Chapter 3. Probability Topics

3.1. Probability Topics*

Student Learning Objectives

By the end of this chapter, the student should be able to:

  • Understand and use the terminology of probability.

  • Determine whether two events are mutually exclusive and whether two events are independent.

  • Calculate probabilities using the Addition Rules and Multiplication Rules.

  • Construct and interpret Contingency Tables.

  • Construct and interpret Venn Diagrams (optional).

  • Construct and interpret Tree Diagrams (optional).

Introduction

It is often necessary to “guess” about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.

You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn to solve probability problems using a systematic approach.

Optional Collaborative Classroom Exercise

Your instructor will survey your class. Count the number of students in the class today.

  • Raise your hand if you have any change in your pocket or purse. Record the number of raised hands.

  • Raise your hand if you rode a bus within the past month. Record the number of raised hands.

  • Raise your hand if you answered “yes” to BOTH of the first two questions. Record the number of raised hands.

Use the class data as estimates of the following probabilities. P(change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. P(bus) means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.

  • Find P(change).

  • Find P(bus).

  • Find P(change and bus) Find the probability that a randomly chosen student in your class has change in his/her pocket or purse and rode a bus within the last month.

  • Find P(change| bus) Find the probability that a randomly chosen student has change given that he/she rode a bus within the last month. Count all the students that rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus.

3.2. Terminology*

Probability measures the uncertainty that is associated with the outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin is an example of an experiment.

The result of an experiment is called an outcome. A sample space is a set of all possible outcomes. Three ways to represent a sample space are to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. For example, if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes.

An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. The probability of an event A is written P(A).

The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between 0 and 1, inclusive (includes 0 and 1 and all numbers between these values). P(A) = 0 means the event A can never happen. P(A) = 1 means the event A always happens. P(A) = 0.5 means the event A is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative fequency of heads approaches 0.5 (the probability of heads).

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head(H) and a Tail(T) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. The sample space has four outcomes. A = getting one head. There are two outcomes {HT, TH}. .

Suppose you roll one fair six-sided die, with the numbers {1,2,3,4,5,6} on its faces. Let event E = rolling a number that is at least 5. There are two outcomes {5, 6}. . If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, 2/6 of the rolls would result an outcome of “at least 5”. The long-term relative frequency of obtaining this result would approach the theoretical probability of 2/6 as the number of repetitions grows larger and larger.

This important characteristic of probability experiments is the known as the Law of Large Numbers: as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes don’t happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.) The Law of Large Numbers will be discussed again in Chapter 7.

It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased . Two math professors in Europe had their statistics students test the Belgian 1 Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos have a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later in this chapter we will learn techniques to use to work with probabilities for events that are not equally likely.

OR” Event:

An outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B . For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}. A OR B  =  {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice.

AND” Event:

An outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let A and B be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}.

The complement of event A is denoted A’ (read “A prime”). A’ consists of all outcomes that are NOT in A . Notice that P(A) + P(A’) = 1. For example, let S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}. Then,

The conditional probability of A given B is written P(A|B). P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B . The formula to calculate P(A|B) is

P(A|B)=

where P(B) is greater than 0.

For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even (2, 4, 6). To calculate P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes in B (and not S ).

We get the same result by using the formula. Remember that S has 6 outcomes.

P(A|B) =

Understanding Terminology and Symbols

It is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.

Exercise 3.2.1. (Go to Solution)

In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events f or parts (a) through (j) below. (Note that you can’t find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)

  • Let F be the event that a student is female.

  • Let M be the event that a student is male.

  • Let S be the event that a student has short hair.

  • Let L be the event that a student has long hair.

a. The probability that a student does not have long hair.
b. The probability that a student is male or has short hair.
c. The probability that a student is a female and has long hair.
d. The probability that a student is male, given that the student has long hair.
e. The probability that a student is has long hair, given that the student is male.
f. Of all the female students, the probability that a student has short hair.
g. Of all students with long hair, the probability that a student is female.
h. The probability that a student is female or has long hair.
i. The probability that a randomly selected student is a male student with short hair.
j. The probability that a student is female.

**With contributions from Roberta Bloom

Solutions to Exercises

Solution to Exercise 3.2.1. (Return to Exercise)

a. P(L’)=P(S)
b. P(M or S)
c. P(F and L)
d. P(M|L)
e. P(L|M)
f. P(S|F)
g. P(F|L)
h. P(F or L)
i. P(M and S)
j. P(F)

Glossary

Conditional Probability

The likelihood that an event will occur given that another event has already occurred.

Equally Likely

Each outcome of an experiment has the same probability.

Experiment

A planned activity carried out under controlled conditions.

Event

A subset in the set of all outcomes of an experiment. The set of all outcomes of an experiment is called a sample space and denoted usually by S. An event is any arbitrary subset in S. It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, etc. Standard notations for events are capital letters such as A, B, C, etc.

Outcome (observation)

A particular result of an experiment.

Probability

A number between 0 and 1, inclusive, that gives the likelihood that a specific event will occur. The foundation of statistics is given by the following 3 axioms (by A. N. Kolmogorov, 1930’s): Let S denote the sample space and A and B are two events in S . Then:

  • 0 ≤ P(A) ≤ 1; .

  • If A and B are any two mutually exclusive events, then P ( A or B ) = P ( A ) + P ( B ) .

  • P ( S ) = 1.

Sample Space

The set of all possible outcomes of an experiment.

3.3. Independent and Mutually Exclusive Events*

Independent and mutually exclusive do not mean the same thing.

Independent Events

Two events are independent if the following are true:

  • P(A|B) = P(A)

  • P(B|A) = P(B)

  • P(A AND B) = P(A) ⋅ P(B)

If A and B are independent, then the chance of A occurring does not affect the chance of B occurring and vice versa. For example, two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.

Sampling may be done with replacement or without replacement.

  • With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.

  • Without replacement:: When sampling is done without replacement, then each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.

If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.

Mutually Exclusive Events

A and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0.

For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}. A AND B = {4, 5}. P(A AND B) = and is not equal to zero. Therefore, A and B are not mutually exclusive. A and C do not have any numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive.

If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise.

The following examples illustrate these definitions and terms.

Example 3.1. 

Flip two fair coins. (This is an experiment.)

The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.

  • Let A = the event of getting at most one tail. (At most one tail means 0 or 1 tail.) Then A can be written as {HH, HT, TH}. The outcome HH shows 0 tails. HT and TH each show 1 tail.

  • Let B = the event of getting all tails. B can be written as {TT}. B is the complement of A . So, B = A’. Also, P(A) + P(B) = P(A) + P(A’) = 1.

  • The probabilities for A and for B are and .

  • Let C = the event of getting all heads. C = {HH}. Since B = {TT}, P(B AND C) = 0. B and C are mutually exclusive. ( B and C have no members in common because you cannot have all tails and all heads at the same time.)

  • Let D = event of getting more than one tail. D = {TT}. .

  • Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = {HT, HH}. .

  • Find the probability of getting at least one (1 or 2) tail in two flips. Let F = event of getting at least one tail in two flips. F = {HT, TH, TT}.


Example 3.2. 

Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}.

  • Find the complement of A , A’. The complement of A , A’, is B because A and B together make up the sample space. P(A) + P(B) = P(A) + P(A’) = 1. Also, and

  • Let event C = odd faces larger than 2. Then C = {3, 5}. Let event D = all even faces smaller than 5. Then D = {2, 4}. P(C and D) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events.

  • Let event E = all faces less than 5. E = {1, 2, 3, 4}.

    Problem (Go to Solution)

    Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?


  • Find P(C|A). This is a conditional. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find P(C|A), find the probability of C using the sample space A . You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So,


Example 3.3. 

Let event G = taking a math class. Let event H = taking a science class. Then, AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H  independent?

If G and H are independent, then you must show ONE of the following:

  • P(G|H) = P(G)

  • P(H|G) = P(H)

  • P(G AND H) = P(G)⋅P(H)

Note

The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.

Problem 1.

Show that P(G|H) = P(G).

Solution



Problem 2.

Show P(G AND H) = P(G)⋅P(H).

Solution

P(G) ⋅ P(H)  =  0.6 ⋅ 0.5  =  0.3  =  P(G AND H)



Since G and H are independent, then, knowing that a person is taking a science class does not change the chance that he/she is taking math. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he/she is taking math. For practice, show that P(H|G) = P(H) to show that G and H are independent events.


Example 3.4. 

In a box there are 3 red cards and 5 blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.

The sample space S  =  R1, R2, R3, B1, B2, B3, B4, B5 . S has 8 outcomes.

  • . . P(R AND B) = 0 . (You cannot draw one card that is both red and blue.)

  • . (There are 3 even-numbered cards, R2, B2, and B4.)

  • . (There are 5 blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are 2 even cards: B2 and B4.)

  • . (There are 3 even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, 2 are blue: B2 and B4.)

  • The events R and B are mutually exclusive because P(R AND B) = 0.

  • Let G = card with a number greater than 3. G = {B4, B5}. . Let H = blue card numbered between 1 and 4, inclusive. H = {B1,B2,B3, B4}. . (The only card in H that has a number greater than 3 is B4.) Since , P(G) = P(G|H) which means that G and H are independent.


Example 3.5. 

In a particular college class, 60% of the students are female. 50 % of all students in the class have long hair. 45% of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that the student is female. Let L be the event that the student has long hair. Are the events of being female and having long hair independent?

  • The following probabilities are given in this example:

  • P(F ) = 0.60 ; P(L ) = 0.50

  • P(F AND L) = 0.45

  • P(L|F) = 0.75

Note

The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(F|L) yet, so you can not use the second condition.

Solution 1

Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) = (0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because P(F and L) does not equal P(F)P(L).

Solution 2

check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.

Interpretation of Results

The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.


**Example 5 contributed by Roberta Bloom

Solutions to Exercises

Solution to Exercise (Return to Problem)

No. C = {3, 5} and E = {1, 2, 3, 4}. . To be mutually exclusive, P(C AND E) must be 0.


Glossary

Independent Events

The occurrence of one event has no effect on the probability of the occurrence of any other event. Events A and B are independent if one of the following is true: (1). P ( A | B ) = P ( A ) ; (2) P ( B | A ) = P ( B ) ; (3) P ( A and B ) = P ( A ) P ( B ).

Mutually Exclusive

An observation cannot fall into more than one class (category). Being in more than one category prevents being in a mutually exclusive category.

3.4. Two Basic Rules of Probability*

The Multiplication Rule

If A and B are two events defined on a sample space, then: P(A AND B) = P(B)⋅P(A|B).

This rule may also be written as : P(A|B)=

(The probability of A given B equals the probability of A and B divided by the probability of B .)

If A and B are independent, then P(A|B) = P(A). Then P(A AND B) = P(A|B) P(B) becomes P(A AND B) = P(A) P(B).

The Addition Rule

If A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) – P(A AND B).

If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) – P(A AND B) becomes P(A OR B) = P(A) + P(B).

Example 3.6. 

Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska

  • Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35.

  • P(A and B) = 0 because Klaus can only afford to take one vacation

  • Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05.


Example 3.7. 

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game.

A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.

Problem 1.

What is the probability that he makes both goals?

Solution

The problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90:

(3.1) P(B AND A) = P(B|A) P(A)  =  0.90 * 0.65 = 0.585

Carlos makes the first and second goals with probability 0.585.



Problem 2.

What is the probability that Carlos makes either the first goal or the second goal?

Solution

The problem is asking you to find P(A OR B).

(3.2)P(A OR B) = P(A) + P(B) – P(A AND B) = 0.65 + 0.65 – 0.585 = 0.715

Carlos makes either the first goal or the second goal with probability 0.715.



Problem 3.

Are A and B  independent?

Solution

No, they are not, because P(B AND A)  =  0.585 .

(3.3) P(B)  ⋅  P(A)  =  (0.65)  ⋅  (0.65)  =  0.423
(3.4) 0.423  ≠  0.585  =  P(B AND A)

So, P(B AND A) is not equal to P(B) ⋅ P(A) .



Problem 4.

Are A and B mutually exclusive?

Solution

No, they are not because P(A and B) = 0.585.

To be mutually exclusive, P(A AND B) must equal 0.




Example 3.8. 

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice 4 times a week. Thirty of the intermediate swimmers practice 4 times a week. Ten of the novice swimmers practice 4 times a week. Suppose one member of the swim team is randomly chosen. Answer the questions (Verify the answers):

Problem 1.

What is the probability that the member is a novice swimmer?

Solution



Problem 2.

What is the probability that the member practices 4 times a week?

Solution



Problem 3.

What is the probability that the member is an advanced swimmer and practices 4 times a week?

Solution



Problem 4.

What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?

Solution

P(advanced AND intermediate)  =  0 , so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.



Problem 5.

Are being a novice swimmer and practicing 4 times a week independent events? Why or why not?

Solution

No, these are not independent events.

(3.5) P(novice AND practices 4 times per week) = 0.0667
(3.6) P(novice) ⋅ P(practices 4 times per week) = 0.0996
(3.7) 0.0667 ≠ 0.0996



Example 3.9. 

Studies show that, if she lives to be 90, about 1 woman in 7 (approximately 14.3%) will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is believed to be negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative.

Problem 1.

What is the probability that a woman develops breast cancer? What is the probability that woman tests negative?

Solution

P(B) = 0.143 ; P(N) = 0.85



Problem 2.

Given that a woman has breast cancer, what is the probability that she tests negative?

Solution

P(N|B) = 0.02



Problem 3.

What is the probability that a woman has breast cancer AND tests negative?

Solution

P(B AND N) = P(B) ⋅ P(N|B) = (0.143) ⋅ (0.02) = 0.0029



Problem 4.

