,5,8 6,8,5,7 6,8,7,5 6,5,7,8 6,5,8,7 7,5,6,8 7,5,8,6 7,6,5,8 7,6,8,5 7,8,5,6 7,8,6,5 8,5,6,7 8,5,7,6 8,6,5,7 8,6,7,5 8,7,5,6 8,7,6,5

Each of these arrangements is a different permutation. To determine the total number of permutations that can be made from four digits using each one only once, we indicate a space for each digit:

______ ______ ______ ______ You can choose any one of the four digits to put in the first space. Then you have any of the remaining three digits to go into the second space. There are now two choices for the third space, and the last digit is placed in the fourth space. The product of the number of choices for each space is the total number of permutations that can be made. So we have 4 3 2 1 24 4!× × × = = different ways to fill the spaces. Now, if we were arranging 6 objects we would have 6 5 4 3 2 1 720 6!× × × × × = =



different ways to fill the spaces. Any arrangement of n objects in a particular order is called a permutation of n objects. We see that there are 24 permutations of 4 different objects and there are 720 permutations of 6 objects. The total number of permutations that can be formed from n objects using all of them without repetition is n! The symbol n! is read n factorial. By definition,

0! 1! 1 2 3 4 ( 1) , if 0.n n n== × × × × × − × >n

Now, we don’t have to arrange all n objects. You are given a set of 6 students who are members of a school club. From these 6 students you have to choose a president, vice-president, secretary and treasurer for the club. In how many ways can you do this? Notice that it makes a difference in the way that we choose them. We could choose the same four students but they could serve in different offices. The symbol nPk represents the number of permutations that can be formed from n objects taken k at a time where k n< . Notice that !n nP n= because we are arranging all n objects. The number of permutations of six objects taken four at a time is 6 5 . 4 3 360× × × =

6 46! 6!6 5 4 3

2 1 (6 4)!P = × × × = =

× −

In general, !( )n k

nPn k

=− !

.

Example 1: In how many different ways can five books be arranged on a shelf? Solution: . 5 5 5! 5 4 3 2 1 120P = = × × × × =

Example 2: How many two digit numbers can be made from the six digits 7, 2, 4, 5, 9, 3 if no digit can be used more than once?

Solution: 6 26! 6! 6 5 4 3 2 1 30

(6 2)! 4! 4 3 2 1P × × × × ×

= = = =− × × ×

Example 3: How many integers of three decimal places can be formed from the digits 5, 1, 8, 4 if repetition is allowed?

Solution: Since we are allowed to use the same digit more than once, we have 4 choices for each digit, which gives us 34 4 4 4 64× × = = three decimal place integers that can be formed from those digits.

Example 4: How many even integers of four places can be formed from the digits 1, 2, 3, 4, 5?

Solution: Here we have a restriction only on the last digit – there are only 2 choices for the last digit if the number is to be even. There are no restrictions on the other 3 digits in the number, so there are . 35 5 5 2 5 2 250× × × = × =

Example 5: How many telephone numbers can exist in a single area code using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9?

Solution: It is not quite as simple as it would at first seem. Have you ever seen a phone number that began in a 0 or a 1? No, however, there is no restriction on the other 8 digits. So the number of different numbers in a single area code should be

67 9 9 9 9 9 9 7 9 3,720,087× × × × × × = × = . Get on the internet and see if this is correct. If it is not, then what assumption did we make that was NOT correct? If this is not correct



then how do they determine the number of phone numbers allowable in a single area code?

3.4.2 Permutations of objects that are not all different

The four numbers 1, 6, 6, 3 can be arranged in 4! ways. However, two of the numbers are the same so several of the arrangements are identical and cannot be distinguished from others. Consider the digits in this way: 1, 6, 6*, 3. The following arrangements can be made:

16*63 1 6*36 166*3 1636* 136*6 1366* 6*631 6*613 6*136 6*163 6*316 6*361 516*3 6136* 6316* 636*1 66*13 66*31 316*6 3166* 36*16 36*61 3616* 366*1

Although we have written down twenty-four permutations, there are only twelve distinct arrangements. The number of distinct permutations of four objects when two are alike may be denoted by:

4! 4 3 2 1 24 122! 2 1 2

× × ×= = =

×

The number of distinct permutations of n objects of which p are alike, q are alike, etc. is !

! !n

p q

Example: How many different permutations can be made using all the letters of the word Connecticut?

Solution: The word Connecticut contains eleven letters including three C’s, two N’s, and two T’s. The number of different permutations of these letters is

11! 1 2 3 4 5 6 7 8 9 10 11 1,663,2003!2!2! (1 2 3)(1 2)(1 2)

× × × × × × × × × ×= =

× × × ×

3.4.3 Combinations

A combination is a group of objects in which the order or arrangement of the objects is NOT important. From the numbers 1, 2, 3, six different permutations can be formed. They are 123, 132, 231, 213, 312, 321. When the order of the digits is not considered, all six of these permutations make up one combination. The number of combinations of three objects, taken three at a time is one. In general, the number of combinations of n objects taken n at a time is one.

1n nC = . When the number of combinations of n objects taken n at a time is multiplied by n! the result is

. n nP

! or!

n nn n n n n n

PC n P Cn

× = = =1.

Now, what if we want to look at the number of combinations that can be made from n objects, taking only k at a time is



!! ( )!

n kn k

P nCk n k

= =− !k

From the formula, it is easy to see that the number of combinations of n things, taken k at a time, is equal to the number of combinations of n things, taken n - k at a time, that is, n k n n kC C −= .

! !( ( ))!( )! !( )!n n k n k

n nC Cn n k n k k n k− = =

− − − −= .

Specifically,

40 38 40 240! 40! 40 39 780

2!(40 2)! 2!38! 2C C ×

= = = = =−

In many cases this fact can be used to simplify computations.

Example 1: In how many ways can a committee of four be chosen from ten people?

Solution: 10 410! 10 9 8 7 10 3 7 2104!6! 4 3 2 1

C × × ×= = = × × =

× × ×

Example 2: How many combinations can be made from seven objects, using them five at a time?

Solution: 7 57! 7 6 21

5!2! 2C ×

= = =

Example 3: Evaluate . 60 57C

Solution: 60 5760! 60 59 58 34,220

57!3! 3 2 1C × ×

= = =× ×

.

3.4.4 Sample Space

To perform experiments scientists must select individuals or samples from a population. A random sample is one in which each member of the population has an equal chance of being selected. If selections favor certain individuals or events the sampling is said to be biased. The result of each experiment is called an event or outcome, and the set of all possible outcomes is known as a sample space.

Example 1: Each positive integer from one to five inclusive is written on a piece of paper, and the pieces of paper are shuffled. List a sample space for the outcome of drawing one piece of paper.

Solution: Any one of the five numbers is equally likely to be drawn. The sample space is {1, 2, 3, 4, 5}.

Example 2: Two jelly beans are to be drawn from a jar known to contain only red jelly beans and green jelly beans. List a sample space for the result of the drawing.

