probabilitybkrein/introductory statistics... · 2020-01-28 · 66 chapter 4. probability 12.a box...

Chapter 4

Probability

61

62 CHAPTER 4. PROBABILITY

4.1 Probability

Is it going to rain tomorrow? Will my favorite sports team win the championship? Will I win thePower Ball lottery? Although we would like to answer these questions, math isn’t that awesome.We can’t give definitive answers but we can try to address the likelihood of one of these eventshappening. That is what probability is about. Before we get to probability we need a little settheory along with some additional terminology.

A Random Variable is a variable whose value is determined by an experiment whose outcomesare random.

Example: Roll a die. In this case, there are 6 outcomes whose outcomes are random. Thepossible values are 1, 2, 3, 4, 5, and 6. The set that contains all outcomes of an experiment is calledSample Space, traditionally denoted with the letter ‘S’, so S={1,2,3,4,5,6}

The Sample Space, S, of a random variable is a set consisting of all the possible outcomes ofthe experiment.

When calculating probabilities we are going to be interested in only some of the possible outcomes.If we are playing a board game, for example, we may only want to roll a ‘2’ or ‘5’. In this case, theoutcomes we are interested in is a subset of the sample space, called an event.

An Event is a subset of the sample space.

An event with only one outcome is called a Simple Event.

An event with more than one outcome is called a Compound Event.

Example: Let S={1,2,3,4,5,6} . List several events. Identify each as simple or compound.It turns out there are dozens of different possible events so here we will give only a few:Some compound events {1,3}, {2,4,6}, {1,5}, {3,6} ,Some simple events: {5},{1}, {2}

Example 4.1.1.

You ask two people whether they have ridden the bus in the last month. Observe how manypeople have ridden the bus in the last month. Find the sample space.

Solution.

In this case we have only 3 possibilities: neither one has taken the bus, exactly one has takenthe bus, or both have taken the bus. So our sample space is S = {0, 1, 2}

4.1. PROBABILITY 63

Example 4.1.2.

You ask two people whether they have ridden the bus in the last month. Observe the order ofresponses obtained. Find the sample space.

Solution.

For this our sample space is not going to consist of numbers. Instead it will consist of a verbaldescription. For example: the first rode the bus and the second person didn’t. Let’s abbreviate thisas ‘BD’, ‘B’ for rode the bus, and ‘D’ for didn’t ride the bus. We could possibly just think aboutit and try and reason it through but this is a good place for a tree diagram.

B

B

D

D

CB

D

BB

BD

DB

DD

In the diagram we start on the left with a branch since the first selection is for the bus or not thebus denoted ‘B’ and ‘D’, respectively. Followed up with the second selection and the final outcome.Our sample space is S = {BB,BD,DB,DD}

We now are ready to calculate some probabilities. The probability of an event is a numericalmeasure of the likelihood of the event occurring. It is a number between 0 and 1, inclusive. We haveall heard the meteorologist on the local news state that there is a ‘30% chance of rain’ tomorrow.This simply means that it the current conditions were to happen over and over again, 30% of thetime it would rain.

Let’s start with the die example from above. We have S={1,2,3,4,5,6} . We want to roll either a‘2’ or a ‘5’. What is the probability of that happening? Before we can do this we need a definition.

Definition: If a sample space consists of equally likely outcomes, then the probability of anevent, A, is given by

P (A) =number of elements in A

number of elements in S

For our die example, A={2,5} so P (A) = 26 = 1

3 .A warning: the events in the sample space must be equally likely. If not, you cannot use the

definition of the probability as stated. If the die were loaded (unfair) then we would not expect theprobabilities to be the same.

Probability is, theoretically, straightforward. All you need to do is count how many are in twosets and divide them. In the last example the counting was very easy. For many cases counting iswhat makes the problem difficult.


For many problems we can’t find, let alone count, the elements in S. We need to estimate theprobabilities in such cases.

To estimate the probability P(A) we repeat our experiment several times and calculate the ratioto get

Relative Frequency Estimate of Probability

P (A) ≈ number of times A occurs

number of times experiment is repeated

Note: the equal sign has been replaced with an approximately equal. The approximation will getbetter the more times the experiment is repeated. We see these estimations whenever we look atpolls. If a poll asks whether or not a person will vote for a particular candidate, the reportedpercentage is really just an estimate of the probability. It would be too expensive to ask all votersso pollsters take a sample to get an estimate.

Often times, we want to calculate the probability of an event that is difficult to do directly butit is much easier to calculate what is called the complement of the event

The Complement of an event A, denoted A, consists of all elements that are in the samplespace, S, that are not in the event A.

Probability of the an event and its complement are related by

P (A) + P (A) = 1

The more useful form is given by

P (A) = 1− P (A)

This last formula we are probably all familiar with, perhaps not written as a formula, but the idea.For example, if 30% of voters voted for a proposition, then the rest, 70% (= 100% − 30%), votedagainst the proposition.

Example 4.1.3.

Let S = {1, 2, 3, 4, 5, 6} from before and A = {2, 5} Find A and P (A)

Solution.

Since A consists of the elements in S that are not in A, we have A = {1, 3, 4, 6}. The probabilityis P (A) = 4/6 = 2/3.

4.1. PROBABILITY 65

4.1.1 Exercises

1. We are to randomly select 2 registered voters and determine if they voted in the last election.Use a tree diagram to determine the sample space.

2. Two students at a large university are to be selected and it is to be determined if they areundergraduates or graduate students. Use a tree diagram to find the sample space of theexperiment.

3. A box contains 12 marbles. 8 of the marbles are green. If you reach in and grab 5 marblesand count how many are green, find the sample space.

4. You flip a coin 4 times. If you are observing the number of heads in the four flips, find thesample space.

5. You are to select 3 people and determine if they have a criminal record. Use a tree diagramto find the sample space where the outcomes note the order of whether or not a person has acrimial record.