What is the probability that a woman has breast cancer or tests negative?

Solution

P(B OR N) = P(B) + P(N) – P(B AND N) = 0.143 + 0.85 – 0.0029 = 0.9901



Problem 5.

Are having breast cancer and testing negative independent events?

Solution

No. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N)



Problem 6.

Are having breast cancer and testing negative mutually exclusive?

Solution

No. P(B AND N) = 0.0020. For B and N to be mutually exclusive, P(B AND N) must be 0.




Glossary

Independent Events

The occurrence of one event has no effect on the probability of the occurrence of any other event. Events A and B are independent if one of the following is true: (1). P ( A | B ) = P ( A ) ; (2) P ( B | A ) = P ( B ) ; (3) P ( A and B ) = P ( A ) P ( B ).

Mutually Exclusive

An observation cannot fall into more than one class (category). Being in more than one category prevents being in a mutually exclusive category.

Sample Space

The set of all possible outcomes of an experiment.

3.5. Contingency Tables*

A contingency table provides a different way of calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.

Example 3.10. 

Suppose a study of speeding violations and drivers who use car phones produced the following fictional data:

Table 3.1.
 Speeding violation in the last yearNo speeding violation in the last yearTotal
Car phone user25280305
Not a car phone user45405450
Total70685755

The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.

Calculate the following probabilities using the table

Problem 1.

P(person is a car phone user) =

Solution



Problem 2.

P(person had no violation in the last year) =

Solution



Problem 3.

P(person had no violation in the last year AND was a car phone user) =

Solution



Problem 4.

P(person is a car phone user OR person had no violation in the last year) =

Solution



Problem 5.

P(person is a car phone user GIVEN person had a violation in the last year) =

Solution

(The sample space is reduced to the number of persons who had a violation.)



Problem 6.

P(person had no violation last year GIVEN person was not a car phone user) =

Solution

(The sample space is reduced to the number of persons who were not car phone users.)




Example 3.11. 

The following table shows a random sample of 100 hikers and the areas of hiking preferred:

Table 3.2. Hiking Area Preference
SexThe CoastlineNear Lakes and StreamsOn Mountain PeaksTotal
Female1816___45
Male______1455
Total___41______

Problem 1. (Go to Solution)

Complete the table.


Problem 2. (Go to Solution)

Are the events “being female” and “preferring the coastline” independent events?

Let F = being female and let C = preferring the coastline.

a. P(F AND C) =
b. P(F)⋅P(C) =

Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.


Problem 3. (Go to Solution)

Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male and let L = prefers hiking near lakes and streams.

a. What word tells you this is a conditional?
b. Fill in the blanks and calculate the probability: P(___|___) = ___.
c. Is the sample space for this problem all 100 hikers? If not, what is it?

Problem 4. (Go to Solution)

Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female and let P = prefers mountain peaks.

a. P(F) =
b. P(P) =
c. P(F AND P) =
d. Therefore, P(F OR P) =


Example 3.12. 

Muddy Mouse lives in a cage with 3 doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is and the probability he is not caught is . If he goes out the second door, the probability he gets caught by Alissa is and the probability he is not caught is . The probability that Alissa catches Muddy coming out of the third door is and the probability she does not catch Muddy is . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is .

Table 3.3. Door Choice
Caught or NotDoor OneDoor TwoDoor ThreeTotal
Caught ____
Not Caught ____
Total____________1
  • The first entry is P(Door One AND Caught).

  • The entry is P(Door One AND Not Caught).

Verify the remaining entries.

Problem 1. (Go to Solution)

Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.


Problem 2.

What is the probability that Alissa does not catch Muddy?

Solution



Problem 3.

What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

Solution



Note

You could also do this problem by using a probability tree. See the Tree Diagrams (Optional) section of this chapter for examples.


Solutions to Exercises

Solution to Exercise 1. (Return to Problem)

Table 3.4. Hiking Area Preference
SexThe CoastlineNear Lakes and StreamsOn Mountain PeaksTotal
Female1816 11 45
Male 16 25 1455
Total 34 41 25 100

Solution to Exercise 2. (Return to Problem)

a.
b.

P(F AND C) ≠ P(F)⋅P(C), so the events F and C are not independent.


Solution to Exercise 3. (Return to Problem)

a. The word ‘given’ tells you that this is a conditional.
b.
c. No, the sample space for this problem is 41.

Solution to Exercise 4. (Return to Problem)

a.
b.
c.
d.

Solution to Exercise 1. (Return to Problem)

Table 3.5. Door Choice
Caught or NotDoor OneDoor TwoDoor ThreeTotal
Caught
Not Caught
Total 1

Glossary

Contingency Table

The method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other. The table provides an easy way to calculate conditional probabilities.

3.6. Venn Diagrams (optional)*

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.

Example 3.13. 

Suppose an experiment has the outcomes 1, 2, 3, … , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then AND B = {6} and OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:

A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.

Example 3.14. 

Flip 2 fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}. A OR B = {TH, TT, HT}.

The sample space when you flip two fair coins is S = {HH, HT, TH, TT}. The outcome HH is in neither A nor B . The Venn diagram is as follows:

Venn diagram with set A containing Tails + Heads and Tails + Tails, and set B containing Tails + Tails and Head + Tails. Head + Heads is contained in neither set, and set A and set B share Tails + Tails.

Example 3.15. 

Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time.

Venn diagram with one set containing students in clubs and students in clubs and working part-time and another set containing C/PT and students working part-time. Both sets share C/PT.

  • The probability that a students belongs to a club is P(C) = 0.40.

  • The probability that a student works part time is P(PT) = 0.50.

  • The probability that a student belongs to a club AND works part time is P(C AND PT) = 0.05.

  • The probability that a student belongs to a club given that the student works part time is:

    (3.8)

  • The probability that a student belongs to a club OR works part time is:

    (3.9) P(C OR PT) = P(C) + P(PT) – P(C AND PT) = 0.40 + 0.50 – 0.05 = 0.85


Glossary

Venn Diagram

The visual representation of a sample space and events in the form of circles or ovals showing their intersections.

3.7. Tree Diagrams (optional)*

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of “branches” that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

Example 3.16. 

In an urn, there are 11 balls. Three balls are red ( R ) and 8 balls are blue ( B ). Draw two balls, one at a time, with replacement. “With replacement” means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

Figure 3.1. 

Tree diagram consisting of the first draw for the first branch and the second draw for the second branch. The first branch consists of 2 lines, 3R and 8B, and the second branch consists of 2 sets of 2 lines of 3R and 8B each. The lines produce 9RR, 24RB, 24BR, and 64BB.
Total = 64 + 24 + 24 + 9 = 121

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the 9 RR outcomes can be written as:

R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, and with replacement. There are 11  ⋅  11  =  121 outcomes, the size of the sample space.

Problem 1. (Go to Solution)

List the 24 BR outcomes: B1R1, B1R2, B1R3, …


Problem 2.