Solution: Since the jelly beans must be red or green, a sample space is { (R,G), (R,R), (G,R), (G,G) }



3.4.5 Probability

Probability expresses the chance or likelihood that a certain event will happen. Mathematical probability is often expressed as a ratio. If it is equally likely that an event may happen in h ways and fail to happen in f ways, where h f n+ = , then the probability that the event will occur may be expressed as the ratio h n , and the probability that it will fail to occur may be expressed as the ratio f n . The fraction that denotes the number of favorable outcomes divided by the total number of possible outcomes represents the probability that an event will happen. The fraction that denotes the number of ways an event will fail to happen divided by the total number of possible outcomes represents the probability that an event will fail to happen.

( ) ( )h fP E and P not En n

= = .

The sum of the probability that an event will happen and the probability that it will fail to happen must be 1. If it is certain that an event will happen, the probability of a favorable outcome is 1. If an event is certain to fail, the probability that it will happen is 0. The probability that an event will happen, P(E), ranges from 0 to 1.

0 ( )P E 1≤ ≤ .

Example 1: What is the probability of throwing a four in one throw of a die? Solution: The die may end up on any one of six faces. Only one of these will be “4”.

The probability of throwing a four is 1/6.

Example 2: The faces of a cube are marked with the letters A, A, B, C, D, E. If the cube is tossed, what is the probability that an A will turn up?

Solution: The cube may turn up six different ways. There are two ways in which an A can turn up. The probability that an A will turn up is 2/6 or 1/3.

Example 3: A committee of three is to be chosen from ten girls. If Ann, Betty and Carol are among the group of ten girls, what is the probability that all three of them will be on the committee?

Solution: The total number of committees of three girls that can be chosen from ten girls is

10 310! 10 9 8 1207!3! 3 2 1

C × ×= = =

× ×.

{Ann, Betty, Carol} forms one of these selections. The probability that the committee will consist of Ann, Betty and Carol is 1/120.

3.4.6 Mutually Exclusive Events

Mutually exclusive events are two or more events that cannot occur at the same time, or simultaneously. If one die is thrown and comes up three, it cannot come up six or any other number at the same time. If a coin is tossed and comes up tails, it cannot come up heads on the same toss. If a person weighs 125 pounds, he cannot have any other weight simultaneously.

The probability of one or the other of two mutually exclusive events happening is the sum of the separate probabilities of these events. If X and Y represent two mutually exclusive events

( ) ( ) (P X or Y P X P Y )= + .



This is known as the Addition Theorem and may be extended to any number of mutually exclusive events.

Example 1: If a bag contains four blue marbles, six yellow marbles, and five green marbles, what is the probability that in one drawing a person will pick either a blue marble or a green marble?

Solution: There are fifteen marbles in the bag. The probability that a blue marble will be selected is 4/15. The probability that a green marble will be drawn is 5/15.

4 5 9( ) ( ) ( )15 15 15 5

P B or G P B P G 3= + = + = = .

The probability that either a blue marble or a green marble will be drawn is 3/5.

Example 2: If a die is thrown, what is the probability that either a two or a six will come up? Solution: The die can come up any one of six ways. The probability that a two will

come up is 1/6. The probability that a six will come up is 1/6. 1 1 2 1(2 6) (2) (6)6 6 6 3

P or P P= + = + = = .

The probability that either a two or a six will come up is 1/3.

3.4.7 Independent Events

Two or more events are independent if the occurrence of one event does not affect the occurrence of any of the others. If one coin is selected at random from a box containing dimes and quarters and is replaced, the result of this selection would not affect the result of a second selection. These two drawings are independent events.

The probability that two independent events will both happen is the product of the separate probabilities. If X and Y are independent events, the probability that both X and Y will happen may be found by the formula

( ) ( )P X and Y P X P Y( )= × .

Example: If a die is thrown twice, what is the probability that a five will come up on the first throw, and a three will come up on the second throw?

Solution: The probability that a five will come up the first time is 1/6. The probability that a three will come up the second time is 1/6.

1 1 1(3 5) (3) (5)6 6 36

P and P P= × = × = .

The probability that both events will happen is 1/36. It is not always the case that events are mutually exclusive. To determine the probability of event X or event Y happening, when the two events are not mutually exclusive:

1. Find the sum of the separate probabilities. 2. From this sum subtract the probability that both events will occur.

( ) ( ) ( ) (P X or Y P X P Y P X and Y )= + − .

Example 1: When a card is drawn at random from a normal deck of 52 cards, what is the probability that it will be either an ace or a spade?



Solution: The probability of drawing an ace from the deck is 4/52. The probability of drawing a spade is 13/52. The probability of drawing the ace of spades is 1/52. The probability of drawing either an ace or a spade is:

( ) ( ) ( ) (4 13 1 16 4

52 52 52 52 13

P ace or spade P ace P spade P ace of spades)= + −

= + − = =

Example 2: If two dice are thrown, what is the probability that one of them will come up less than five?

Solution: The dice can come up the following ways:

{1,1} {2,1} {3,1} {4,1} {5,1} {6,1} {1,2} {2,2} {3,2} {4,2} {5,2} {6,2} {1,3} {2,3} {3,3} {4,3} {5,3} {6,3} {1,4} {2,4} {3,4} {4,4} {5,4} {6,4} {1,5} {2,5} {3,5} {4,5} {5,5} {6,5} {1,6} {2,6} {3,6} {4,6} {5,6} {6,6}

The probability that the first die will come up less than 5 is 24/36. The probability that the second die will come up less than 5 is 24/36. The probability that both dice will come up less than 5 is 16/36. The probability that one of them will come up less than 5 is

24 24 16 32 836 36 36 36 9

+ − = = .

3.4.8 Dependent Events

Two events are dependent if the occurrence of one of the events affects the probability of the occurrence of the other. Consider a box that contains seven white, five green, and four blue marbles. If two marbles are drawn from the box, and the first marble is not replaced before the second marble is drawn, the outcome of the first selection affects the outcome of the second drawing. The probability of drawing a blue marble the first time is 4/16 or 1/4. If a blue marble is drawn first, then three of the remaining marbles are blue, and the probability of getting a blue marble on the second draw is 3/15 or 1/5.

The probability that two dependent events will happen may be expressed as follows: If Pl is the probability that event X will happen and, after X has happened, P2 is the probability that event Y will happen, then the probability that the events will occur in the order X,Y is , the product of their respective probabilities.

1 2P P

3.4.9 Expected Value and Fair Games

In probability (and especially gambling), the expected value (or expectation) of a random variable is the sum of the probability of each possible outcome of the experiment multiplied by its payoff (value). Thus, it represents the average amount one expects to win per bet if bets with



identical odds are repeated many times. Note that the value itself may not be expected in the general sense, it may be unlikely or even impossible.