6. A pollster wishes to ask 3 workers if they use mass transit to get to work. Use a tree diagramto find the sample space. Where the pollster observes the order of the responses, e.g. Yes-Yes-No.

7. Let S = {a, b, c, d, e} be the sample space that consists of equally likely outcomes.

(a) let A = {a, b, e} Find P (A)

(b) let B = {c, d} Find P (B)

(c) let C = {d} Find P (C)

(d) let D = {b}. Find P (D)

(e) Which pair of events listed above are complements.

8. Let S = {1, 2, 3, 4, 5} be the sample space that consists of equally likely outcomes.

(a) let A = {1, 2, 4, 5} Find P (A)

(b) let B = {2, 4} Find P (B)

(c) let C = {3} Find P (C)

(d) let D = {2}. Find P (D)

(e) Which pair of events listed above are complements.

9. A company consists of 26 employees. 12 of them are full time employees and 14 are part timeemployees. If you randomly select a worker find the probability that the worker is a full timeemployee.

10. A parking lot has 35 cars that have California license plates and 12 that are out of statelicense plates. If one car is selected at random, find the probability that the car has out ofstate plates.

11. A bag of candies contains 13 red candies, 17 blue candies, and 25 white candies. A candy isto be selected. Find the probability that you will get a blue candy.


12. A box of apples contains 3 granny smith apples, 6 golden delicious apples, and 5 pink ladyapples. One of the apples is to be selected. Find the probability that the apple will be a pinklady apple.

13. A bag of candies contains 13 red candies, 17 blue candies, and 25 white candies. A candy isto be selected. Find the probability that you don’t get a blue candy.

14. A box of apples contains 3 granny smith apples, 6 golden delicious apples, and 5 pink ladyapples. One of the apples is to be selected. Find the probability that the apple will not be apink lady apple.

15. At a blood drive 40 people have O+ blood, 31 have A+ blood, 8 have B+, 2 have AB+, 9have O-, 7 have A-, 2 have B-, and 1 has AB-. A person is to be selected.

(a) Find the probability the person has O- blood.

(b) Find the probability the person has A blood (+ or -).

(c) Find the probability that the person is Rh +. (has a ‘+’ blood type).

(d) Find the probability that the person does not have A+ blood.

16. At a car show, there are 23 Ford convertibles, 12 Ford hard tops, 18 Chevy convertibles, 25Chevy hard tops, 9 Chrysler convertibles, and 16 Chrysler hard tops. You are to randomlyselect one car.

(a) Find the probability of selecting a Chevy convertible.

(b) Find the probability of getting a convertible.

(c) Find the probability that the car is a Ford.

(d) Find the probability that the car is not a Chevy hard top.

4.2. CONDITIONAL PROBABILITY AND INDEPENDENCE OF EVENTS 67

4.2 Conditional Probability and Independence of Events

When we are calculating probabilities, we sometimes need to restrict our attention to a subpopula-tion of our population. We may want to examine the differences in disease rates for different ethnicgroups, for example. This is where conditional probability is applied.

The probability of an event A occurring given that another event, B, has already occurred iscalled the Conditional Probability of A given B and is written P (A | B)

Example 4.2.1.

Define events and write the following as conditional probabilities: ‘twelve percent of womenwill develop breast cancer’, ‘forty-five percent of people in Argentina have type O+ blood’, ‘fifteenpercent of men are left-handed’

Solution.

Let W be the event ‘Woman’ and C be the event ‘Breast Cancer’. Then the conditional proba-bility is P (C |W ) = .12

Note that the order is important. P (C | W ) 6= P (W | C) = .12 . We expect P (W | C) ≈ 1.Why?

Similarly, if we define A to be the event ‘from Argentina’ and O as the event ‘has O+ blood’then we get P (O | A) = .45

Lastly, let M be the event ‘Man’ and L be the event ‘left-handed’ then we get P (L |M) = .15

We will use the idea of independence quite a bit as we proceed. The idea behind independenceis that two events are independent if one event occurring does not affect the probability of anotherevent occurring. Consider the events ‘Breast Cancer’ and ‘Woman’ that we looked at before. Ifyou choose a woman, then the likelihood of that woman developing breast cancer is greater thanthe population as a whole. If we randomly select an adult from the population, the probability isabout .06 that they will develop breast cancer. If the person we chose is a woman, the probabilityis about .12, different from the entire population. If we were to choose a man, the probability theywill develop breast cancer drops to .001.1 Whether or not a person is a woman and the likelihoodof developing cancer are related or dependent. We can generalize

The events A and B are independent if

P (A) = P (A | B) and P (B) = P (B | A)

It turns out that if one of the statements is true then the other one is true. This tells us we onlyneed to check if one is true.

1nationalbreastcancer.org


Example 4.2.2.

224 employees at a company were asked about their full/part time status and whether they area college graduate or not. The results are summarized in the table that follows.

Full Time Part TimeCollege Graduate 59 15

Not a College Graduate 97 53

1. One of these employees is to be selected. Find the probability that . . .

(a) A college graduate is chosen

(b) A part time employee is chosen

(c) A college graduate given that they a part-time employee

(d) A full-time employee given that they are not a college graduate

2. Are the events ‘Part Time’ and ‘College Graduate’ independent?

Solution.

In order to use the definition of probability from before we will need the totals. Also notice wehave given letters to the events: F = Full Time, etc.

Full Time(F ) Part Time(PT ) TotalCollege Graduate(G) 59 15 74

Not a College Graduate(N) 97 53 150Total 156 68 224

We can restate the problem

1. P (G)

2. P (PT )

3. P (G | PT )

4. P (F | G)

5. Are PT and G independent?

To determine the probability of a college graduate(G) chosen we see that there are 74 collegegraduates(G) and a total of 224 employees. So we get P (G) = 74/224 = 0.3304

For the probability of PT , we have 68 total PT s and a total of 224 employees. So P (PT ) =68/224 = 0.3036

Both of these require the use of the definition of probability given before. For P (G | PT ), weare restricting our attention to just the part-time employees. So we see the table as . . .