Using the tree diagram, calculate P(RR).

Solution



Problem 3.

Using the tree diagram, calculate P(RB OR BR).

Solution



Problem 4.

Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).

Solution



Problem 5.

Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw).

Solution

This problem is a conditional. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24  +  64  =  88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. .



Problem 6. (Go to Solution)

Using the tree diagram, calculate P(BB).


Problem 7. (Go to Solution)

Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).



Example 3.17. 

An urn has 3 red marbles and 8 blue marbles in it. Draw two marbles, one at a time, this time without replacement from the urn. “Without replacement” means that you do not put the first ball back before you select the second ball. Below is a tree diagram. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, .

Figure 3.2. 

Tree diagram consisting of the first draw for the first branch and the second draw for the second branch. The first branch consists of 2 lines, B 8/11 and R 3/11, and the second branch consists of 2 sets of 2 lines with B 7/10 and R 3/10 extending from line B 8/11 and B 8/10 and R 2/10 coming from line R 3/11. These 4 lines produce BB 56/110, BR 24/110, RB 24/110, and RR 6/10.

Note

If you draw a red on the first draw from the 3 red possibilities, there are 2 red left to draw on the second draw. You do not put back or replace the first ball after you have drawn it. You draw without replacement, so that on the second draw there are 10 marbles left in the urn.

Calculate the following probabilities using the tree diagram.

Problem 1.

P(RR) =

Solution



Problem 2. (Go to Solution)

Fill in the blanks:


Problem 3. (Go to Solution)

P(R on 2d | B on 1st) =


Problem 4. (Go to Solution)

Fill in the blanks:


Problem 5. (Go to Solution)

P(BB) =


Problem 6.

P(B on 2nd | R on 1st)  =

Solution

There are 6  +  24 outcomes that have R on the first draw (6 RR and 24 RB). The 6 and the 24 are frequencies. They are also the numerators of the fractions and . The sample space is no longer 110 but 6  +  24  =  30 . Twenty-four of the 30 outcomes have B on the second draw. The probability is then . Did you get this answer?



If we are using probabilities, we can label the tree in the following general way.

Tree diagram consisting of a first branch and a second branch. The first branch consists of 2 lines, P(R) and P(B), and the second branch consists of 2 sets of 2 lines with one set of P(B)(B) and P(R)(B) from line P(B) and one set of P(B)(R) and P(R)(R) from line P(R). P(B)(B) and P(R)(B) produce P(B and B)=P(BB) and P(B and R)=P(BR) and P(B)(R) and P(R)(R) produce P(R and B)=P(RB) and P(R and R)=P(RR).
  • P(R|R) here means P(R on 2nd | R on 1st)

  • P(B|R) here means P(B on 2nd | R on 1st)

  • P(R|B) here means P(R on 2nd | B on 1st)

  • P(B|B) here means P(B on 2nd | B on 1st)


Solutions to Exercises

Solution to Exercise 1. (Return to Problem)

B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3


Solution to Exercise 6. (Return to Problem)


Solution to Exercise 7. (Return to Problem)

There are 9  +  24 outcomes that have R on the first draw (9 RR and 24 RB ). The sample space is then 9  +  24  =  33 . Twenty-four of the 33 outcomes have B on the second draw. The probability is then .


Solution to Exercise 2. (Return to Problem)


Solution to Exercise 3. (Return to Problem)


Solution to Exercise 4. (Return to Problem)

P(R on 1st and B on 2nd)  =  P(RB)  = 


Solution to Exercise 5. (Return to Problem)


Glossary

Sample Space

The set of all possible outcomes of an experiment.

Tree Diagram

The useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes simultaneously with associated probabilities (frequencies, relative frequencies).

3.8. Summary of Formulas*

Formula 3.1. Complement

If A and A’ are complements then P(A) + P(A’ ) = 1


Formula 3.2. Addition Rule

P(A OR B) = P(A) + P(B) – P(A AND B)


Formula 3.3. Mutually Exclusive

If A and B are mutually exclusive then P(A AND B) = 0 ; so P(A OR B) = P(A) + P(B).


Formula 3.4. Multiplication Rule

  • P(A AND B) = P(B) P(A|B)

  • P(A AND B) = P(A) P(B|A)


Formula 3.5. Independence

If A and B are independent then:

  • P(A|B) = P(A)

  • P(B|A) = P(B)

  • P(A AND B) = P(A) P(B)


3.9. Practice 1: Contingency Tables*

Student Learning Objectives

  • The student will practice constructing and interpreting contingency tables.

Given

An article in the New England Journal of Medicine (by Haiman, Stram, Wilkens, Pike, et al., 1/26/06), reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most 10 cigarettes per day, there were 9886 African Americans, 2745 Native Hawaiians, 12,831 Latinos, 8378 Japanese Americans, and 7650 Whites. Of the people smoking 11-20 cigarettes per day, there were 6514 African Americans, 3062 Native Hawaiians, 4932 Latinos, 10,680 Japanese Americans, and 9877 Whites. Of the people smoking 21-30 cigarettes per day, there were 1671 African Americans, 1419 Native Hawaiians, 1406 Latinos, 4715 Japanese Americans, and 6062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2305 Japanese Americans, and 3970 Whites.

Complete the Table

Complete the table below using the data provided.

Table 3.6. Smoking Levels by Ethnicity
Smoking LevelAfrican AmericanNative HawaiianLatinoJapanese AmericansWhiteTOTALS
1-10      
11-20      
21-30      
31+      
TOTALS      

Analyze the Data

Suppose that one person from the study is randomly selected.

Exercise 3.9.1. (Go to Solution)

Find the probability that person smoked 11-20 cigarettes per day.


Exercise 3.9.2. (Go to Solution)

Find the probability that person was Latino.


Discussion Questions

Exercise 3.9.3. (Go to Solution)

In words, explain what it means to pick one person from the study and that person is “Japanese American AND smokes 21-30 cigarettes per day.” Also, find the probability.


Exercise 3.9.4. (Go to Solution)

In words, explain what it means to pick one person from the study and that person is “Japanese American OR smokes 21-30 cigarettes per day.” Also, find the probability.


Exercise 3.9.5. (Go to Solution)

In words, explain what it means to pick one person from the study and that person is “Japanese American GIVEN that person smokes 21-30 cigarettes per day.” Also, find the probability.


Exercise 3.9.6.

Prove that smoking level/day and ethnicity are dependent events.


Solutions to Exercises

Solution to Exercise 3.9.1. (Return to Exercise)


Solution to Exercise 3.9.2. (Return to Exercise)


Solution to Exercise 3.9.3. (Return to Exercise)


Solution to Exercise 3.9.4. (Return to Exercise)


Solution to Exercise 3.9.5. (Return to Exercise)


3.10. Practice 2: Calculating Probabilities*

Student Learning Objectives

  • Students will define basic probability terms.

  • Students will practice calculating probabilities.

  • Students will determine whether two events are mutually exclusive or whether two events are independent.