For example, an American Roulette wheel has 38 equally possible outcomes. A bet placed on a single number pays 35-to-1 (this means that he is paid 35 times his bet, while also his bet is returned, together he gets 36 times his bet). So the expected value of the profit resulting from a $1 bet on a single number is, considering all 38 possible outcomes:

37 11 35 0.052638 38

⎛ ⎞ ⎛ ⎞− × + × ≈ −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

.

Therefore one expects, on average, to lose over 5 cents for every dollar bet.

Significance of Expected Value There is a lottery of 1000 tickets; every ticket costs $1. The lottery has the following prizes: one ticket wins $500, 5 tickets win $50 and 20 tickets win a $10 prize.

Number Probability Prize1 1/1000 $5005 5/1000 $50 20 20/1000 $10 974 974/1000 $0

Therefore, total prize amount is 1 $500 5 $50 20 $10 $950× + × + × = , or the average is $950 1000 $0.95= for every ticket. This index is called Expected Return.

Now deduct the ticket price from expected return: $0.95 – $1 = –$0.05. This index is called the player edge or profit. It measures the average game result. Note that this lottery is not a profitable game; you lose about $0.05 on every ticket.

The example we used above is easy and the calculations are evident. However, there are many other games where we can’t apply the same methods. How do you find a general principle for expected return or edge calculation?

In general, the expected return can be computed as expected value of winning, and the player edge is the expected value of the game results.

Expected return is:

500 × 1/1000 +50 × 5/1000 + 10 × 20/1000 + 0 × 974/1000 = 0.95

Player edge is:

(500 – 1) × 1/1000 + (50 – 1) × 5/1000 + (10 – 1) × 20/1000 + (0 – 1) × 974/1000 = –0.05

For convenience sake the expected return and edge are regularly computed by gamblers as a percent of the wager:

Expected return = 100% × ($0.95/$1.00) = 95%

Player edge = 100% × (–$0.05/$1.00) = –5%

Therefore, we can see the relation of the expected return and edge in percentage:

Edge = Expected Return – 100%



Usually the edge is used for a basic rating of gamble benefit; but sometimes the expected return is considered as the main index in video poker.

If you know the player edge you can forecast the gambling productivity for a long period of playing the game. For example, roulette profit is –2.7%. Therefore, if you bet $1 for 1000 times, the game balance will be about –2.7% × $1000 = –$27.

Moreover, the expected return affords a player to compare the gambling efficiency of different games and to choose the best one for his purpose. It is evident that a game with a higher expected return is more preferable for gambling to the gambler. Moreover, you can earn money when expected return is positive! Don’t expect to see these games being played at the casinos!

Example: Flip two coins. For each head that appears you receive $2 from your rich uncle. For each tail you pay him $1. The outcomes, or points in the sample space, of the experiment are

Outcome X = amount you receive

HH 4 HT 1 TH 1 TT –2

As an astute player you should be interested in the probability of a particular outcome of the game. The game assigns a dollar value to each of the four possible outcomes. Since two of them are monetarily the same we get the following random variable and distribution:

X P(X = x) $4 1/4 $1 1/2

–$2 1/4

After explaining the above game, your uncle asks how much you are willing to pay in order to play.

You are not interested in your expected winnings on any given pair of throws, but you are interested in your long run winnings, or what you could expect to win if you played the game many times. We could repeat the experiment an infinite number of times and calculate the average payoff per trial. There is an easier way to determine this average than flipping coins for the rest of our lives.

Expected value is merely a weighted average. Your grade point average is a weighted average. The weights in the coin toss game are the probabilities that an outcome will occur.



For the coin game X P(X = x) X×P(X)$4 1/4 1.00$1 1/2 0.50

–$2 1/4 –0.50so our long run average winnings will be $1 per toss. We would never pay more than that to play.

The expected value of a random variable gives one a sense of what the average value of the random variable should be, after taking into account all the values with all their probabilities.

Example 1: A coin is flipped. Heads, you win $1. Tails, you lose $1. a) What are the possible outcomes? Solution: The probability is ½ that you will win $1, and ½ that you will lose $1. Hence

the possible outcomes are { ($1,½), (−$1,½) }. b) What are your expected winnings?

Solution: 1 1( ) (1) ( 1) 02 2

E X = + − =

This means that if you play this game over and over again, over the long haul on average you should break even. A game whose expected winnings are 0 is called a fair game.

Example 2: Again, a coin is flipped. Heads, you win $1. Tails, you lose $2. a) What are the possible outcomes? Solution: { ($1,½), (−$2,½) }. b) What are your expected winnings?

Solution: 1 1( ) (1) ( 2) $0.502 2

E X = + − = −

So, on average, you should expect to lose 50 cents every time you play this game.

Example 3: The Mathematics of Death A life insurance policy for a 40-year-old woman will pay $10,000 if she dies within 1 year. The policy costs $300. Statistics (namely, mortality tables) indicate that the relative frequency of a 40-year-old woman dying within 1 year is 0.02.

a) What is the expected value of the policy to the woman? Solution: Using relative frequency as an estimate for probability, there is a 2% chance

that the woman will die within 1 year and the policy will pay the woman (well, her beneficiary) $10,000. There is a 98% chance that the woman will live for another year and the policy will pay the woman nothing. Hence the outcomes for the value of the policy to the woman are

( ){ }$10000,.02 , ($0,.98) . So the expected value of the policy is $10000(.02) + $0(.98) = $200.

b) Why would the woman pay $300 for a policy whose expected value is only $200? Solution: Normally this would be a stupid thing to do. However, the peace of mind she

will receive by knowing that her family will be taken care of in the event of



her death is worth the $100 she is paying over and above the expected value of the policy. The point of insurance is security and peace of mind.

c) What is the expected profit of this policy to the insurance company? Solution: If the woman dies (probability .02), the company loses $10,000. Since they

were paid $300, their actual loss would be $9,700. If the woman doesn’t die (probability 0.98), the company will profit $300. So the set of outcomes for the profit of the policy to the company is {(−$9700, 0.02), ($300, 0.98) } So the expected profit of the policy to the company is

−$9700(.02) + $300(.98) = −$194 + $294 = $100. (This is the $100 that the woman is paying over and above the expected value of the policy to her.) So, on average, the insurance company can expect to make $100 for every policy that they sell.

Example 3: A two-card hand is dealt. What is the expected number of aces? Solution: First, let’s construct the set of outcomes for the number of aces. We can

draw 0 aces, 1 ace, or 2 aces, so the sample space is {0, 1, 2}. Now, what is the probability of each outcome? There are

52 252 52! 13262 50!2!

C ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

ways to draw two cards; this will be the denominator of all our probabilities. To draw 0 aces, we need to draw 2 cards from the 48 non-aces. Hence the

numerator of this probability is

48 248 48! 11282 46!2!

C ⎛ ⎞= = =⎜ ⎟

⎝ ⎠.

So 1128(0)1326

P = .

To draw 1 ace, we need to draw 1 card from the 4 aces and 1 card from the 48 non-aces. So the numerator of this probability is

4 1 48 1

4 48192

1 1C C ⎛ ⎞⎛ ⎞

= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

.