Full Time(F) Part Time(PT) TotalCollege Graduate(G) 15

Not a College Graduate(N) 53Total 68

From this we see that there are 15 college graduates out of a total of 68. So we get P (G | PT ) =1568 = 0.2206

Likewise, for P (F | N), we see the table as . . .

Full Time(F ) Part Time(PT ) TotalCollege Graduate(G)

Not a College Graduate(N) 97 53 150Total

There are 97 full time employees out of 150 non college graduates (N). So we get P (F | N) =97150 = .6467

To see if PT and G independent, we need to check if either P (PT ) = P (PT | G) or P (G) =P (G | PT ). We have already determined the probabilities for the second equality. We haveP (G) = .3304 and P (G | PT ) = .2206. So we can state that G and PT are not independent (ordependent)

4.2.1 Exercises

1. According to the CDC 18 of every 100 adult men are smokers. Define appropriate events andwrite the statement as a conditional probability.

2. According to the CDC 14 of every 100 adult women are smokers. Define appropriate eventsand write the statement as a conditional probability.

3. According to the CDC, in 28% of all traffic-related deaths in the US were in alcohol-impaireddriving crashes. Define appropriate events and write the statement as a conditional probabil-ity.

4. According to cpcstrategy.com, 67% of millennials shop online. Define appropriate events andwrite the statement as a conditional probability.

5. Several people who wanted to have children were asked how many siblings they have and howmany children they wanted to have. See the table for the results.

0 siblings 1+ siblingsWant 1 child 49 94

Want 2+ children 68 106

(a) One of these persons is to be selected. Find the probability that . . .

i. They have no siblings

ii. They only want one child

iii. They want 2+ children given that they have 1+ siblings


iv. They are an only child given that they only want one child.

(b) are the events ‘1+sibling’ and ‘Want 2+ children’ dependent?

6. Several people with smart phones were asked if they use it to check their email and their age.The results are in the table that follows.

Check email Don’t check emailunder 30 years old 49 94

30+ years old 68 106

(a) One of these persons is to be selected. Find the probability that . . .

i. They don’t check email

ii. They are under 30 years old

iii. They check their email given that they are 30+ years old

iv. They are under 30 years old given that they check their email

(b) are the events ‘Check email’ and ‘30+ years old’ dependent?

7. An airline is interested in how late planes arrive. Several of the companies flights were selectedand it was determined how late, if at all, the planes were and the size of the plane. The resultsare in the table that follows.

On Time or Early Late: < 15 minutes 15+ minutes lateSmall aircraft 302 198 58Large aircraft 83 123 91

(a) One of these flights is to be selected. Find the probability that . . .

i. A large aircraft is chosen

ii. A flight that is 15+ minutes late is chosen

iii. A small aircraft is chosen given that it is on time

iv. A flight is on time or early given that a large aircraft was chosen

(b) are the events ‘Small Aircraft’ and ‘15+ Minutes Late’ independent?

8. Several students at a high school were asked their class and whether or not they planned ongoing to college. The results follow.

Frosh Soph Junior SeniorCollege:Yes 302 201 186 166College:No 136 124 105 86

(a) One of these students is to be selected. Find the probability that . . .

i. A freshman is chosen

ii. A person planning on going to college is chosen

iii. A junior given that they are planning on going to college

iv. A person not planning on going to college given that they are a sophomore

(b) are the events ‘Senior’ and ‘College:Yes’ independent?


9. Several movie-goers were asked to rate a movie that had a lot of violence prior to release.They were also asked if they had children. The results are in the table

Thumbs Up Thumbs DownHave Children 26 75

Don’t Have Children 67 96

(a) We are to select one of these movie-goers. Find the probability that . . .

i. They gave the movie a thumbs down

ii. They have children

iii. A person without children given that they gave it a thumbs up?

(b) What percent of people with children gave it a thumbs down?

(c) are the events ‘Have Children’ and ‘Thumbs Up’ independent?

10. A biologist is examining salmon carcasses on a stretch of river and notes the gender andwhether or not the salmon was hatchery raised or not. The results follow in the table.

Female MaleHatchery 147 127

Wild 116 107

(a) One of these salmon carcasses, one is to be selected at random. Find the probabilitythat the salmon . . .

i. Is a female.

ii. Is hatchery raised.

iii. Is a male given that it is wild.

iv. Is hatchery raised given that it is a female.

(b) Are the events ‘Female’ and ‘Wild’ independent?

11. A survey asked several people what type of phone and computer they own. The results aresummarized below.

iPhone Other PhoneApple Computer 89 32Other Computer 167 216

(a) One of these persons are to be selected at random. Find the probability of selecting . . .

i. Someone who uses a non-Apple computer.

ii. Someone who has an iPhone.

iii. Someone who uses a non-apple phone given that they use a non-Apple computer.

iv. Someone who uses an Apple computer given that they use a non-Apple phone.

(b) Are the events ‘iPhone’ and ‘Apple Computer’ independent?


12. At a company meeting, several donuts were consumed. The results of the donut by type andtopping are summarized in the table.

Chocolate Glazed MapleCake 11 24 9Yeast 18 33 11

One of these donuts is to be selected. Find the probability that . . .

(a) A glazed donut is selected.

(b) A cake donut is selected.

(c) A cake donut is selected given it was maple.

(d) A chocolate donut is selected given it was a yeast donut.

(e) Are the events ‘Cake’ and ‘Chocolate’ independent?

13. At a business luncheon, three entrees were served: a beef dish, chicken dish, and a vegetariandish. Additionally guests were given the option of water, tea, or lemonade. The results aresummarized in the table below.

