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7. Recognizing Independence and Making Decisions
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Chapter 6
7. 

Recognizing Independence and Making Decisions

This lesson explores the concept of independence and conditional probability in the context of everyday situations. It discusses how these mathematical principles can be applied to understand and make decisions in various scenarios, such as voting behavior, work habits, and even food preferences. For example, it examines whether taking a nap after lunch affects being late for work. The lesson also delves into how probabilities can be conditional, based on prior events or conditions. Understanding these concepts can help in making informed decisions in daily life, from choosing a political candidate to deciding whether to take that afternoon nap.
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7 Exercises - Grade E - A
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Recognizing Independence and Making Decisions
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In daily life, some situations can be analyzed using probability concepts like independence or conditional probability. In this lesson, various everyday situations will be presented to study these concepts.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Guessing Answers in a Test

In a multiple-choice test, Davontay randomly selected the answers to all five questions. Each question had two options to choose from.

A paper with 5 questions and each question has two options
Let A be the event of guessing the answer to the second question correctly. Let B be the event of guessing correctly on the fifth question. Are these events independent? If so, what does it mean for the events to be independent?
Example

Presidential Election in 2016

To find out about the voting behavior of people according to their age, a survey was conducted outside of a polling station. The following table shows some data about the presidential election in 2016 between Donald Trump and Hillary Clinton.

Clinton Trump
Young Adults (18-29 years old) 55 36
Adults (30-44 years old) 51 41
Middle-Aged Adults (45-64 years old) 44 52
Seniors (65+ years old) 45 52

Find the following conditional probabilities and describe their meaning in everyday words. Round each answer to two decimal places.

a
b
c
d

Answer
a Probabilities:
  • P(Clinton|Young Adult) ≈ 0.60
  • P(Trump|Young Adult) ≈ 0.40

Descriptions:

  • The chances that a young adult voted for Clinton are about 60 %.
  • The chances that a young adult voted for Trump are about 40 %, the complement of those voting for Clinton.

Alternative description: If a person is selected at random among the surveyed young adults, there is a 60 % chance they voted for Clinton and a 40 % chance they voted for Trump.

b Probabilities:
  • P(Adult|Clinton) ≈ 0.26
  • P(Middle-Aged Adult|Clinton) ≈ 0.23

Descriptions:

  • Knowing that someone voted for Clinton, the probability that they are an adult is about 26 %.
  • Knowing that someone voted for Clinton, the probability that they are a middle-aged adult is about 23 %.

Alternative description: If a person that voted for Clinton is selected at random, there is a 26 % chance that they are aged between 30 and 44 years old and a 23 % chance that they are aged between 45 and 64.

c Probabilities:
  • P(Clinton|Middle-Aged Adult) ≈ 0.46
  • P(Middle-Aged Adult|Trump) ≈ 0.29

Descriptions:

  • Knowing that a person is a middle-aged adult, the probability that they voted for Clinton is about 46 %.
  • Knowing that someone voted for Trump, there is a 29 % probability that they are between 45 and 64 years old.
d Probabilities:
  • P(Senior|Clinton) ≈ 0.23
  • P(Trump|Senior) ≈ 0.54

Descriptions:

  • Knowing that a person voted for Clinton, the probability that they are older than 64 is about 23 %.
  • Knowing that someone is older than 64, there is a 54 % probability that they voted for Trump.

Hint
a Find the marginal frequencies of the given table.
b Since both probabilities rely on the fact that a person voted for Clinton, use the first column of the given table.
c For the first probability, look at the third row. For the second probability, look at the second column.
d The first probability relies in the fact that the person voted for Clinton, while the second probability relies on the fact that the person is older than 64 years old. Which parts of the table give this information?

Solution
a Before finding each of the required probabilities, the marginal frequencies will be found.
Clinton Trump Total
Young Adults (18-29 years old) 55 36 91
Adults (30-44 years old) 51 41 92
Middle-Aged Adults (45-64 years old) 44 52 96
Seniors (65+ years old) 45 52 97
Total 195 181 376

To find P(Clinton|Young Adult) and P(Trump|Young Adult), focus on the first row. Of the total 91 young adults who participated in the survey, 55 voted for Clinton and 36 voted for Trump. Knowing this, the desired conditional probabilities can be calculated. P(Clinton|Young Adult) &= 55/91 ≈ 0.60 [0.3cm] P(Trump|Young Adult) &= 36/91 ≈ 0.40 Consequently, the following two conclusions can be made.