So 192(1)1326

P = .

To draw 2 aces, we need to draw 2 cards from the 4 aces. So the numerator of this probability is

4 2

46

2C ⎛ ⎞

= =⎜ ⎟⎝ ⎠

. So 6(2)1326

P = .

So the set of outcomes for the number of aces is 1128 192 60, , 1, , 2,1326 1326 1326

⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝⎩ ⎭

⎟⎠

The expected number of aces is



1128 192 6 192 12 2040 1 2 0 .1541326 1326 1326 1326 1326 1326

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + = + + = ≈⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

.

Therefore, on average, we will draw 0.154 aces every time we draw a 2-card hand. If we win $1 for every ace drawn, then the odds are in our favor IF it costs less than 15.4¢ to draw a 2-card hand.

Example 4: A quiz has four questions: a True-False question, two multiple choice questions each with three choices and a multiple choice question with four choices. If a student randomly guesses at each of the questions, what is the expected number of questions he will answer correctly? Solution: He could possibly get 0, 1, 2, 3, or 4 questions right.

P(X = 0) = P(all questions wrong) = 1 2 2 3 122 3 3 4 72

⋅ ⋅ ⋅ = .

( 1) = (RWWW) + (WRWW) + (WWRW) + (WWWR) 1 2 2 3 1 1 2 3 1 2 1 3 1 2 2 1=2 3 3 4 2 3 3 4 2 3 3 4 2 3 3 412 6 6 472 72 72 722872

P X P P P P=

⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅

= + + +

=

( 2) ( ) ( ) ( ) ( ) ( ) (1 1 2 3 1 2 1 3 1 2 2 1 1 1 1 3 1 1 2 1 1 2 1 1=2 3 3 4 2 3 3 4 2 3 3 4 2 3 3 4 2 3 3 4 2 3 3 46 6 4 3 2 272 72 72 72 72 722372

P X P RRWW P RWRW P RWWR P WRRW P WRWR P WWRR= = + + + + +

⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅

= + + + + +

=

)

( 3) ( ) ( ) ( ) ( )

1 1 1 1 1 2 1 1 1 1 2 1 1 1 1 3=2 3 3 4 2 3 3 4 2 3 3 4 2 3 3 41 2 2 372 72 72 72872

P X P WRRR P RWRR P RRWR P RRRW= = + + +

⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅

= + + +

=

1 1 1 1 1(4) ( )2 3 3 4 72

P P RRRR= = ⋅ ⋅ ⋅ =

So the expected number of questions answered correctly is 12 28 23 8 1 28 46 24 4 1020 1 2 3 4 0 1.472 72 72 72 72 72 72 72 72 72

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + = + + + + = ≈⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

17



For the first question (True-False), P(0 right) = P(wrong) = ½ and P(1 right) = P(right) = ½. So the expected number is 0⋅½ + 1⋅½ = ½.

For the second question (multiple choice with 3 choices), P(0) = 23

and P(1) = 13

, so the

expected number is 13

.

The third question is like the second (multiple choice with 3 choices), so the expected

number is again 13

.

The fourth question is multiple choice with 4 choices, so the expected number (you can

work it out) is 14

.

So, the expected number of questions answered correctly on the entire quiz will be 1 1 1 1 36 24 24 18 102 1.417.2 3 3 4 72 72 72 72 72

+ + + = + + + = ≈

Example 5: What is the expected score on this quiz for someone who guesses at every question? Solution: There are 4 questions on the quiz, and a guesser is expected to answer 1.417

of them correctly, so his expected score is 1.417 (100%) 35%4

≈ .

One word of caution. We cannot break up an expected value into parts unless each of the parts is independent of each other.



3.5 Activity 1: Coin Toss

In groups, have students predict that if they tossed the coin 10 times, how many of those times the coin will land “heads up”.

Students then perform the task starting with tossing the coin 10 times and increase tosses by 10 each trail. Each group should perform 5 trials recording outcomes.

Collect data from each group to begin class discussion of the Law of Large Numbers and Empirical and Theoretical Probability.



Coin Toss Worksheet 1. If you will toss a coin 10 times, how many times should it come up heads?

__________

2. If you will toss a coin 20 times, how many times should it come up heads? __________




6. Take a “fair coin” (no two headed coins or such) and toss it as fairly as you can 10 times. Record the number of heads in the table below.

7. Toss this same coin 20 times and record the number of heads in the appropriate spot in the table below.




# of tosses 10 20 30 40 50

# of Heads

11. Report your findings to the class by recording your results at the board, or on the overhead.

12. What is the average number of heads for all of the groups when you look at the 10 tosses? Is this closer to what you would expect?

13. Do the same for all of the groups of tosses? As you do more does the average look more like what you would expect?

14. Use the simulation capability of your calculator to simulate a large (huge) number of coin tosses.



Activity 2: Empirical Probability

Objectives: The students will derive a general formula for the probability of a given event given each possible outcome is equally likely.

Apparatus Needed: 6 dice 30 colored balls (1 each of five different colors) 6 coins 6 paper plate holders 6 plastic cups 6 paper bags 19 ping pong balls (numbered from 0 to 18)

Recommended strategy: The students were told they had an opportunity to win a prize in today's class by playing the lottery. They were to pick a 3 digit number, a 4 digit number (repeats are allowed) and a combination of 4 numbers from 1 to 18 (no number could be chosen twice).

The students were divided into groups of 3 or 4. Each group received a paper bag containing 1 die, 5 colored balls (1 of each color-blue, green, yellow, orange and pink), 1 coin, 1 paper plate holder and 1 plastic cup. Each group had to conduct three experiments.

1. Flip the coin 50 times into the paper plate holder, record the results (heads or tails). 2. Pull a ball from the bag, record its color, replace the ball and repeat this process a total of

50 times. 3. Use the cup to shake the die. Roll the die into the paper plate holder. Record the result

and repeat the process 60 times.

As your class collects the data from each experiment, ask the students questions such as: Were the results of each experiment what you would have expected? What would you expect if we were to repeat the process 1000 times? 10,000 times? 10,000,000 times?

Have the class generate the formula for the probability of an event P(E)

number of favorable outcomes( )number of possible outcomes

P E =

Discuss whether or not the formula will tell us exactly what will happen for a given event.



3.6 Introductory Activity

Is this game fair? Why or Why not? The Game: Pair up with a partner. Designate one of you as Player A and one of you as Player B. Each of you will have a die. You and your partner will roll your dice simultaneously and find the absolute value of the differences of the numbers that you roll. Player A gets one (1) point (and Player B gets 0) if the value is 0, 1, or 2. Player B gets one (1) point (and Player A gets 0) if the value is 3, 4, or 5. After 10 rolls of the dice, the player with the most points wins the game.

1. Make a conjecture about the fairness of the game. Write the arguments that support your conjecture.

2. Test your conjecture by playing the game six times with your partner. Record your results. Organize your data. Be prepared to share your results. Do your data support your conjecture?

