Probability Topics

19 Venn Diagrams

Venn Diagrams

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. Venn diagrams also help us to convert common English words into mathematical terms that help add precision.

Venn diagrams are named for their inventor, John Venn, a mathematics professor at Cambridge and an Anglican minister. His main work was conducted during the late 1870’s and gave rise to a whole branch of mathematics and a new way to approach issues of logic. We will develop the probability rules just covered using this powerful way to demonstrate the probability postulates including the Addition Rule, Multiplication Rule, Complement Rule, Independence, and Conditional Probability.

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 A intersect B = A\cap B=\text{{6}} and A union B = A\cup B=\text{{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.

(Figure) shows the most basic relationship among these numbers. First, the numbers are in groups called sets; set A and set B. Some number are in both sets; we say in set A \cap in set B. The English word “and” means inclusive, meaning having the characteristics of both A and B, or in this case, being a part of both A and B. This condition is called the INTERSECTION of the two sets. All members that are part of both sets constitute the intersection of the two sets. The intersection is written as A\cap B where \cap is the mathematical symbol for intersection. The statement A\cap B is read as “A intersect B.” You can remember this by thinking of the intersection of two streets.

There are also those numbers that form a group that, for membership, the number must be in either one or the other group. The number does not have to be in BOTH groups, but instead only in either one of the two. These numbers are called the UNION of the two sets and in this case they are the numbers 1-5 (from A exclusively), 7-9 (from set B exclusively) and also 6, which is in both sets A and B. The symbol for the UNION is \cup, thus A\cup B= numbers 1-9, but excludes number 10, 11, and 12. The values 10, 11, and 12 are part of the universe, but are not in either of the two sets.

Translating the English word “AND” into the mathematical logic symbol \cap, intersection, and the word “OR” into the mathematical symbol \cup, union, provides a very precise way to discuss the issues of probability and logic. The general terminology for the three areas of the Venn diagram in (Figure) is shown in (Figure).

Try It

Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C\cap P=\text{{blue}} and C\cup P=\text{{green, blue, purple, red, yellow}}. Draw a Venn diagram representing this situation.

Flip two 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\cap B=\text{{TT}}. A\cup B=\text{{TH, TT, HT}}.

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

This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.
Try It

Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A\cap B=\text{{3, 5}}. A\cup B=\text{{1, 2, 3, 5}}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.

A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.

This is an empty Venn diagram showing two overlapping circles. The left circle is labeled O and the right circle is labeled RH-.

The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.

We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor.

  1. P(O) = ___________
  2. P(R) = ___________
  3. P\left(O\cap R\right)= ___________
  4. P\left(O\cup R\right)= ____________
  5. In the Venn Diagram, describe the overlapping area using a complete sentence.
  6. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.

a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.

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.

This is a venn diagram with one set containing students in clubs and another set containing students working part-time. Both sets share students who are members of clubs and also work part-time. The universe is labeled S.

If a student is selected at random, find

  • the probability that the student belongs to a club. P(C) = 0.40
  • the probability that the student works part time. P(PT) = 0.50
  • the probability that the student belongs to a club AND works part time. P\left(C\cap \mathrm{PT}\right)=0.05
  • the probability that the student belongs to a club given that the student works part time. P\text{(}C|PT\text{)}=\frac{P\text{(}C\cap PT\text{)}}{P\text{(}PT\text{)}}=\frac{0.05}{0.50}=0.1
  • the probability that the student belongs to a club OR works part time. P\left(C\cup \mathrm{PT}\right)=P\left(C\right)+P\left(\mathrm{PT}\right)-P\left(C\cap \mathrm{PT}\right)=0.40+0.50-0.05=0.85

In order to solve (Figure) we had to draw upon the concept of conditional probability from the previous section. There we used tree diagrams to track the changes in the probabilities, because the sample space changed as we drew without replacement. In short, conditional probability is the chance that something will happen given that some other event has already happened. Put another way, the probability that something will happen conditioned upon the situation that something else is also true. In (Figure) the probability P(C|PT) is the conditional probability that the randomly drawn student is a member of the club, conditioned upon the fact that the student also is working part time. This allows us to see the relationship between Venn diagrams and the probability postulates.

Try It

Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works.

This is a venn diagram with one set containing students in clubs and another set containing students working part-time. Both sets share students who are members of clubs and also work part-time.
Try It

In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.

  1. Draw a Venn diagram representing the situation.
  2. Find the probability that the customer buys either a novel or a non-fiction book.
  3. In the Venn diagram, describe the overlapping area using a complete sentence.
  4. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.

a. and d. In the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks.

b. P(novel or non-fiction) = P(Blue \cup Red) = P(Blue) + P(Red) – P(Blue \cap Red) = 0.6 + 0.4 – 0.2 = 0.8.
c. The overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book.