Water Iced Tea LemonadeBeef 17 35 16

Chicken 23 33 10Vegetarian 12 11 8

One of these meals is to be selected. Find the probability that . . .

(a) A meal with iced tea is selected.

(b) A meal who has the vegetarian entree is selected.

(c) A meal with water is selected given that they had the beef.

(d) A meal with chicken is chosen given that they had lemonade.

(e) Are the events ‘Lemonade’ and ‘Beef’ independent?

14. Some roses are bred to have no, or very little, scent. Others have a noticeable scent. Severalroses are individually sold. The color and if they have a scent are in the following table.

Red White YellowScented 75 44 23

Unscented 52 26 10

(a) What percent of roses sold were red?

(b) What percent of roses sold were scented?

(c) What percent of red roses had no scent?

(d) What percent of scented roses were yellow?

(e) Are the events ‘Scented’ and ‘Red’ independent?


15. In a triathlon, participants swim, cycle, and run. At a local triathlon, several participantswere asked their favorite part of the race and if they are a ‘local’. The summary of the resultsfollow.

Swim Cycle RunLocal 56 43 61

Not Local 153 162 208

(a) What percent of participants prefer the swim?

(b) What percent of participants are local?

(c) What percent of local prefer the run?

(d) What percent of participants who prefer the cycle are not local?

(e) Are the events ‘Local’ and ‘Run’ independent?

16. A pollster is taking a poll about the support of a new tax law and whether or not the personpaid taxes with their federal tax returns or got a refund in the last year.

Refund Paid Some Paid a LotIn Favor 68 268 267Opposed 164 316 218

(a) What percent of those polled are in favor of the new tax law?

(b) What percent of those who got a refund are opposed to the new law?

(c) What percent of people polled who paid a lot of taxes in favor of the new law?

(d) What percent of people who are in favor of the new law are in favor of the new bill?

(e) Are the events ‘Paid Some’ and ‘Opposed’ independent?

17. Go online, find a percentage that is a conditional probability, define the events, and write theprobability.


4.3 Intersection of Events

We often want to find the probability of two events occurring at the same time. For example, if weare selecting a person where the information is summarized in the following table:

Full Time(F ) Part Time TotalCollege Graduate(G) 59 15 74

Not a College Graduate 97 53 150Total 156 68 224

We might what the probability of selecting someone who is a Full Time College Graduate thenwe need to select someone who is in both the Full Time group and the College Graduate group.Where the row and column intersect we see that there are 59 people in there out of a total of 224.So the probability of selecting someone who is a full time college graduate is given by

P (F ∩G) = 59224 = .2634

Definition: The Intersection of two events, A and B, consists of all elements that are in bothA and B and is denoted as P (A ∩B)

If A={1,2,4,5,8 } and B={2,5,8,9} then the intersection of A and B is A ∩B = {2, 5, 8}

With a two way table it is a simple matter to use the definition of probability given before. Ifwe don’t have a two way table we need to find another way.

When we deal with continuous probabilities later we will think about probabilities in terms ofareas. That’s the approach we will take here. Consider the Venn diagram below. It shows the eventA and the sample space S. The probability of A would be the fraction of the area that A takes upthe box representing the sample space, S.

S

A

A very crude estimate for P (A) would be around .2. This is the area of the circle divided bythe area of the entire rectangle. (Yes, I measured it and calculated the areas).

Let’s look at the probability of P (A ∩B) .

A ∩B

S

BA

4.3. INTERSECTION OF EVENTS 75

P (A | B) =Area of the shaded area

Area of B

If we divide the numerator and denominator by the area of S we get

P (A | B) =Area of the shaded area/Area of S

Area of B/Area of S

The fraction in the numerator is P (A ∩B) and the fraction in the denominator is P (B), so weget

P (A | B) =P (A ∩B)

P (B)

Equivalently,P (A ∩B) = P (B)P (A | B)

orP (A ∩B) = P (A)P (B | A)

Example 4.3.1.

Use the formula above to find the probability of selecting someone who is a Full Time CollegeGraduate from the first example.

Solution.

In this case we have the events F and G as defined before and we want P (F∩G) = P (F )P (G | F ). We get P (F ) = 156

224 and P (G | F ) = 59156 from the table and this yields P (F ∩ G) = P (F )P (G |

F ) = 156224 × 59

156 = 59224 = .2634 which is what we obtained before. Clearly, this way is more time

consuming than the way we solved the problem before.

Example 4.3.2.

In a group of 8 people, 3 are college graduates. 2 are to be chosen. What is the probability ofselecting two college graduates?

Solution.

In this problem we have events occurring one after another. Therefore, let G1 be the event: acollege graduate was chosen on the first pick. Similarly, let G2 be the event: a college graduatewas chosen on the second pick. In order to get two college graduates you have to get a collegegraduate on the first pick and get a college graduate on the second pick. So to say that we selectedtwo college graduates is the same thing as G1 ∩G2 . So we want P (G1 ∩G2) = P (G1)P (G2 | G1).


Since there are 3 college graduates and 8 people, P (G1 = 38 . For P (G2 | G1) , the event G1 means

we picked a college graduate so we now have only 2 college graduates left and 7 people left. (Oncesomeone is picked, they can’t be picked again). so P (G2 | G1) = 2

7 . If we put this together we getP (G1 ∩G2) = 3

8 × 27 = .1071. We can organize it by using a tree diagram:

G1P (G1)

= 3/8G2

P (G2 | G2) = 2/7

G1

G1

CG1

G2

P (G1 ∩G2) = 3/8× 2/7 = 3/28

Example 4.3.3.

In a large city, 35% of adults are married. Two adults from this city are to be selected atrandom. Find the probability of selecting two that are married.

Solution.