3. What mathematics is involved in this investigation?

4. How might you extend this investigation?



Alternative Introductory Activity –

Is this game fair? Why or Why not? The Game: Pair up with a partner. Designate one of you as Player A and one of you as Player B. Roll a pair of dice. If the sum of the two is 2, 3, 4, 10, 11, or 12, Player A gets one (1) point (and Player B gets0). If the sum is 5, 6, 7, 8, or 9, Player B gets one (1) point (and Player A gets 0). Continue rolling the dice. The player with the most points after 10 rolls wins the game.

1. Make a conjecture about the fairness of the game. Write the arguments that support your conjecture.

2. Test your conjecture by playing the game six times with your partner. Record your results. Organize your data. Be prepared to share your results. Do your data support your conjecture?

3. What mathematics is involved in this investigation?

4. How might you extend this investigation?



3.7 Genetics

Be certain to put the name of this activity on your paper. Sickle cell anemia is an inherited disease that occurs in about 1 in every 500 African-American births and about 1 in every 160,000 non-African-American births. Unlike cystic fibrosis, in which the cystic fibrosis gene is recessive, sickle cell anemia is codominant. In other words a person inheriting two sickle cell genes will have sickle cell anemia, whereas a person inheriting only one of the sickle cell genes will have a mild version of sickle cell anemia, called sickle cell trait. Let’s call the disease-free genes S and the sickle cell gene s. If both parents have Ss genes, determine the following:

1. The probability that an offspring will have sickle cell anemia.

2. The probability that an offspring will have sickle cell trait.

3. The probability that an offspring will have neither sickle cell anemia nor the sickle cell trait.



3.8 Hazardous Waste In Who’s Backyard?

The chart at the right shows the 8 states in the United States with the greatest number of hazardous waste sites.

State Total New Jersey 110 Pennsylvania 100 California 96 New York 80 Michigan 74 Florida 55 Washington 47 Illinois 41

1. If one state from the list is selected at random, determine the probability the state has exactly 80 hazardous waste sites.

2. If one state from the list is selected at random, determine the probability the state has greater than 80 hazardous waste sites.

3. If one state from the list is selected at random, determine the probability the site is from Florida.

4. If one state from the list is selected at random, determine the probability the site is from New Jersey or New York.



B

A

3.9 Geometric Probability Sheet 1

1. Suppose you are playing a game with a friend. Your probability of winning is 0.4. a. You play the game once. What do you expect to happen? b. You play the game 50 times. What do you expect to happen? What could happen? c. You play the game 10 times and only win once. Does this contradict above? Why or

why not? d. You are playing for money and it costs you $1 every time you lose. How much money

would you need to gain each time you win in order for it to be worth your while to play?

2. Darts are thrown randomly at the dartboard shown to the left.

10

a. What is the area of the rectangle? b. What is the area of the circle? c. Calculate the probability that a dart hits inside the circle.

15 d. How many points should each area be worth in order to make the game fair?

3. Two real numbers are chosen at random from 0 to 5, inclusive. What is the probability that their sum is greater than 6?

4. What is the probability of landing inside of A in the maze to the left? In B?



5. If a basketball player is a 65% free throw shooter, what percentage of the time will a shooter, in a one and one situation, score: a. 0 points b. 1 point c. 2 points

6. If a parachutist is trying to find the best landing plot, which of the two below would be the safest? a. A field of size 25 feet by 30 feet; trees of radius 5 feet. b. A field of size 35 feet by 45 feet; trees of diameter 16 feet.

7. What is the probability of throwing a silver dollar (d = 38 mm) into a square whose sides are 75 mm long?



3.9 Geometric Probability - Darts Anyone? Sheet 2

1. Pat and Enn are playing a game with the board shown below. A dart is thrown at random at the board. Pat scores a point if the dart lands in an area marked A. Enn scores a point if the dart lands in an area marked B. Is this a fair game?

P(A) = _____ P(B) = _____

2. Find probabilities for the board below. Would this board make a fair dart game?

P(A) = _____ P(B) = _____

3. If a dart is thrown at random at this dartboard, what is the probability it will land in Area A? Area B? Area C?

A

B

C

4. 4. What is the probability that a dart is thrown at random at this board will land in Area A? What is the probability it will land in Area B? How would you assign points so the game would be fair?

B A

B A A

A B

A B B

B A B

ABB

B BA

BB A

B A AB

BB A A

B

B

B A

A A



3.9 Geometric Probability Sheet 3

The Paratrooper’s Quandary & Fairground Game

Ever wanted to jump out of a perfectly good airplane and go parachuting? What about those annoying games at the State Fair where you have to throw a coin onto a square to win a big prize? Try these two problems: 1. The Lucky Coin game at the fair works on a simple premise: Toss a nickel onto a square

without it landing on any edge. Sounds simple enough! Try to see if you can! Here is what it looks like:

The size of the nickel and square are of major importance aren’t they? In fact, the square is proportioned to a unique size for a reason! Let’s say the coin has a radius of 3 mm and the square has a side of length of 10 mm. What is the probability of throwing the coin into the square?

2. The parachute problem is similar in concept. Assuming the parachutist has little to no control over his/her direction. What is the probability that he or she will land in the illustrated field without landing in the trees?

The field is 10 × 20 ft and the trees have a radius of 5 ft.

3. Sketch a diagram and then find the probability of throwing a coin with a radius of 8 mm onto a square with a length of 35 mm.



4. Calculate the probability of a skydiver landing safely in the following field:

The field is 75 × 40 ft. and the trees have a radius of 14 ft.



3.10 Mutually Exclusive/ Independent/Dependent

Textbook – Stewart Algebra and Trigonometry, p. 899 - 903 Use manipulatives – students can draw items (cards, M&Ms, Skittles, marbles, multiple

colors of similar objects) from a bag. Discuss replacement/non-replacement of items after each draw of the item. Students can record their outcome. Have students create their own problem using the items in the bag.

3.11 Permutations

Students can physically arrange themselves to discover the concept of permutations. Include reference points, circular permutations, and repetitions

3.12 Combinations

Extend activities from the previous day to explore the concept of combinations. Show students how to simplify permutations and combinations on the calculator. Odds – Lottery discussion



3.13 Probability Paradox

As a diversion I was working on some of the Lewis Carroll Pillow Problems. They are mostly geometry or probability. One of them seems to have a reasonable solution from one standpoint, but seems impossible to determine from another. Are you game?

A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colors without taking them out of the bag.

Solution: We know that, if a bag contained 3 counters, 2 being black and one white, the chance of

drawing a black one would be 2/3; and that any OTHER state of things would NOT give this chance.

Now the chances that the given bag contains

(a) BB, (b) BW, (c) WW are, respectively, 1/4, 1/2, 1/4.

Add a black counter.

Then the chances that it contains

(a) BBB, (b) BWB, (c) WWB

are, as before, 1/4, 1/2, 1/4.