A set of 20 German Shepherd dogs is observed. 12 are male, 8 are female, 10 have some brown coloring, and 5 have some white sections of fur. Answer the following using Venn Diagrams.

Draw a Venn diagram simply showing the sets of male and female dogs.

The Venn diagram below demonstrates the situation of mutually exclusive events where the outcomes are independent events. If a dog cannot be both male and female, then there is no intersection. Being male precludes being female and being female precludes being male: in this case, the characteristic gender is therefore mutually exclusive. A Venn diagram shows this as two sets with no intersection. The intersection is said to be the null set using the mathematical symbol ∅.

Draw a second Venn diagram illustrating that 10 of the male dogs have brown coloring.

The Venn diagram below shows the overlap between male and brown where the number 10 is placed in it. This represents \text{Male}\cap \text{Brown}: both male and brown. This is the intersection of these two characteristics. To get the union of Male and Brown, then it is simply the two circled areas minus the overlap. In proper terms, \text{Male}\cup \text{Brown}=\text{Male}+\text{Brown}-\text{Male}\cap \text{Brown} will give us the number of dogs in the union of these two sets. If we did not subtract the intersection, we would have double counted some of the dogs.

Now draw a situation depicting a scenario in which the non-shaded region represents “No white fur and female,” or White fur′ \cap Female. the prime above “fur” indicates “not white fur.” The prime above a set means not in that set, e.g. A\prime means not A. Sometimes, the notation used is a line above the letter. For example, \overline{A} = A\prime.

The Addition Rule of Probability

We met the addition rule earlier but without the help of Venn diagrams. Venn diagrams help visualize the counting process that is inherent in the calculation of probability. To restate the Addition Rule of Probability:

P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)

Remember that probability is simply the proportion of the objects we are interested in relative to the total number of objects. This is why we can see the usefulness of the Venn diagrams. (Figure) shows how we can use Venn diagrams to count the number of dogs in the union of brown and male by reminding us to subtract the intersection of brown and male. We can see the effect of this directly on probabilities in the addition rule.

Let’s sample 50 students who are in a statistics class. 20 are freshmen and 30 are sophomores. 15 students get a “B” in the course, and 5 students both get a “B” and are freshmen.

Find the probability of selecting a student who either earns a “B” OR is a freshmen. We are translating the word OR to the mathematical symbol for the addition rule, which is the union of the two sets.

We know that there are 50 students in our sample, so we know the denominator of our fraction to give us probability. We need only to find the number of students that meet the characteristics we are interested in, i.e. any freshman and any student who earned a grade of “B.” With the Addition Rule of probability, we can skip directly to probabilities.

Let “A” = the number of freshmen, and let “B” = the grade of “B.” Below we can see the process for using Venn diagrams to solve this.

The P\left(A\right)=\frac{20}{50}=0.40, P\left(B\right)=\frac{15}{50}=0.30, and P\left(A\cap B\right)=\frac{5}{50}=0.10.

Therefore, P\left(A\cap B\right)=0.40+0.30-0.10=0.60.

If two events are mutually exclusive, then, like the example where we diagram the male and female dogs, the addition rule is simplified to just P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-0. This is true because, as we saw earlier, the union of mutually exclusive events is the null set, ∅. The diagrams below demonstrate this.

The Multiplication Rule of Probability

Restating the Multiplication Rule of Probability using the notation of Venn diagrams, we have:

P\left(A\cap B\right)=P\left(A|B\right)\cdot P\left(B\right)

The multiplication rule can be modified with a bit of algebra into the following conditional rule. Then Venn diagrams can then be used to demonstrate the process.

The conditional rule: P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}

Using the same facts from (Figure) above, find the probability that someone will earn a “B” if they are a “freshman.”

P\left(A|B\right)=\frac{0.10}{0.30}=\frac{1}{3}

The multiplication rule must also be altered if the two events are independent. Independent events are defined as a situation where the conditional probability is simply the probability of the event of interest. Formally, independence of events is defined as P\left(A|B\right)=P\left(A\right) or P\left(B|A\right)=P\left(B\right). When flipping coins, the outcome of the second flip is independent of the outcome of the first flip; coins do not have memory. The Multiplication Rule of Probability for independent events thus becomes:

P\left(A\cap B\right)=P\left(A\right)\cdot P\left(B\right)

One easy way to remember this is to consider what we mean by the word “and.” We see that the Multiplication Rule has translated the word “and” to the Venn notation for intersection. Therefore, the outcome must meet the two conditions of freshmen and grade of “B” in the above example. It is harder, less probable, to meet two conditions than just one or some other one. We can attempt to see the logic of the Multiplication Rule of probability due to the fact that fractions multiplied times each other become smaller.