This is similar to the last example with one major difference. When we select an adult fromthe population, the probability of getting someone that is married on the next pick changes solittle, we can ignore the difference (It is a large city). In practical terms we treat it as if theprobabilities don’t change. Let M1 be the event: a married person was selected on the first pick,and M2 the event: a married person was selected on the second pick. So in this case, we haveP (M1) = P (M2 |M1) = .35 giving P (M1)P (M2 |M1) = .35× .35 = .1225

Using a tree diagram, we have

M1P (M1)

= .35 M2

P (M2 |M2) = .35

M1

M1

MM1

M2

P (M1 ∩M2) = .35× .35 = .1225


4.3.1 Exercises

1. A fair die is to be tossed twice. What is the probability of throwing a ‘5’ followed by a ‘3’?

2. 20% of residents in a large city smoke. If you are to randomly select 2 residents from the city,find the probability that the first one is a smoker and the second one isn’t.

3. A hat contains the names of 8 students. 3 of the students haven’t completed their homeworkand the remaining 5 have completed their homework. If you are to randomly select 2 namesat random, find the probability that . . .

(a) Both have done their homework

(b) The first has completed their homework and the second hasn’t.

(c) Both haven’t done their homework

4. In a standard deck of cards, there are 12 face cards and 40 non-face cards. If two cards areto be selected at random find the probability of

(a) Both being face cards if the first card is returned and the deck is shuffled before thesecond card is drawn

(b) Both being face cards if the first card is not returned to the deck before the second cardis drawn

5. In a standard deck of cards, there are 13 hearts, 13 diamonds, 13 clubs, and 13 spades. If twocards are to be selected at random find the probability of

(a) Both being diamonds if the first card is returned and the deck is shuffled before thesecond card is drawn

(b) Both being diamonds if the first card is not returned to the deck before the second cardis drawn

6. For a research paper, a student listed 10 sources. 4 of the sources were online and theremaining 6 were hard copies. If you randomly select 2 of the sources find the probabilitythat . . .

(a) Both are online sources

(b) The first was online and the second was a hard copy

(c) Neither were online sources

7. 22% of all college students at a large university are seniors. 64% of all seniors are full-timestudents. What percent of all students are full-time seniors?

8. 37% of cars on a highway are driving over the speed limit. 29% of cars that are speeding havegotten at least one ticket in the past year. What percent of all cars are speeding with at leastone ticket in the last year?


9. Several people who wanted to have children in their family were asked how many siblings theyhave and how many children they wanted to have. See the table for the results.



One of these persons is to be selected. Find the probability that . . .

(a) They have no siblings and want to have 2 or more children.

(b) They weren’t an only child and don’t want to have an only child.

10. Several people with smart phones were asked if they use it to check their email and their age. The results are in the table that follows.




(a) They don’t check email on their smart phone and are under 30 years old.

(b) They are 30+ years old and they check their email on their smart phone.


On Time or Early Late: < 15 minutes late 15+ minutes lateSmall aircraft 302 198 58Large aircraft 83 123 91

One of these flights is to be selected. Find the probability that . . .

(a) A large aircraft is chosen and it is 15+ minutes late.

(b) A flight that wasn’t late and the aircraft was small.



One of these students is to be selected. Find the probability that . . .

(a) A freshman planning on going to college is chosen.

(b) A senior not planning on going to college is chosen.





We are to randomly select one of these movie-goers. Find the probability that the movie-goer. . .

(a) is a childless movie-goer that gave it a thumbs up.

(b) is a movie-goer with children that gave it a thumbs down.



Wild 116 107

One of these salmon carcasses, one is to be selected at random. Find the probability that thesalmon . . .

(a) Is a hatchery raised female.

(b) Is a wild male.



One of these persons are to be selected at random. Find the probability of selecting . . .

(a) An iPhone owner that has an Apple computer.

(b) A non-Apple computer owner that has an Apple phone.




(a) A chocolate cake donut is selected.

(b) A yeast donut with maple is selected.






(a) A meal with beef and water is selected.

(b) A meal with lemonade and chicken is selected.



Unscented 52 26 10

(a) What percent of roses sold were scented yellow roses?

(b) What percent of roses sold were unscented red roses?



Not Local 153 162 208

(a) What percent of participants preferred to run and were local?

(b) What percent of participants aren’t local and don’t prefer to swim?



One of these people are to be selected at random. Find the probability of selecting . . .

(a) A person in favor of the law that also got a refund.

(b) A person who paid a lot that is opposed to the bill.

4.4. UNION OF EVENTS 81

4.4 Union of Events

We have seen the probability of the intersection of events, i.e. A and B. Now we would like toexamine the probability of A or B. Technically, we are looking at the union of events.

Definition: The Union of two events, A and B, consists of all elements that are in either A orB and is denoted as A ∪B

In the Venn diagram below, A ∪B is shaded.

S

A B

Recall the following table from a previous section

Full Time Part Time TotalCollege Graduate 59 15 74

Not a College Graduate 97 53 150Total 156 68 224

Let’s say we want to find the probability that someone selected is either a college graduate or afull time employee. We are looking to get P (F ∪G) . We need to find how many people are in theunion of the two events F and G. We can simply add the numbers in the cells that are in F or G(or both) and then divide by the total number of employees.

P (F ∪G) =59 + 15 + 97

224=

171

224= .7634

Lets look at deriving a formula for the union of events. In the Venn diagram we think of theprobabilities as ratios of areas. In the diagram we need to subtract off the area of the intersectionbecause it is counted in both A and B.

A ∪B

=

A

+

B

−

A ∩B

Dividing by the area of the sample space, S, we get the following


P (A ∪B) =

=

P (A) +

+

P (B) −

−

P (A ∩B)

Finally, we getP (A ∪B) = P (A) + P (B)− P (A ∩B)

If A and B are mutually exclusive then

P (A ∩B) = 0 and P (A ∪B) = P (A) + P (B)

We can use this to find the probability in the opening example of this section. Substituting infor A and B we get

P (F ∪G) = P (F ) + P (G)− P (F ∩G)

P (F ∪G) =156

224+

74

224− 59

224=

171

224= .7634

As expected, we get the same answer as before.