Hence the chance of now drawing a black one, is

1 1 2 1 114 2 3 4 3

23

× + × + × = .

Hence the bag now contains BBW (since any OTHER state of things would NOT give this chance).

Hence, before the black counter was added, it contained BW, i.e. one black counter and one white.

Does this mean it isn’t possible to put two white counters in a bag? Jane R. Barnett

Scotland County High School



3.14 A Guaranteed Millionaire

Goals 1. Students will determine how many possible tickets exist in the lottery using the concepts of

permutations and combinations. 2. Students will work in groups to develop a scheme to ensure winning the lottery and be able

to explain and justify their scheme to the class.

Abstract This activity focuses on the application of permutations and combinations to and actual lottery. Students are asked to individually determine the total possible tickets and then work in groups to develop a scheme to guarantee winning the lottery. It is intended to be used after the students have been introduced to the concepts of counting, permutations and combinations.

Problem Statement Tell the students that even though there are many people who play the lottery because they feel “lucky”, few actually examine the chances that they will actually win. Let them know that they will be determining a method for guaranteeing that they will win the lottery regardless of the jackpot.

Instructor Suggestions 1. Introduce the activity by discussing the Problem Statement above with the students. 2. Distribute the A Guaranteed Millionaire activity sheet and allow the students to individually

read and complete part (a). 3. Divide the students into groups and have each group complete the rest of the activity. Be sure

to remind them that they have to be able to explain and justify their plan. 4. When the small groups are finished, have the groups present their plans. 5. Discuss the plans as they relate to probability and counting.

Materials A Guaranteed Millionaire activity sheet, chalk or overhead materials

Time Introduction 5 minutes Individual work 5 minutes Group work 30 – 45 minutes Presentations and Discussion 20 minutes

Mathematics Concepts Counting, Permutations, Combinations, Probability, Fundamental Counting Principle



Further Investigation This activity could be extended by asking the students to develop a plan that uses the least number of people to win any lottery or how they might win Power Ball lotteries that involve multiple states.

Variations/Comments This activity may take more than one traditional class period to complete. One variation would be to have the students complete the last part as homework or a project. Also, this activity assumes that there is only one winning ticket chosen by one person, with a variation being what happens if two people pick the winning numbers.



A Guaranteed Millionaire You decide that you've had it with playing the lottery because you never win anything. You hear about a company in Lizard Lick that tried to buy every possible ticket to ensure that they would win. Since this seems to be a full-proof way to be a guaranteed millionaire, you decide to give it a shot.

First, you must determine how many tickets are possible. In South Carolina, the lottery balls have the numbers 1 – 53 on them and you must pick 5 numbers. For a PowerBall play you have to pick a sixth number between 1 and 42. For the Carolina 5 game you have to pick 5 numbers between 1 and 36 for a possible payout of $100,000 if you match all 5. If you match 4 of 5 you win $100 and if you match 3 of 5 you win $10.

a. Determine how many possible tickets exist for the Carolina 5 lottery game and for the PowerBall game.

b. Now, you need to come up with a plan to buy all those tickets. With your group, develop a scheme to do this. Be sure to include all costs, number of people involved and time it would take to carry out and any other information crucial to the plan. You will also be asked to explain and justify the method to the rest of the class. You may assume that the entire jackpot is yours if you win in immediate cash. (Even though we know this is not true)

A research company out of California hears about your operation and wants to know what the lowest jackpot would have to be to guarantee at least $1 dollars for each person involved in the plan.

c. Determine the lowest jackpot for your plan. Would you revise your plan given this idea?



3.15 Group Project

In groups of three or four, students are to design two carnival midway games based on the concepts of probability that we discuss in class. For example, you may design a dart board game or a card game similar to those discussed in class.

Calculations should include expected value, probability for outcomes, odds for winning, and empirical data, etc…

A description of each game is required. Include rules for playing the game, what constitutes winning or possible outcomes (small prize vs. large prize). Also, include a list of materials needed to create the game and a visual of each game.

Each group is expected to create a flyer describing both games to promote carnival attendance.

Each group should submit calculations and designs prior to their oral presentation.

Each group must have prior approval for the choice of midway games that they include in their carnival.



Project Submission Date

Preliminary Calculations and Designs Date

Presentation Date



Midway Game Rubric

Items Possible Points Points Earned

Title Page (names, date, etc…) 5

Introduction 5

Descriptions Materials Lists Rules of the games Object of the games

10

Calculations 25

Flyer (separate page) 10

Visual 10

Oral Presentation (informative, interesting) 20

Written Presentation (grammar, spelling, neatness, organization)

5

Level of Difficulty 10



3.16 Conditional Probabilities and False Positives1

When looking at probabilities we consider the ratio of the number of favorable outcomes to the total number of outcomes; graphically, if the large rectangle represents our sample space S,

then (if |T| denotes size of a set T) the probability of the event A is

( )A

P AS

= .

In discussing conditional probabilities we use the notation to denote the probability of the event A happening given that we know event B has occurred already. This restricts the sample space under consideration to the set B so that is the ratio of the number of outcomes in to the number of outcomes in B, i.e.

( | )P A B

( | )P A BA B∩

( | )A B

P A BB∩

= ,

now this may be rewritten as ( ) ( )( | )P A B A B S B S= ∩ , equivalently

( | ) ( ) ( )P A B P A B P B= ∩ or

( ) ( | ) ( )P A B P A B P B∩ = .

A typical use of conditional probabilities is in the testing for disease. Tests for disease are not 100% accurate and we need to be aware that a positive test result may not in fact mean that the disease is present, thus requiring invasive or expensive procedures. Such a result is called a false positive. Of course it is desirable to minimize false positives which we do by retesting or by using alternate tests.

The following example is interesting in that when asked of a group of 60 students and staff at the Harvard Medical School, only 11 answered correctly.

We know that the prevalence of a particular disease is 1/1000 in the general population. A test for this ailment has a false positive rate of 5%, that is, 5% of the time the test will erroneously indicate that the disease is present when in fact it is not. We are also told that 98% of the people with the disease will in fact test positive. Assuming that you know nothing about particular individuals or their symptoms, what is the probability that a person that has tested positive does in fact have the disease?

What's the problem here? Why do so many people not even estimate the answer very well? Why do they often guess more than a 50% chance? Why is it that most people would be immediately worried unnecessarily? The problem lies in the relatively small probability of any one having the disease.

1 Roberta LaHaye (University of Regina) and Penny Nom



First let's model this: we will use D to denote presence of the disease, to denote it is not present, we will use P for positive tests and N for negative tests. Our problem then translates as

D∼

Given that 1( )1000

P D = , 5( | )100

P P D =∼ , and that 98( | )100

P P D = , find . ( | )P D P

A convenient way to present this is with a tree diagram

95

100

5

100

2

100

98

100

1

1000

999

1000

~D

D

P

N

P

N

From these branches we see that

1 98 98( )1000 100 100,000

P D P∩ = =

and

1 98 999 5 98 4995 5093( ) ( ) (~ )1000 100 1000 100 100,000 100,000

P P P D P P D P += ∩ + ∩ = + = =

Thus ( ) ( )( | ) ( ) / ( ) 98 100,000 5093 100,000 98 5093 0.01924P D P P D P P P= ∩ = = = … That is a little less than 2% chance that the individual has the disease! Graphically the picture is something like

and the crucial point is that D is only 1/1000 of S!