Note

Use probability rules to solve the problems below. Show your work.

Given

68% of Californians support the death penalty. A majority of all racial groups in California support the death penalty, except for black Californians, of whom 45% support the death penalty (Source: San Jose Mercury News, 12/2005). 6% of all Californians are black (Source: U.S. Census Bureau).

In this problem, let:

  • C = Californians supporting the death penalty

  • B = Black Californians

Suppose that one Californian is randomly selected.

Analyze the Data

Exercise 3.10.1. (Go to Solution)

P ( C )  =


Exercise 3.10.2. (Go to Solution)

P ( B )  =


Exercise 3.10.3. (Go to Solution)

P ( C | B )  =


Exercise 3.10.4.

In words, what is ” C | B “?


Exercise 3.10.5. (Go to Solution)

P ( B  AND  C )  =


Exercise 3.10.6.

In words, what is “ B and C ”?


Exercise 3.10.7. (Go to Solution)

Are B and C independent events? Show why or why not.


Exercise 3.10.8. (Go to Solution)

P ( B  OR  C )  =


Exercise 3.10.9.

In words, what is “ B or C ”?


Exercise 3.10.10. (Go to Solution)

Are B and C mutually exclusive events? Show why or why not.


Solutions to Exercises

Solution to Exercise 3.10.1. (Return to Exercise)

0.68


Solution to Exercise 3.10.2. (Return to Exercise)

0.06


Solution to Exercise 3.10.3. (Return to Exercise)

0.45


Solution to Exercise 3.10.5. (Return to Exercise)

0.027


Solution to Exercise 3.10.7. (Return to Exercise)

No


Solution to Exercise 3.10.8. (Return to Exercise)

0.713


Solution to Exercise 3.10.10. (Return to Exercise)

No


3.11. Homework*

Exercise 3.11.1. (Go to Solution)

Suppose that you have 8 cards. 5 are green and 3 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 3 yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

  • G = card drawn is green

  • E = card drawn is even-numbered

a. List the sample space.
b. P(G) =
c. P(G|E) =
d. P(G AND E) =
e. P(G OR E) =
f. Are G and E mutually exclusive? Justify your answer numerically.

Exercise 3.11.2.

Refer to the previous problem. Suppose that this time you randomly draw two cards, one at a time, and with replacement.

  • G 1 = first card is green

  • G 2 = second card is green

a. Draw a tree diagram of the situation.
b.
c. P ( at least one green ) =
d. P ( G 2 G 1 ) =
e. Are G 2 and G 1 independent events? Explain why or why not.

Exercise 3.11.3. (Go to Solution)

Refer to the previous problems. Suppose that this time you randomly draw two cards, one at a time, and without replacement.

  • G 1 = first card is green

  • G 2 = second card is green

a. Draw a tree diagram of the situation.
b>. P( G 1  AND  G 2 ) =
c. P(at least one green) =
d. P( G 2 | G 1 ) =
e. Are G 2 and G 1 independent events? Explain why or why not.

Exercise 3.11.4.

Roll two fair dice. Each die has 6 faces.

a. List the sample space.
b. Let A be the event that either a 3 or 4 is rolled first, followed by an even number. Find P(A).
c. Let B be the event that the sum of the two rolls is at most 7. Find P(B).
d. In words, explain what “P(A|B)” represents. Find P(A|B).
e. Are A and B mutually exclusive events? Explain your answer in 1 - 3 complete sentences, including numerical justification.
f. Are A and B independent events? Explain your answer in 1 - 3 complete sentences, including numerical justification.

Exercise 3.11.5. (Go to Solution)

A special deck of cards has 10 cards. Four are green, three are blue, and three are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin.

a. List the sample space.
b. Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P(A).
c. Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in 1 - 3 complete sentences, including numerical justification.
d. Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive? Explain your answer in 1 - 3 complete sentences, including numerical justification.

Exercise 3.11.6.

An experiment consists of first rolling a die and then tossing a coin:

a. List the sample space.
b. Let A be the event that either a 3 or 4 is rolled first, followed by landing a head on the coin toss. Find P(A).
c. Let B be the event that a number less than 2 is rolled, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in 1 - 3 complete sentences, including numerical justification.

Exercise 3.11.7. (Go to Solution)

An experiment consists of tossing a nickel, a dime and a quarter. Of interest is the side the coin lands on.

a. List the sample space.
b. Let A be the event that there are at least two tails. Find P(A).
c. Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in 1 - 3 complete sentences, including justification.

Exercise 3.11.8.

Consider the following scenario:

  • Let P(C) = 0.4

  • Let P(D) = 0.5

  • Let P(C|D) = 0.6

a. Find P(C AND D) .
b. Are C and D mutually exclusive? Why or why not?
c. Are C and D independent events? Why or why not?
d. Find P(C OR D) .
e. Find P(D|C).

Exercise 3.11.9. (Go to Solution)

E and F mutually exclusive events. P(E) = 0.4; P(F) = 0.5. Find P(EF).


Exercise 3.11.10.

J and K are independent events. P(J | K) = 0.3 . Find P(J) .


Exercise 3.11.11. (Go to Solution)

U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find:

a. P(U AND V) =
b. P(U | V) =
c. P(U OR V) =

Exercise 3.11.12.

Q and R are independent events. P(Q) = 0.4 ; P(Q AND R) = 0.1 . Find P(R).


Exercise 3.11.13. (Go to Solution)

Y and Z are independent events.

a. Rewrite the basic Addition Rule P(Y OR Z) = P(Y) + P(Z) – P(Y AND Z) using the information that Y and Z are independent events.
b. Use the rewritten rule to find P(Z) if P(Y OR Z) = 0.71 and P(Y) = 0.42 .


Exercise 3.11.14.

G and H are mutually exclusive events. P(G) = 0.5; P(H) = 0.3

a. Explain why the following statement MUST be false: P ( HG ) = 0 . 4 .
b. Find: P(H OR G).
c. Are G and H independent or dependent events? Explain in a complete sentence.

Exercise 3.11.15. (Go to Solution)

The following are real data from Santa Clara County, CA. As of March 31, 2000, there was a total of 3059 documented cases of AIDS in the county. They were grouped into the following categories (Source: Santa Clara County Public H.D.):

Table 3.7. * includes homosexual/bisexual IV drug users
 Homosexual/Bisexual IV Drug User*Heterosexual ContactOtherTotals
Female07013649____
Male214646360135____
Totals____________________

Suppose one of the persons with AIDS in Santa Clara County is randomly selected. Compute the following:

a. P(person is female) =
b. P(person has a risk factor Heterosexual Contact) =
c. P(person is female OR has a risk factor of IV Drug User) =
d. P(person is female AND has a risk factor of Homosexual/Bisexual) =
e. P(person is male AND has a risk factor of IV Drug User) =
f. P(female GIVEN person got the disease from heterosexual contact) =
g. Construct a Venn Diagram. Make one group females and the other group heterosexual contact.