The development of the Rules of Probability with the use of Venn diagrams can be shown to help as we wish to calculate probabilities from data arranged in a contingency table.

(Figure) is from a sample of 200 people who were asked how much education they completed. The columns represent the highest education they completed, and the rows separate the individuals by male and female.

Less than high school grad High school grad Some college College grad Total
Male 5 15 40 60 120
Female 8 12 30 30 80
Total 13 27 70 90 200

Now, we can use this table to answer probability questions. The following examples are designed to help understand the format above while connecting the knowledge to both Venn diagrams and the probability rules.

What is the probability that a selected person both finished college and is female?

This is a simple task of finding the value where the two characteristics intersect on the table, and then applying the postulate of probability, which states that the probability of an event is the proportion of outcomes that match the event in which we are interested as a proportion of all total possible outcomes.

P(College Grad\capFemale) = \frac{30}{200}=0.15

What is the probability of selecting either a female or someone who finished college?

This task involves the use of the addition rule to solve for this probability.

P(College Grad\cupFemale) = P(F) + P(CG)− P(F \cap CG)

P(College Grad\cupFemale) = \frac{80}{200}+\frac{90}{200}-\frac{30}{200}=\frac{140}{200}=0.70

What is the probability of selecting a high school graduate if we only select from the group of males?

Here we must use the conditional probability rule (the modified multiplication rule) to solve for this probability.

P(HS Grad|Male = \frac{P\left(\text{HS Grad}\phantom{\rule{0.2em}{0ex}}\cap \phantom{\rule{0.2em}{0ex}}\text{Male}\right)}{\text{P}\left(\text{Male}\right)}=\frac{\left(\frac{15}{200}\right)}{\left(\frac{120}{200}\right)}=\frac{15}{120}=0.125

Can we conclude that the level of education attained by these 200 people is independent of the gender of the person?

There are two ways to approach this test. The first method seeks to test if the intersection of two events equals the product of the events separately remembering that if two events are independent than P(A)*P(B) = P(A \cap B). For simplicity’s sake, we can use calculated values from above.

Does P(College Grad \cap Female) = P(CG) ⋅ P(F)?

\frac{30}{200}\ne \frac{90}{200}\cdot \frac{80}{200}because 0.15 ≠ 0.18.

Therefore, gender and education here are not independent.

The second method is to test if the conditional probability of A given B is equal to the probability of A. Again for simplicity, we can use an already calculated value from above.

Does P(HS Grad | Male) = P(HS Grad)?

\frac{15}{120}\ne \frac{27}{200}because 0.125 ≠ 0.135.

Therefore, again gender and education here are not independent.

Chapter Review

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 or universe of the objects of interest together with circles or ovals. The circles or ovals represent groups of events called sets. A Venn diagram is especially helpful for visualizing the \cup event, the \cap event, and the complement of an event and for understanding conditional probabilities. A Venn diagram is especially helpful for visualizing an Intersection of two events, a Union of two events, or a Complement of one event. A system of Venn diagrams can also help to understand Conditional probabilities. Venn diagrams connect the brain and eyes by matching the literal arithmetic to a picture. It is important to note that more than one Venn diagram is needed to solve the probability rule formulas introduced in Section 3.3.

Use the following information to answer the next four exercises.(Figure) shows a random sample of musicians and how they learned to play their instruments.

Gender Self-taught Studied in school Private instruction Total
Female 12 38 22 72
Male 19 24 15 58
Total 31 62 37 130

Find P(musician is a female).

Find P(musician is a male \cap had private instruction).

P(musician is a male \cap had private instruction) = \frac{15}{130} = \frac{3}{26} = 0.12

Find P(musician is a female \cup is self taught).

Are the events “being a female musician” and “learning music in school” mutually exclusive events?

P(being a female musician \cap learning music in school) = \frac{38}{130} = \frac{19}{65} = 0.29

P(being a female musician)P(learning music in school) = \left(\frac{72}{130}\right)\left(\frac{62}{130}\right) = \frac{4,464}{16,900} = \frac{1,116}{4,225} = 0.26

No, they are not independent because P(being a female musician \cap learning music in school) is not equal to P(being a female musician)P(learning music in school).

The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man 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. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation.