Example 4.4.1.

Bob lives in a rural area. His only source of transportation into the city is his only car. 23% ofdays his car won’t start. 18% of the days the roads are impassible. 5% of the days both the roadsare impassible and his car won’t start. What is the probability that Bob won’t be able to get totown on a randomly selected day?

Solution.

Solution: The problem is asking for the union of two events. Let’s start by defining a few events.Let C = ”the car won’t start”Let R = ”the roads are impassible”Translating, we are looking for P (C∪R) . We are given P (C) = .23, P (R) = .18, and P (C∩R) =

.05 .Putting into the formula we get

P (C ∪R) = P (C) + P (R)− P (C ∩R) = .23 + .18− .05 = .36


So the probability is .36 that Bob won’t be able to get to town. Put in percent form, 36% ofdays Bob can’t get to town.

4.4.1 Exercises

1. What is the probability of throwing a pair of dice and getting at least one 3 showing?

2. A die has 3 sides that are blue, 2 sides that are red , and 1 side that is black. What is theprobability of throwing the die twice and getting at least one red face up?

3. In a large city, 52% of households own dogs, 38% of households own cats, and 19% own bothcats and dogs. What percent of households own a pet (cat or dog)?

4. A baseball fan is considering going to a game on the weekend. The fan is broke and will onlybe able to buy tickets after getting paid on Friday. The fan’s favorite team has a game onboth Saturday and Sunday. The fan estimates that the probability that Saturday’s game willhave tickets available Friday is .67. The fan also estimates that the Sunday game has a .84probability that the game will have tickets available on Friday. Their is a .53 probability thaton Friday there will be tickets available for both days. What is the probability that the fanwill be able to see the team play this weekend?

5. Several people who wanted to have children in their family were asked how many siblings theyhave and how many children they wanted to have. See the table for the results.




(a) They have no siblings or want to have 2 or more children.

(b) They weren’t an only child or don’t want to have an only child.

6. Several people with smart phones were asked if they use it to check their email and their age. The results are in the table that follows.




(a) They don’t check email on their smart phone or are under 30 years old.

(b) They are 30+ years old or they check their email on their smart phone.


On Time or Early Late: < 15 minutes 15+ minutes lateSmall aircraft 302 198 58Large aircraft 83 123 91


One of these flights is to be selected. Find the probability that . . .

(a) A large aircraft is chosen or it is 15+ minutes late.

(b) A flight that wasn’t late or the aircraft was small.



One of these students is to be selected. Find the probability that . . .

(a) A freshman or someone planning on going to college is chosen.

(b) A senior or a student not planning on going to college is chosen.




We are to select one of these movie-goers. Find the probability that . . .

(a) A childless movie-goer or someone that gave it a thumbs up.

(b) A movie-goer that gave it a thumbs down or someone that has children.



Wild 116 107

One of these salmon carcasses, one is to be selected at random. Find the probability that thesalmon . . .

(a) Is a hatchery raised or is a female.

(b) Is a wild salmon or is a male.



One of these persons are to be selected at random. Find the probability of selecting . . .

(a) An iPhone owner or someone that has an Apple computer.

(b) A non-Apple computer owner or someone that has an ‘other’ phone.





(a) A yeast or glazed donut is selected.

(b) A maple or chocolate donut is selected.





(a) A meal with chicken or water is selected.

(b) A meal with water or iced tea is selected.



Unscented 52 26 10

(a) What percent of roses sold were red or unscented?

(b) What percent of roses sold were not red?



Not Local 153 162 208

(a) What percent of participants were local or preferred the swim?

(b) What percent of participants weren’t local or preferred the run?



One of these people are to be selected at random. Find the probability of selecting . . .


(a) A person in favor of the law or someone that got a refund.

(b) A person who paid a lot or someone that is opposed to the bill.

4.5. COUNTING TECHNIQUES 87

4.5 Counting Techniques

If we are interested in playing the Power Ball Lottery and want to know the probability of winning,how do we do this? We have, conceptually, the idea already. We simply need to divide one(thereis only one way to pick the correct numbers) by the total number of ways of picking the numbers.How do we count the number of ways? In this section we will address this.

When we are counting, we need to distinguish when we are sampling without replacement orwith replacement. We also need to determine if the order the objectes are selected matters or not.

Example 4.5.1.

A vacationer is picking shorts, a shirt, and a hat to go explore. The vacationer brought 3 shorts,4 shorts and 2 hats. How many different possible outfits are possible?

Solution.

This problem is really a tree diagram problem. We will see that the tree needs some ‘pruning’.When our vacationer goes to get dressed, they need to pick out shorts: there are 3 different waysto do this. This is seen in the tree diagram where we have three branches. Next, we need a shirt.There are 4 ways to pick out a shirt for each way we pick a pair of shorts. In the tree diagram thereare 4 branches coming off each of the 3 branches from before. This gives us 12 ways to pick out apair of shorts and a shirt. Lastly, we need to pick a hat. There are 2 branches for each short/shirtcombination. This gives us a total of 24 different possible outfits.

In the problem we made the assumption that there were no restrictions of combinations ofoutfits. (as if our vacationer had no sense of fashion) If there were restrictions, e.g. this pair ofshorts does not look good with this shirt, then the counting becomes more difficult.


Counting Principle:

If an experiment consists of 2 steps where each step has n1 and n2 possible outcomes thenthe total number of outcomes of the experiment is n = n1 × n2

For three steps we get n = n1 × n2 × n3.

This can be extended to any number of steps in the experiment.

Example 4.5.2.

Our vacationer has stopped to get lunch. There are 8 different sandwiches to choose from, 6different chips possible, 12 drinks possible, and 3 different cookies. The vacationer will choose onesandwich, one type of chip, one drink, and one cookie. How many different lunches are possible.

Solution.