Exercise 3.11.16.

Solve these questions using probability rules. Do NOT use the contingency table above. 3059 cases of AIDS had been reported in Santa Clara County, CA, through March 31, 2000. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact.

a. P(person is female) =
b. P(person obtained the disease through heterosexual contact) =
c. P(female GIVEN person got the disease from heterosexual contact) =
d. Construct a Venn Diagram. Make one group females and the other group heterosexual contact. Fill in all values as probabilities.

Exercise 3.11.17. (Go to Solution)

The following table identifies a group of children by one of four hair colors, and by type of hair.

Table 3.8.
Hair TypeBrownBlondBlackRedTotals
Wavy20 15343
Straight8015 12 
Totals 20  215
a. Complete the table above.
b. What is the probability that a randomly selected child will have wavy hair?
c. What is the probability that a randomly selected child will have either brown or blond hair?
d. What is the probability that a randomly selected child will have wavy brown hair?
e. What is the probability that a randomly selected child will have red hair, given that he has straight hair?
f. If B is the event of a child having brown hair, find the probability of the complement of B.
g. In words, what does the complement of B represent?

Exercise 3.11.18.

A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into the following table.

Table 3.9.
Shirt#≤ 210211-250251-290290≤
1-3321500
34-6661874
66-99612225

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

a. Find the probability that his shirt number is from 1 to 33.
b. Find the probability that he weighs at most 210 pounds.
c. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.
d. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.
e. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.
f. If having a shirt number from 1 to 33 and weighing at most 210 pounds were independent events, then what should be true about P(Shirt# 1-33 | ≤ 210 pounds) ?

Exercise 3.11.19. (Go to Solution)

Approximately 249,000,000 people live in the United States. Of these people, 31,800,000 speak a language other than English at home. Of those who speak another language at home, over 50 percent speak Spanish. (Source: U.S. Bureau of the Census, 1990 Census)

Let: E = speak English at home; E’ = speak another language at home; S = speak Spanish at home

Finish each probability statement by matching the correct answer.

Table 3.10.
Probability StatementsAnswers
a. P(E’) =i. 0.8723
b. P(E) =ii. > 0.50
c. P(S) =iii. 0.1277
d. P(S|E’) =iv. > 0.0639

Exercise 3.11.20.

The probability that a male develops some form of cancer in his lifetime is 0.4567 (Source: American Cancer Society). The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51 (Source: USA Today). Some of the questions below do not have enough information for you to answer them. Write “not enough information” for those answers.

Let: C = a man develops cancer in his lifetime; P = man has at least one false positive

a. Construct a tree diagram of the situation.
b. P(C) =
c. P(P|C) =
d. P(P|C’ ) =
e. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not.

Exercise 3.11.21. (Go to Solution)

In 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let G = won Green Card.

a. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement.
b. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Let F = was a finalist. Write your answer as a conditional probability statement.
c. Are G and F independent or dependent events? Justify your answer numerically and also explain why.
d. Are G and F mutually exclusive events? Justify your answer numerically and also explain why.

Note

P.S. Amazingly, on 2/1/95, Renate learned that she would receive her Green Card — true story!


Exercise 3.11.22.

Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with $10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned. (Source: Wall Street Journal)

Let: R = money returned; E = economics classes; O = other classes

a. Write a probability statement for the overall percent of money returned.
b. Write a probability statement for the percent of money returned out of the economics classes.
c. Write a probability statement for the percent of money returned out of the other classes.
d. Is money being returned independent of the class? Justify your answer numerically and explain it.
e. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer.

Exercise 3.11.23. (Go to Solution)

The chart below gives the number of suicides estimated in the U.S. for a recent year by age, race (black and white), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population. (Source: The National Center for Health Statistics, U.S. Dept. of Health and Human Services)

Table 3.11.
Race and Sex1 - 1415 - 2425 - 64over 64TOTALS
white, male210336013,610 22,050
white, female805803380 4930
black, male104601060 1670
black, female040270 330
all others     
TOTALS310465018,780 29,760

Note

Do not include “all others” for parts (f), (g), and (i).

a. Fill in the column for the suicides for individuals over age 64.
b. Fill in the row for all other races.
c. Find the probability that a randomly selected individual was a white male.
d. Find the probability that a randomly selected individual was a black female.
e. Find the probability that a randomly selected individual was black
f. Find the probability that a randomly selected individual was male.
g. Out of the individuals over age 64, find the probability that a randomly selected individual was a black or white male.
h. Comparing “Race and Sex” to “Age,” which two groups are mutually exclusive? How do you know?
i. Are being male and committing suicide over age 64 independent events? How do you know?

The next two questions refer to the following: The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20 - 64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20 - 64; 13.53% are age 65 or over. (Source: Federal Highway Administration, U.S. Dept. of Transportation)

Exercise 3.11.24.

Complete the following:

a. Construct a table or a tree diagram of the situation.
b. P(driver is female) =
c. P(driver is age 65 or over | driver is female) =
d. P(driver is age 65 or over AND female) =
e. In words, explain the difference between the probabilities in part (c) and part (d).
f. P(driver is age 65 or over) =
g. Are being age 65 or over and being female mutually exclusive events? How do you know

Exercise 3.11.25. (Go to Solution)

Suppose that 10,000 U.S. licensed drivers are randomly selected.

a. How many would you expect to be male?
b. Using the table or tree diagram from the previous exercise, construct a contingency table of gender versus age group.
c. Using the contingency table, find the probability that out of the age 20 - 64 group, a randomly selected driver is female.

Exercise 3.11.26.

Approximately 86.5% of Americans commute to work by car, truck or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. (Source: Bureau of the Census, U.S. Dept. of Commerce. Disregard rounding approximations.)

a. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.
b. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?
c. Suppose that 1000 workers are randomly selected. How many would you expect to travel alone to work?
d. Suppose that 1000 workers are randomly selected. How many would you expect to drive in a carpool?

Exercise 3.11.27.

Explain what is wrong with the following statements. Use complete sentences.

a. If there’s a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there’s a 130% chance of rain over the weekend.
b. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit.

Try these multiple choice questions.

The next two questions refer to the following probability tree diagram which shows tossing an unfair coin FOLLOWED BY drawing one bead from a cup containing 3 red ( R ), 4 yellow ( Y ) and 5 blue ( B ) beads. For the coin, and where H = “heads” and T = “tails”.

Figure 3.3. 

Tree diagram with 2 branches. The first branch consists of 2 lines of H=2/3 and T=1/3. The second branch consists of 2 sets of 3 lines each with the both sets containing R=3/12, Y=4/12, and B=5/12.

Exercise 3.11.28. (Go to Solution)

Find P(tossing a Head on the coin AND a Red bead)

A.
B.
C.
D.

Exercise 3.11.29. (Go to Solution)

Find P(Blue bead).

A.
B.
C.
D.

The next three questions refer to the following table of data obtained from www.baseball-almanac.com showing hit information for 4 well known baseball players. Suppose that one hit from the table is randomly selected.