This is a tree diagram with two branches. The first branch, labeled Cancer, shows two lines: 0.4567 C and 0.5433 C'. The second branch is labeled False Positive. From C, there are two lines: 0 P and 1 P'. From C', there are two lines: 0.51 P and 0.49 P'.

Bringing It Together

Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, 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 ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites.

Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

Smoking Levels by Ethnicity
Smoking level African American Native Hawaiian Latino Japanese Americans White TOTALS
1–10
11–20
21–30
31+
TOTALS

Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

\frac{35,065}{100,450}

Find the probability that the person was Latino.

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

To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is \frac{4,715}{100,450}.

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

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

To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is \frac{4715}{15,273}.

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

Use the following information to answer the next two exercises. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.

Suppose that you randomly draw two cards, one at a time, with replacement.
Let G1 = first card is green
Let G2 = second card is green

  1. Draw a tree diagram of the situation.
  2. Find P(G1 \cap G2).
  3. Find P(at least one green).
  4. Find P(G2|G1).
  5. Are G2 and G1 independent events? Explain why or why not.
  1. This is a tree diagram with branches showing probabilities of each draw. The first branch shows two lines: 5/8 Green and 3/8 Yellow. The second branch has a set of two lines (5/8 Green and 3/8 Yellow) for each line of the first branch.
  2. P(GG) = \left(\frac{5}{8}\right)\left(\frac{5}{8}\right) = \frac{25}{64}
  3. P(at least one green) = P(GG) + P(GY) + P(YG) = \frac{25}{64} + \frac{15}{64} + \frac{15}{64} = \frac{55}{64}
  4. P(G|G) = \frac{5}{8}
  5. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.

Suppose that you randomly draw two cards, one at a time, without replacement.
G1 = first card is green
G2 = second card is green

  1. Draw a tree diagram of the situation.
  2. Find P(G1 \cap G2).
  3. Find P(at least one green).
  4. Find P(G2 | G1).
  5. Are G2 and G1 independent events? Explain why or why not.

Use the following information to answer the next two exercises. 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.

Complete the following.

  1. Construct a table or a tree diagram of the situation.
  2. Find P(driver is female).
  3. Find P(driver is age 65 or over | driver is female).
  4. Find P(driver is age 65 or over \cap female).
  5. In words, explain the difference between the probabilities in part c and part d.
  6. Find P(driver is age 65 or over).
  7. Are being age 65 or over and being female mutually exclusive events? How do you know?
  1. <20 20–64 >64 Totals
    Female 0.0244 0.3954 0.0661 0.486
    Male 0.0259 0.4186 0.0695 0.514
    Totals 0.0503 0.8140 0.1356 1
  2. P(F) = 0.486
  3. P(>64 |F) = 0.1361
  4. P(>64 and F) = P(F) P(>64|F) = (0.486)(0.1361) = 0.0661
  5. P(>64 |F) is the percentage of female drivers who are 65 or older and P(>64 \capF) is the percentage of drivers who are female and 65 or older.
  6. P(>64) = P(>64 \capF) + P(>64 \capM) = 0.1356
  7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 \cap F) = 0.0661.

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

  1. How many would you expect to be male?
  2. Using the table or tree diagram, construct a contingency table of gender versus age group.
  3. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.

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.

  1. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.
  2. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?
  3. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work?
  4. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool?
  1. Car, truck or van Walk Public transportation Other Totals
    Alone 0.7318
    Not alone 0.1332
    Totals 0.8650 0.0390 0.0530 0.0430 1
  2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = 0.7318 + 0.0390 = 0.7708.
  3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771
  4. (0.1332)(1,000) = 133

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

  1. Based on the given data, find P(H) and P(T).
  2. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.
  3. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.
  4. Use the tree to find the probability of obtaining at least one head.

Homework

Use the information in the (Figure) to answer the next eight exercises. The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection.

Up for reelection: Democratic party Republican party Other Total
November 2014 20 13 0
November 2016 10 24 0
Total

What is the probability that a randomly selected senator has an “Other” affiliation?

0

What is the probability that a randomly selected senator is up for reelection in November 2016?

What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016?

\frac{10}{67}

What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014?

Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat?

\frac{10}{34}

Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican?

The events “Republican” and “Up for reelection in 2016” are ________

  1. mutually exclusive.
  2. independent.
  3. both mutually exclusive and independent.
  4. neither mutually exclusive nor independent.

d

The events “Other” and “Up for reelection in November 2016” are ________

  1. mutually exclusive.
  2. independent.
  3. both mutually exclusive and independent.
  4. neither mutually exclusive nor independent.