Imagine drawing a tree diagram here! We will use the above counting priciple. We have8× 6× 12× 3 = 1728 different ways to pick a lunch.

4.5.1 Factorials

We now look at what are called factorials. These come from our counting priciple.

Example 4.5.3.

Our vacationer has brought four books to read while vacationing. In how many different ways(orders) can the vacationer do this?

Solution.

From our multiplication principle it should be: the number of ways to pick the first book ×the number of ways to pick the second book × the number of ways to pick the third book × thenumber of ways to pick the fourth book.

There are 4 books so there are 4 ways to pick the first book.There are now 3 books left so there are 3 ways to pick the second book.There are now 2 books left so there are 2 ways to pick the third book.There is only 1 book left so there is 1 way to pick the fourth book.

So there are 4× 3× 2× 1 = 24 ways to pick the order to read the books.

n! is read ‘n factorial’. And represents the number of ways to arrange n items. By definition,0! = 1.

n! = n× (n− 1) · · · 3× 2× 1


To do this on your calculator, input the number first then go to MATH > PRB > 4:! then hitENTER.

Example 4.5.4.

In how many ways can we arrange the cards in a standard deck of cards?

Solution.

There are 52 cards in a standard deck of cards so there are 52! ways to arrange them. We get8.07× 1067 from our calculator.2

4.5.2 Permutations

Permuatations describe the number of ways of selecting items when order matters. Unlike factorials,where order also matters, we are not selecting all items.

nPr Represents the number of ways of selecting r items from n items when order matters.

Example 4.5.5.

From 12 members of a council, 3 members are to be selected: one will serve as the chair, anotheras secretary, and a third as treasurer. How many ways can the assignments be made.

Solution.

We are selecting 3 people out of 12. Order matters here: being picked as the secretary is differentthan being selected as treasurer, etc. This is a permuation problem. Although we haven’t figuredout how to do this yet as a permutation, we can use the counting principle from before. There are12 ways to pick the first, 11 ways to pick the second and 10 ways to pick the third. There are atotal of 12× 11× 10 = 1320 different ways to make the assignments.

nPr can be accessed from you calculator.

� Input the first number (n)

� MATH>PRB>2:nPr

� Input the second number (r)

� ENTER

For the above example first input 12 then MATH > PRB > 2:nPr, input the 3 then ENTER.If you aren’t selecting more than a few items, it is probably easier to use the counting principle.

2Most calculators will display this as 8.0658E67


4.5.3 Combinations

Combinations are just like permutations with one major difference: with combinations order doesnot matter.

nCr or, equivalently,(nk

)read ‘n choose k’ is the number of ways of selecting k items out of n

items if order does not matter.

To calculate the combinations, we proceed exactly as we do with permuations but instead ofchoosing nPr on our calculator, we choose nCr.

Example 4.5.6.

A yougnster is standing in front of a soda dispensing machine that has 8 different flavors. Theyouth will dispense equal amounts of 5 different sodas into their cup, mix and drink. How manydifferent ways can our adventurous youth do this?

Solution.

A few subtleties we need to see: equal amounts of different drinks are being mixed together.Since they are equal amounts and different, we can think of this as choosing 5 items out of 8. Sincethey are being mixed together, it doen’t matter which beverage is first, second, etc. So order doesnot matter.

We get 8C5 =(

85

)= 56 different possible beverages.

The following example uses more than one of the previous methods.

Example 4.5.7.

Find the probabiltiy of winning the top prize in the Powerball Lottery.

Solution.

We need to do some research. Here we go: There are 69 white balls, numbered 1 to 69 and 26Powerballs numbered 1 to 26. 5 white balls are picked and 1 Powerball. To play, you bubble in thenumbers you want for the white balls and bubble in one Powerball number3. To win, you need tomatch the correct 5 white balls and the Powerball.

The order the balls are selected does not matter. Think of this as a two step experiment: firstpick the white balls then pick the Powerball.

For the white balls, we are picking 5 balls out of 69 balls. There are(

695

)= 11, 238, 513 ways to

choose the white balls. If we got all of these, we need to match the Powerball as well.

There are 26 Powerballs so there are 26 ways to pick one Powerball.

3Just like the bubble-in scantron form you use for multiple choice tests.


Putting it together, there are 11, 238, 513× 26 = 292, 201, 3384 different ways for the Powerballlottery to turn out. There is only one way to match all the numbers chosen. So, using the classicalapproach to probability we get P (winning) = 1

292,201,338

Calculating Combinations and Permutations by hand. (Optional)

For small values of r, we can calculate combinations and permutations just as easily by hand aswith the nCr or nPr functions on our calculator. The formulas for them follow presented for thesake of completion. We will never use these formulas as written. To do so, we would need to findfactorials. To find the factorials, we use our calculator which has nPr and nCr as neighbors. Sowe will use nPr and nCr directly in those cases.

nPr =n!

(n− r)! and nCr =

(n

r

)=

n!

(n− r)!r!To calculate permutations by hand, nPr, we start with the first number, n, multiply by one less

than the first number, mutliply by two less than the first number etc., until the number of factorsequals r.

12P3 = 12× 11× 10 = 1320

3 items

Start with

To calculate combintations,(nr

), start with a fraction with n in the numerator and r in the

denominator. Then multiply by one less than each number, continue until the the last factor in thedenominator is 1.

12C3 =(123)= 12× 11× 10

3× 2× 1 = 220

4.5.4 Exercises

1. After having their computer stolen, a computer user is looking purchasing a new system.There are 6 different computers that are acceptable, 8 different monitors, and 6 differentprinters. How many ways can the user select a new system?

2. A car purchaser is considering purchasing a new car. They have the model picked out but nowneed to decide on options. There are 5 different colors that the purchaser likes, 4 different trimoptions, and 4 different stereo options. How many different ways can the car be specified?