Table 3.12.
NAMESingleDoubleTripleHome RunTOTAL HITS
Babe Ruth15175061367142873
Jackie Robinson1054273541371518
Ty Cobb36031742951144189
Hank Aaron2294624987553771
TOTAL84711577583172012351

Exercise 3.11.30. (Go to Solution)

Find P(hit was made by Babe Ruth).

A.
B.
C.
D.

Exercise 3.11.31. (Go to Solution)

Find P(hit was made by Ty Cobb | The hit was a Home Run)

A.
B.
C.
D.

Exercise 3.11.32. (Go to Solution)

Are the hit being made by Hank Aaron and the hit being a double independent events?

A. Yes, because P(hit by Hank Aaron | hit is a double) = P(hit by Hank Aaron)
B. No, because P(hit by Hank Aaron | hit is a double) ≠ P(hit is a double)
C. No, because P(hit is by Hank Aaron | hit is a double) ≠ P(hit by Hank Aaron)
D. Yes, because P(hit is by Hank Aaron | hit is a double) = P(hit is a double)

Exercise 3.11.33. (Go to Solution)

Given events G and H: P(G) = 0.43 ; P(H) = 0.26 ; P(H and G) = 0.14

A. Find P(H or G)
B. Find the probability of the complement of event (H and G)
C. Find the probability of the complement of event (H or G)


Exercise 3.11.34. (Go to Solution)

Given events J and K: P(J) = 0.18 ; P(K) = 0.37 ; P(J or K) = 0.45

A. Find P(J and K)
B. Find the probability of the complement of event (J and K)
C. Find the probability of the complement of event (J or K)

Exercise 3.11.35. (Go to Solution)

United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, http://www.unitedbloodservices.org/humanbloodtypes.html, a person with type O blood and a negative Rh factor (Rh−) can donate blood to any person with any bloodtype. Their data show that 43% of people have type O blood and 15% of people have Rh− factor; 52% of people have type O or Rh− factor.

A. Find the probability that a person has both type O blood and the Rh− factor
B. Find the probability that a person does NOT have both type O blood and the Rh− factor.

Exercise 3.11.36. (Go to Solution)

At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.

A. Find the probability that a course has a final exam or a research project.
B. Find the probability that a course has NEITHER of these two requirements.

Exercise 3.11.37. (Go to Solution)

In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.

A. Find the probability that a cookie contains chocolate or nuts (he can’t eat it).
B. Find the probability that a cookie does not contain chocolate or nuts (he can eat it).

Exercise 3.11.38. (Go to Solution)

A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student

A. Find P(D and E)
B. Find P(E | D)
C. Find P(D or E)
D. Using an appropriate test, show whether D and E are independent.
E. Using an appropriate test, show whether D and E are mutually exclusive.

Exercise 3.11.39. (Go to Solution)

At a certain store the manager has determined that 30% of customers pay cash and 70% of customers pay by debit card. (No other method of payment is accepted.) Let M = event that a customer pays cash and D= event that a customer pays by debit card.

A. Suppose two customers (Al and Betty) come to the store. Explain why it would be reasonable to assume that their choices of payment methods are independent of each other.
B. Draw the tree that represents the all possibilities for the 2 customers and their methods of payment. Write the probabilities along each branch of the tree.
C. For each complete path through the tree, write the event it represents and find the probability.
D. Let S be the event that both customers use the same method of payment. Find P(S)
E. Let T be the event that both customers use different methods of payment. Find P(T) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.
F. Let U be the event that the second customer uses a debit card. Find P(U)

Exercise 3.11.40. (Go to Solution)

A box of cookies contains 3 chocolate and 7 butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it also. (How many cookies did he take?)

A. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection, or do the probabilities depend on the type of cookie that Miguel selected first? Explain.
B. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
C. For each complete path through the tree, write the event it represents and find the probabilities.
D. Let S be the event that both cookies selected were the same flavor. Find P(S).
E. Let T be the event that both cookies selected were different flavors. Find P(T) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.
F. Let U be the event that the second cookie selected is a butter cookie. Find P(U).

Exercise 3.11.41. (Go to Solution)

When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian 1 Euro coin was a fair coin. They spun the coin rather than tossing it, and it was found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). Therefore, they claim that this is not a fair coin.

A. Based on the data above, find P(H) and P(T).
B. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.
C. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.
D. Use the tree to find the probability of obtaining at least one head.

**Exercises 33 - 41 contributed by Roberta Bloom

Solutions to Exercises

Solution to Exercise 3.11.1. (Return to Exercise)

a. {G1, G2, G3, G4, G5, Y1, Y2, Y3}
b.
c.
d.
e.
f. No

Solution to Exercise 3.11.3. (Return to Exercise)

b.
c.
d.
e. No

Solution to Exercise 3.11.5. (Return to Exercise)

a.
b.
c. Yes
d. No

Solution to Exercise 3.11.7. (Return to Exercise)

a.
b.
c. Yes

Solution to Exercise 3.11.9. (Return to Exercise)

0


Solution to Exercise 3.11.11. (Return to Exercise)

a. 0
b. 0
c. 0.63

Solution to Exercise 3.11.13. (Return to Exercise)

b. 0.5

Solution to Exercise 3.11.15. (Return to Exercise)

The completed contingency table is as follows:

Table 3.13. * includes homosexual/bisexual IV drug users
 Homosexual/Bisexual IV Drug User*Heterosexual ContactOtherTotals
Female07013649 255
Male214646360135 2804
Totals 2146 533 196 184 3059
a.
b.
c.
d. 0
e.
f.

Solution to Exercise 3.11.17. (Return to Exercise)

b.
c.
d.
e.
f.

Solution to Exercise 3.11.19. (Return to Exercise)

a. iii
b. i
c. iv
d. ii


Solution to Exercise 3.11.21. (Return to Exercise)

a. P ( G ) = 0 . 008
b. 0.5
c. dependent
d. No


Solution to Exercise 3.11.23. (Return to Exercise)

c.
d.
e.
f.
g.
h. Black females and ages 1-14
i. No

Solution to Exercise 3.11.25. (Return to Exercise)

a. 5140
c. 0.49

Solution to Exercise 3.11.28. (Return to Exercise)

C


Solution to Exercise 3.11.29. (Return to Exercise)

A


Solution to Exercise 3.11.30. (Return to Exercise)

B


Solution to Exercise 3.11.31. (Return to Exercise)

B


Solution to Exercise 3.11.32. (Return to Exercise)

 C


Solution to Exercise 3.11.33. (Return to Exercise)

A. P(H or G) = P(H) + P(G) − P(H and G) = 0.26 + 0.43 − 0.14 = 0.55
B. P( NOT (H and G) ) = 1 − P(H and G) = 1 − 0.14 = 0.86
C. P( NOT (H or G) ) = 1 − P(H or G) = 1 − 0.55 = 0.45