(Figure) gives the number of suicides estimated in the U.S. for a recent year by age, race (black or white), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population.

Race and sex 1–14 15–24 25–64 Over 64 TOTALS
White, male 210 3,360 13,610 22,050
White, female 80 580 3,380 4,930
Black, male 10 460 1,060 1,670
Black, female 0 40 270 330
All others
TOTALS 310 4,650 18,780 29,760

Do not include “all others” for parts f and g.

  1. Fill in the column for the suicides for individuals over age 64.
  2. Fill in the row for all other races.
  3. Find the probability that a randomly selected individual was a white male.
  4. Find the probability that a randomly selected individual was a black female.
  5. Find the probability that a randomly selected individual was black
  6. Find the probability that a randomly selected individual was male.
  7. Out of the individuals over age 64, find the probability that a randomly selected individual was a black or white male.
  1. Race and sex 1–14 15–24 25–64 Over 64 TOTALS
    White, male 210 3,360 13,610 4,870 22,050
    White, female 80 580 3,380 890 4,930
    Black, male 10 460 1,060 140 1,670
    Black, female 0 40 270 20 330
    All others 100
    TOTALS 310 4,650 18,780 6,020 29,760
  2. Race and sex 1–14 15–24 25–64 Over 64 TOTALS
    White, male 210 3,360 13,610 4,870 22,050
    White, female 80 580 3,380 890 4,930
    Black, male 10 460 1,060 140 1,670
    Black, female 0 40 270 20 330
    All others 10 210 460 100 780
    TOTALS 310 4,650 18,780 6,020 29,760
  3. \frac{\text{22,050}}{\text{29,760}}
  4. \frac{\text{330}}{\text{29,760}}
  5. \frac{\text{2,000}}{\text{29,760}}
  6. \frac{\text{23,720}}{\text{29,760}}
  7. \frac{\text{5,010}}{\text{6,020}}

Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected.

Name Single Double Triple Home run TOTAL HITS
Babe Ruth 1,517 506 136 714 2,873
Jackie Robinson 1,054 273 54 137 1,518
Ty Cobb 3,603 174 295 114 4,189
Hank Aaron 2,294 624 98 755 3,771
TOTAL 8,471 1,577 583 1,720 12,351

Find P(hit was made by Babe Ruth).

  1. \frac{1518}{2873}
  2. \frac{2873}{12351}
  3. \frac{583}{12351}
  4. \frac{4189}{12351}

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

  1. \frac{4189}{12351}
  2. \frac{114}{1720}
  3. \frac{1720}{4189}
  4. \frac{114}{12351}

b

(Figure) identifies a group of children by one of four hair colors, and by type of hair.

Hair type Brown Blond Black Red Totals
Wavy 20 15 3 43
Straight 80 15 12
Totals 20 215
  1. Complete the table.
  2. What is the probability that a randomly selected child will have wavy hair?
  3. What is the probability that a randomly selected child will have either brown or blond hair?
  4. What is the probability that a randomly selected child will have wavy brown hair?
  5. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?
  6. If B is the event of a child having brown hair, find the probability of the complement of B.
  7. In words, what does the complement of B represent?

In 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 were compiled into the following table.

Shirt # ≤ 210 211–250 251–290 > 290
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5

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

  1. Find the probability that his shirt number is from 1 to 33.
  2. Find the probability that he weighs at most 210 pounds.
  3. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.
  4. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.
  5. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.
  1. \frac{26}{106}
  2. \frac{33}{106}
  3. \frac{21}{106}
  4. \left(\frac{26}{106}\right) + \left(\frac{33}{106}\right)\left(\frac{21}{106}\right) = \left(\frac{38}{106}\right)
  5. \frac{21}{33}

Use the following information to answer the next two exercises. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads. For the coin, P(H) = \frac{2}{3} and P(T) = \frac{1}{3} where H is heads and T is tails.

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.

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

  1. \frac{2}{3}
  2. \frac{5}{15}
  3. \frac{6}{36}
  4. \frac{5}{36}

Find P(Blue bead).

  1. \frac{15}{36}
  2. \frac{10}{36}
  3. \frac{10}{12}
  4. \frac{6}{36}

a

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

  1. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
  2. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain.
  3. For each complete path through the tree, write the event it represents and find the probabilities.
  4. Let S be the event that both cookies selected were the same flavor. Find P(S).
  5. Let T be the event that the 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.
  6. Let U be the event that the second cookie selected is a butter cookie. Find P(U).

Glossary

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

License

Icon for the Creative Commons Attribution 4.0 International License

Introductory Business Statistics by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book