3. A burger joint only sells burgers, fries, and drinks. If there are 4 different burger options, 3different size fries, and 12 different drinks what is the probability of someone guessing whatthe next customer orders, assuming they order one of each?

4Compare to the US population in 2020: 329,174,929


4. A school has 10 classrooms in a building. Over summer break there are enough funds toremodel 4 of the classrooms. In how many different ways can the rooms be chosen for remod-eling?

5. A third-grade teacher is casting a play about animals in the woods for the class. The teacherneed to cast an owl, a bear, a wolf, and an eagle. In how many ways can the teacher cast theplay is there are 25 students in the class?

6. A music teacher has only four instruments but 20 students. There is one guitar, one tamborine,one ukelele, and one tom-tom drum.

(a) In how many ways can the instruments be assigned?

(b) If it turns out that 7 of the students only know how to play guitar, 3 can only play theukelele, and the remaining can play both the the tamborine and the tom tom, but notthe guitar or ukelele. How many ways can the instruments be assigned if only peoplethat know how to play the instrument get assigned to a particular instrument.

7. A child is planning on coloring while on vacation. At the last minute they realize that theyforgot the crayons. There are 30 different crayons but the child is in a hurry and can onlytake 6 colors. How many ways can the child make the choices?

8. From biology we learned that there are 4 different bases for the DNA structure abbreviatedT, A, C, and G. According to wikipedia, the first DNA sequencing was done in 1977 of abacteriophage that has 5386 bases. (Sequencing is finding the order fo the bases in the DNA)Find an expression that gives the number of ways to pick the bases. (Your calculator will notbe able to do the calculation)

9. A quiz consists of 10 multiple choice questions. Each question has four possible answers butonly one answer is correct for each question. If a student guesses, what is the probability ofgetting all questions correct?

10. A student is picking classes for the upcoming semester and wants to take 3 classes: a mathclass, an English class, and an art class. There are 5 different math classes that the studentcan take, 9 different English classes, and 4 art classes. None of the times confilict for theclasses. How many ways can the student make their schedule?

11. A baseball team has 16 players but only 11 can play at a time.

(a) How many different ways can the coach pick the players to start the game, includingwhat postion they will play?

(b) The assistant coach will decide the order that the players bat. In how many ways canthis be done once the starters have been chosen?

12. A manufacturer of chips is planning on a multipack. There are 12 different varieties of chipsfrom which to choose. The multipack will have 4 different types of chips.

(a) How many different ways can the chips be selected if the multipack contains 20 bags ofchips with equal numbers of the different types of chips?


(b) How many different ways can the chips be selected if the multipack contains 20 bags ofchips with 8 of one kind, 6 of another kind, 4 of another kind, and 2 of the last kind?

13. In a department of 20 people, 5 are to be selected to evaluate their supervisor. In how manydifferent ways can the selection be made?

14. A combination lock is opened by rotating a wheel clockwise to a number, then counterclockwiseto a different number, then finally clockwise to a number different from the second number.The same number can’t be consecutive: 34-5-34 is ok, but 34-34-5 is not. If the wheel has 35numbers on it, how many different combinations are possible?

15. In California, the standard licence plate for a car is a digit, followed by 3 letters, and followedby 3 digits. How many different licence plates are possible?

16. A state offers personalized plates for automobiles. Residents can pick any 6 characters (eithera letter or digit). How many different licence plates are possible. (In reality, the actual numberis a little less than what you will calculate because you can’t create a licence plate that saysanything offensive.)

17. A computer system at a company has been hacked so a user needs to change their password.The user is lazy so they are going to create a password that consists of only letters from thehome row of their keyboard: ASDFGHJKL. The user needs to pick a password that is exactly6 characters long.

(a) How many passwords are possible if they only use lowercase letters?

(b) How many passwords are possible if they are able to use uppercase or lowercase letters?

(c) If the user needs a password that contains at least one uppercase letter and at leastone lowercase letters how many passwords are possible? Hint: there are three distinctpossibilities: all uppercase, all lowercase, and at least one of each.

18. A movie has 5 famous performers in it. The producer has the task of listing the big stars,with big egos to match, on the credits. How many different ways can the names be arrangedin the credits?

19. A chocolate lover has 8 different types of chocolate and is planning on melting equal amountsof 4 of the chocolates together for a fondue. How many different flavors are possible?

20. For a coworker’s birthday, a worker is planning on bringing cupcakes for the festivities. Thereare 12 different flavors of cake that are possible and 8 different frosting flavors.

(a) How many different options are there if each cupcake needs frosting?

(b) How many different options are there if not all cupcakes need to have frosting?

21. After receiving gifts for their birthday, a grandparent writes 6 thank-yous. The cards arewritten and the envelopes addressed when they ask you to mail them. Unfortunately, thecards and envelopes fall to the ground scattered. You decide to guess which thank-you goesin which envelope. What is the probability that you guess all correctly?

22. A phone has a code needed to get in. The user needs to enter 4 digits to gain access.


(a) What is the probability of you guessing the code on the first guess?

(b) What is the probability of you needing no more than two guesses to get in?

(c) What is the probability that 4 guesses aren’t enough?

23. In the game of Clue, players need to figure out who the murderer is, what room it happened in,and the weapon used. Three cards are hidden from view with the solution. If they correctlyfigure it out, they win the game. There are 9 rooms, 6 suspects, and 6 weapons.

(a) What is the probability of you guessing the three cards that are to be selected?

(b) You have been dealt 2 suspect cards and 1 weapon card. What is the probability ofwinning after looking at your cards and making one guess?

(c) What is the probability of that 4 guesses aren’t enough?

24. At a party, the party planner wants two cakes and three types of ice cream available to guestsfor dessert. There are 8 different cakes available and 6 different ice creams. In how manydifferent ways can the planner select the desserts for the party?

probabilitybkrein/introductory statistics... · 2020-01-28 · 66 chapter 4. probability 12.a box...

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