Solution to Exercise 3.11.34. (Return to Exercise)

A. P(J or K) = P(J) + P(K) − P(J and K); 0.45 = 0.18 + 0.37 − P(J and K) ; solve to find P(J and K) = 0.10
B. P( NOT (J and K) ) = 1 − P(J and K) = 1 − 0.10 = 0.90
C. P( NOT (J or K) ) = 1 − P(J or K) = 1 − 0.45 = 0.55

Solution to Exercise 3.11.35. (Return to Exercise)

A. P(Type O or Rh−) = P(Type O) + P(Rh−) − P(Type O and Rh−) 0.52 = 0.43 + 0.15 − P(Type O and Rh−); solve to find P(Type O and Rh−) = 0.06 6% of people have type O Rh− blood
B. P( NOT (Type O and Rh−) ) = 1 − P(Type O and Rh−) = 1 − 0.06 = 0.94 94% of people do not have type O Rh− blood

Solution to Exercise 3.11.36. (Return to Exercise)

A. P(R or F) = P(R) + P(F) − P(R and F) = 0.72 + 0.46 − 0.32 = 0.86
B. P( Neither R nor F ) = 1 − P(R or F) = 1 − 0.86 = 0.14

Solution to Exercise 3.11.37. (Return to Exercise)

Let C be the event that the cookie contains chocolate. Let N be the event that the cookie contains nuts.
A. P(C or N) = P(C) + P(N) − P(C and N) = 0.36 + 0.12 − 0.08 = 0.40
B. P( neither chocolate nor nuts) = 1 − P(C or N) = 1 − 0.40 = 0.60

Solution to Exercise 3.11.38. (Return to Exercise)

A. P(D and E) = P(D|E)P(E) = (0.20)(0.40) = 0.08
B. P(E|D) = P(D and E) / P(D) = 0.08/0.10 = 0.80
C. P(D or E) = P(D) + P(E) − P(D and E) = 0.10 + 0.40 − 0.08 = 0.42
D. Not Independent: P(D|E) = 0.20 which does not equal P(D) = .10
E. Not Mutually Exclusive: P(D and E) = 0.08 ; if they were mutually exclusive then we would need to have P(D and E) = 0, which is not true here.

Solution to Exercise 3.11.39. (Return to Exercise)

Contact your instruction.


Solution to Exercise 3.11.40. (Return to Exercise)

Contact your instruction.


Solution to Exercise 3.11.41. (Return to Exercise)

Contact your instructor.


3.12. Review*

The first six exercises refer to the following study: In a survey of 100 stocks on NASDAQ, the average percent increase for the past year was 9% for NASDAQ stocks. Answer the following:

Exercise 3.12.1. (Go to Solution)

The “average increase” for all NASDAQ stocks is the:

A. Population
B. Statistic
C. Parameter
D. Sample
E. Variable


Exercise 3.12.2. (Go to Solution)

All of the NASDAQ stocks are the:

A. Population
B. Statistic
C. Parameter
D. Sample
E. Variable


Exercise 3.12.3. (Go to Solution)

9% is the:

A. Population
B. Statistic
C. Parameter
D. Sample
E. Variable


Exercise 3.12.4. (Go to Solution)

The 100 NASDAQ stocks in the survey are the:

A. Population
B. Statistic
C. Parameter
D. Sample
E. Variable


Exercise 3.12.5. (Go to Solution)

The percent increase for one stock in the survey is the:

A. Population
B. Statistic
C. Parameter
D. Sample
E. Variable


Exercise 3.12.6. (Go to Solution)

Would the data collected be qualitative, quantitative – discrete, or quantitative – continuous?


The next two questions refer to the following study: Thirty people spent two weeks around Mardi Gras in New Orleans. Their two-week weight gain is below. (Note: a loss is shown by a negative weight gain.)

Table 3.14.
Weight GainFrequency
-23
-15
02
14
413
62
111

Exercise 3.12.7. (Go to Solution)

Calculate the following values:

a. The average weight gain for the two weeks
b. The standard deviation
c. The first, second, and third quartiles

Exercise 3.12.8.

Construct a histogram and a boxplot of the data.


Solutions to Exercises

Solution to Exercise 3.12.1. (Return to Exercise)

C. Parameter


Solution to Exercise 3.12.2. (Return to Exercise)

A. Population


Solution to Exercise 3.12.3. (Return to Exercise)

B. Statistic


Solution to Exercise 3.12.4. (Return to Exercise)

D. Sample


Solution to Exercise 3.12.5. (Return to Exercise)

E. Variable


Solution to Exercise 3.12.6. (Return to Exercise)

quantitative - continuous


Solution to Exercise 3.12.7. (Return to Exercise)

a. 2.27
b. 3.04
c. -1, 4, 4

3.13. Lab: Probability Topics*

Class time:

Names:

Student Learning Outcomes:

  • The student will use theoretical and empirical methods to estimate probabilities.

  • The student will appraise the differences between the two estimates.

  • The student will demonstrate an understanding of long-term relative frequencies.

Do the Experiment:

Count out 40 mixed-color M&M’s® which is approximately 1 small bag’s worth (distance learning classes using the virtual lab would want to count out 25 M&M’s®). Record the number of each color in the “Population” table. Use the information from this table to complete the theoretical probability questions. Next, put the M&M’s in a cup. The experiment is to pick 2 M&M’s, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of the empirical table. Do this 24 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of the “Empirical Results” table. After you record the pick, put both M&M’s back. Do this a total of 24 times, also. Use the data from the “Empirical Results” table to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions.

Table 3.15. Population
ColorQuantity
Yellow (Y) 
Green (G) 
Blue (BL) 
Brown (B) 
Orange (O) 
Red (R) 
Table 3.16. Theoretical Probabilities
Note: G2 = green on second pick; R1 = red on first pick; B1 = brown on first pick; B2 = brown on second pick; doubles = both picks are the same colour.
 With ReplacementWithout Replacement
P(2 reds)   
  
  
  
P(no yellows)   
P(doubles)   
P(no doubles)   
Table 3.17. Empirical Results
With ReplacementWithout Replacement
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
( __ , __ ) ( __ , __ )( __ , __ ) ( __ , __ )
Table 3.18. Empirical Probabilities
Note:
 With ReplacementWithout Replacement
P(2 reds)   
  
  
  
P(no yellows)   
P(doubles)   
P(no doubles)   

Discussion Questions

  1. Why are the “With Replacement” and “Without Replacement” probabilities different?

  2. Convert P(no yellows) to decimal format for both Theoretical “With Replacement” and for Empirical “With Replacement”. Round to 4 decimal places.

    a. Theoretical “With Replacement”: P(no yellows) =
    b. Empirical “With Replacement”: P(no yellows) =
    c. Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why?

  3. If you increased the number of times you picked 2 M&M’s to 240 times, why would empirical probability values change?

  4. Would this change (see (3) above) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know?

  5. Explain the differences in what and represent. Hint: Think about the sample space for each probability.