It may be instructive to think of the population S having size 100,000 (see below), then only 100 would be expected to have the disease. If you tested the whole population, all 100,000 of them, then the test would accurately pick up 98 diseased people but the 5% false positives



amount to 4995 people. That is, only 98 of the total 5093 people showing positive are indeed ailing with the disease.



3.17 Do You Feel Lucky? Julie Anna Hartwell

Richland One School District, SC

Core Curriculum Objective: How do you use probability to predict the chance of an event occurring?

Overview: In this lesson the learner will define and use probability to predict the chance of an event occurring. Students will play a game of chance and decide whether statistics are appropriately used in real-world examples.

Resources/ Materials : Rocket Instruction Sheet (straws, tape, colored typing paper, pencil, scissors) Masking Tape Instructional Activities:

Launch Activity: The teacher will use a commonly known statistic and discuss its applicability. Misuses of the statistic will also be discussed to stimulate student's thinking about probability and statistics. (Ex: 4/10 American are Hispanic) Cognitive Teaching Strategies: 1. The teacher will define probability as it relates to the statistic discussed in the

activator. (Ex: What does 4/10 mean in South Carolina? Is this true?) 2. The teacher will model finding probability and students will complete similar

problems from the text. (teacher choice) 3. The teacher will introduce probability notation P(occurrence) = ? The teacher will

also explain probabilities of 1 and 0, supported by examples of each. Students will cite examples of zero probability using items or students in the classroom. For instance, P(student having one eyebrow)?

4. The teacher will model how to determine the probability. Several guided examples will be completed. Students will then complete a worksheet on probability (teacher choice). The teacher will circulate around the room, giving assistance when needed.

5. The teacher will choose several problems to review with the class and answer any resulting questions.

6. Students will then receive the homework and be placed in cooperative pairs for the Summarizing/Culminating Activity.

Summarizing/Culminating Assessment: 1. Students will construct a rocket following the directions on the Rocket Construction sheet. 2. Students will then set an appropriate distance for the rocket to travel. Mark the distance with

masking tape on the floor. Also mark the halfway point with tape. 3. Students will take turns firing the rocket at least 20 times. Students will record the distance

each rocket traveled and whether it exceeded the distance marked by the masking tape. 4. Each student will then determine the following probabilities:

P( rocket traveling past the marked distance) = P( rocket traveling past halfway point) = P( misfiring rocket) =



5. Ask students to remain in pairs and write a paragraph response to the following question. All papers are due before the end of the period.

Make a prediction of how far your rocket would have traveled on the (nth) firing (full distance, half distance, misfire)? Explain your prediction.

Do you think you performed the experiment enough times to make an educated guess? Why or Why not?

What could you do to ensure that your rocket would travel past the marked distance each time?



ROCKET CONSTRUCTION SHEET

Rocket Assembly Items:

Pencil Colored typing paper Scotch tape Scissors

Construction 1. Each student will be given the assembly items. 2. Students will cut the paper lengthwise into equal halves. 3. Students will wrap one half of the paper snugly around the pencil and tape the outer end.

Students should not tape the inner wrapping to the pencil. Only the outer end should be secured with a small piece of tape. This will form the body of the rocket.

4. After carefully removing the pencil from the body of the rocket it should resemble a drinking straw.

5. Cut both ends of the body of the rocket so that they are level. 6. Cut three "wings" should the remaining half of the paper. Students are encouraged to cut

their wings in any shape that they believe will allow their rocket to travel farther. The length of the wings should be no longer than 3 inches

7. Use tape to affix the wings to one end of the body of the rocket in a circular pattern. 8. Press the other end of the rocket together and secure with a piece of tape. This end should

be airtight. 9. Insert the straw into the wing end (open end) of the rocket and breath forcibly into the

straw. 10. Record the distance the rocket traveled.



3.18 Exercises:

1. In how many ways can the offices of president, secretary and treasurer be filled from a group of nine people?

2. In how many ways can five girls be arranged in a straight line?

3. In how many ways can seven boys be arranged in a straight line if one particular boy is to be at the beginning of the line, one particular boy is to be in the middle of the line, and one particular boy is to be at the end of the line?

4. How many integers between 10 and 100 can be formed by the digits 1, 2, 3, 4, 5 if no repetition is allowed? How many can be formed if repetition is allowed? 5.) How many odd-numbered integers can be formed by the digits 2, 3, 6, 5, 9, 8 if each digit may be used only once?

5. In how many different ways can the letters of the word number be arranged if each arrangement begins with a vowel?

6. A theater has five entrances. In how many ways can you enter and leave by a different entrance?

7. In how many ways can you mail three letters in six letter boxes if no two are mailed in the same box?

8. Milltown has eight grocery stores and six meat markets. In how many ways can you buy a pound of hot dogs and a bag of flour?

9. Four people enter a bus in which there are six empty seats. In how many ways can the people be seated?

10. How many different permutations can be made using all the letters of the word dinner?

11. How many distinct permutations can be made using all the letters of the word (a) challenge (b) banana (c) staff (d) tuition (e) assassination (f) committee?

12. How many different seven digit numbers can be made using all the seven digits 3, 3, 3, 4, 4, 5, 5?

13. In how many ways can five nickels, three dimes, four pennies and a quarter be distributed among thirteen people so that each person may receive one coin?

14. How many signals can be made by raising four red flags, two green flags, and one white flag on a pole at the same time?

15. Find the number of combinations of five objects taken from a group of nine objects.

16. How many combinations of four items are there in a given set of six items?

17. How many diagonals can be drawn in an octagon?

18. In how many ways can seven questions out of ten be chosen on an examination?

19. In how many ways can three books be chosen from five books?



20. From a group of twelve ladies a committee of three is to be selected. In how many ways can this committee be formed with Mrs. Adams on the committee, but with Mrs. Jones excluded, if these two are part of the group of twelve?

21. How many committees can be formed from a group of eight men, if one particular member of the group is to be included and two other members of the group are to be excluded from the committee?

Complete each statement in exercises 22 – 25. 22. A ___ ___ is a sample in which each member of a population is equally likely to be selected.

23. A sample is ___ when certain individuals are favored in a selection.

24. The result of an experiment is called an ___ or an ___

25. The set of all possible outcomes is known as a ___.

26. Each letter in the word flower is written on a card and the cards are shuffled. List a sample space for the outcome of drawing one card.

27. Two balls are to be drawn successively from a bag known to contain only yellow balls and purple balls. List a sample space for the experiment.

28. List a sample space that indicates all possible outcomes when two dice are thrown.

29. List a sample space to show all possible outcomes when a family has three children.

30. A bag contains 24 balls. Five of the balls are red, four are green, seven are blue, and eight are yellow. What is the probability that a ball picked at random will be (a) red? (b) green? (c) blue? (d) yellow?

31. There are twenty-eight students in a class. Sixteen are girls, and twelve are boys. Find the probability that a student selected at random will be a girl.

32. Find the probability that a number selected at random from the set of numbers 5, 6, 7, 10, 12, 14, 17, 21, 28, 30 will be divisible by 7.

33. If you select a letter at random from the alphabet, what is the probability that it will be a consonant?

34. If a number is selected at random from the set of numbers 1, 3, 17, 25, 71, what is the probability that the number is (a) an odd digit? (b) an even digit? (c) divisible by 3? (d) a prime number? (e) a composite number?

35. If two dice are thrown, what is the probability of getting a sum of eight?

36. Four marbles are drawn at random from a bag containing five orange marbles and seven brown marbles. What is the probability that (a) all four marbles are orange? (b) all four marbles are brown?

37. If six cards are drawn at random from a deck of 52 cards, what is the probability that they are all spades?

38. If a coin is thrown, what is the probability that it will turn up “tails”?



39. In Hillcross High School there are 300 freshmen, 280 sophomores, 275 juniors, and 256 seniors. What is the probability that a student selected at random will be (a) a freshman? (b) a sophomore? (c) a junior? (d) a senior?

40. Are the following pairs of events mutually exclusive? a) Living in Cary and working in Raleigh. b) Being a freshman and being a junior in high school. c) Being a professor and being an author of a book. d) Drawing a red card and drawing the ace of spades. e) Drawing a face card and drawing the six of hearts from a normal deck of cards.

41. If the probabilities that Joan, Beverly and Evelyn will be elected secretary of a ski club are 1/8, 2/5, and 1/3 respectively, find the probability that one of the three will be elected.

42. If the probabilities that John and Harry will be valedictorian of a high school class are 1/4 and 3/7 respectively, what is the probability that either John or Harry will be valedictorian?

43. Chris and Janet are among twenty girls who enter a tennis tournament. What is the probability that either one of these two girls will win the tournament?

44. In a drawer are six white gloves, four black gloves, and eight brown gloves. If a glove is picked at random, what is the probability that it will be either white or brown?

45. If the probabilities that Mary and Sue will receive awards in a contest are 3/5 and 1/3 respectively, what is the probability that one or the other will receive an award?

46. If five coins are tossed, what is the probability that all five coins will turn up heads?

47. Find the probability that a person will throw 4, 8, and 10 on the first, second, and third tosses of a pair of dice.

48. If two dice are thrown, what is the probability that one of them will come up greater than four?

49. A bag contains six white balls, four green balls, and three brown balls. If three balls are drawn, one at a time, and the ball is replaced after each drawing, what is the probability that the balls drawn will be green, white and brown?

50. A box contains four spools of black thread, six spools of brown thread, and ten spools of white thread. A spool is drawn, replaced, then a second spool is drawn. What is the probability that either a black or a brown spool is drawn?

51. A card is drawn from a standard deck of 52 cards, replaced, and a second card is drawn. What is the probability that both cards are tens?

52. The probability that Joe will solve a certain problem is 3/5, that Jane will solve it is 5/6, and that Sam will solve it is 1/4, What is the probability that Joe and Jane will solve it, and Sam will not solve it?

53. A bag contains five green marbles, four yellow marbles, and nine white marbles. If two marbles are drawn in succession, and the first marble is not replaced before the second is drawn, what is the probability that: a) the second marble is yellow, if the first marble drawn is green? b) the second marble is white, if the first marble drawn is yellow? c) both marbles are green?



d) both marbles are yellow? e) both marbles are white?

54. A box contains ten slips of paper. Three slips are marked with the letter G, two slips are marked M, and five slips are marked K. If two slips of paper are drawn in succession, and the first is not replaced before the second is drawn, what is the probability that: a) the first slip is marked G, and the second is marked K? b) the first slip is marked G, and the second is marked H? c) the first is marked H, and the second is marked G? d) the first is marked K, and the second is marked G? e) the first is marked H, and the second is marked K? f) the first slip is marked K, and the second slip is marked M? g) both slips are marked G? h) both slips are marked M? i) both slips are marked K?

55. The probability that Mr. Smith will be elected president is 5/7, and if he is elected, the probability that he will appoint Mr. Jones attorney general is 2/3. Find the probability that Mr. Jones will be attorney general.



BIBLIOGRAPHY FOR TEACHERS

Modules on CD: Lets Take a Chance What is Chance? CMC HIV Handout Explorations with Chance Facing the Odds

Textbooks Coxford, Arthur P., and Payne, Joseph N. ABJ Algebra 2 with Trigonometry. New York:

Harcourt, Brace, Jovanovich Publishers, 1983.

Kline, William R., et al. Foundations of Advanced Mathematics. New York: American Book Company, 1965.

Mayor, John R., and Wilcox, Marie S. Algebra Second Course. Second Edition. Englewood Cliffs, New Jersey: Prentice— Hall, Incorporated, 1961.

Morgan, Frank M., and Paige, Burnham L. Algebra 2. New York: Henry Holt and Company, 1958.

Rees, Paul K., and Sparks, Fred W. Algebra and Trigonometry. New York: McGraw-Hill Book Company, Incorporated, 1962.

Runyon, Richard P., and Haber, Audrey. Fundamentals of Behavioral Statistics. Fifth Edition. Reading, Massachusetts: Addison-Wesley Publishing Company, 1984.

Vannatta, Glen D., et al. Advanced High School Mathematics. Expanded Edition. Columbus, Ohio: Charles E. Merrill Books, Incorporated, 1965.

Willoughby, Stephen S., and Vogeli, Bruce R. Probability and Statistics. Morristown, New Jersey: Silver Burdett Company, 1968.

READING LIST FOR STUDENTS Corcoran, Clyde L., et al. Algebra One. Glenview, Illinois: Scott Foresman and Company, 1977.

Dilley, Clyde A., et al. Heath Algebra 1. Lexington, Massachusetts: D. C. Heath and Company, 1987.

Dolciani, Mary P. et al. Algebra Structure and Method Book 1. Boston: Houghton-Mifflin Company, 1986.

———. Modern Algebra and Trigonometry Book 2. Boston: Houghton-Mifflin Company, 1965.

Greenberg, Herbert J., et al. SRA Mathematics. Chicago: Science Research Associates, Incorporated, 1978.

Troutman, Andria P., et al. Laidlaw Mathematics Brown Book. River Forest, Illinois: Laidlaw Brothers Publishers, 1978.

Troutman, Andria P., et al. Laidlaw Mathematics Gold Book. River Forest, Illinois: Laidlaw Brothers Publishers, 1978.

White, Myron R. Advanced Algebra. Boston: Allyn and Bacon, Incorporated, 1960.


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