  • Knowing that a person is a young adult, the chance that they voted for Clinton is about 60 %.
  • Knowing that a person is a young adult, the chance that they voted for Trump is about 40 %, the complement of those voting for Clinton.

In other words, if a person is selected at random among the surveyed young adults, there is a 60 % chance they voted for Clinton and a 40 % chance they voted for Trump.

b To find P(Adult|Clinton) and P(Middle-Aged Adult|Clinton) the first column of the table will be used. A total of 195 people voted for Clinton. Of those, 51 are adults and 44 are middle-aged adults. With this information, the desired conditional probabilities can be calculated.

P(Adult|Clinton) &= 51/195 [0.2cm] &≈ 0.26 [0.3cm] P(Middle-Aged Adult|Clinton) &= 44/195 [0.2cm] &≈ 0.23 Consequently, the following two conclusions can be made.

  • Knowing that someone voted for Clinton, the probability that they are an adult is about 26 %.
  • Knowing that someone voted for Clinton, the probability that they are a middle-aged adult is about 23 %.

In other words, if a person that voted for Clinton is selected at random, there is a 26 % probability that they are between 30 and 44 years old and a 23 % probability that they are between 45 and 64.

c To find P(Clinton|Middle-Aged Adult), the third row of the table will be analyzed. A total of 96 middle-aged adults were surveyed. Of those, 44 voted for Clinton. Therefore, the conditional probability is the quotient of these two numbers.

P(Clinton|Middle-Aged Adult) &= 44/96 [0.2cm] &≈ 0.46 Consequently, knowing that a person is a middle-aged adult, the probability that they voted for Clinton is about 46 %. Next, to find P(Middle-Aged Adult|Trump), focus on the second column of the table. A total of 181 people voted for Trump, and 52 of those are middle-aged adults. P(Middle-Aged Adult|Trump) &= 52/181 [0.2cm] &≈ 0.29 In conclusion, knowing that someone voted for Trump, there is a 29 % probability that they are between 45 and 64 years old.

d To find P(Senior|Clinton), the first column will be studied. A total of 195 people voted for Clinton. Of those, 45 are seniors. Therefore, the conditional probability is the quotient of these two numbers.

P(Senior|Clinton) &= 45/195 ≈ 0.23 Consequently, knowing that a person voted for Clinton, the probability that they are a senior is about 23 %. Next, to find P(Trump|Senior), the fourth row will be analyzed. Of the total 97 seniors that were surveyed, 52 voted for Trump. P(Trump|Senior) &= 52/97 ≈ 0.54 In conclusion, knowing that someone is older than 64 years old, there is a 54 % chance that they voted for Trump.

Example

Interpreting Independent Events

Last month, Ignacio got a part-time job working from 4:00 P.M. to 8:00 P.M. during weekdays. Ignacio, who knows statistics, said to his peers that the events of taking a nap after lunch and being late for work are independent. However, Tadeo, who does not know much about statistics, does not understand what Ignacio meant.

Ignacio speaking with Tadeo
a Explain what Ignacio meant using everyday words so that someone who does not know statistics would understand.
b The following table records Ignacio's last 20 working days.
Late On Time Total
Nap 2 6 8
No Nap 3 9 12
Total 5 15 20
According to this data, is it true that the events claimed by Ignacio are independent?

Answer
a Ignacio is saying that the probability that he is late for work is the same whether or not he takes a nap after lunch. Therefore, if Ignacio is late for work, napping is not the cause.
b The probability that Ignacio is late is 0.25. If he takes a nap, the probability that he is late still is 0.25. Therefore, the events are independent.

Hint
a Remember the definition of independent events.
b Find and compare the probability that Ignacio is late and the conditional probability that Ignacio is late given that he takes a nap.

Solution
a By definition, two events are independent when the occurrence of one event does not affect or influence the other event. Knowing this, review the claim made by Ignacio again.

The events of taking a nap after lunch and being late for work are independent.

Ignacio is saying that the probability that he is late for work is the same whether or not he takes a nap after lunch. Therefore, on the days that Ignacio is late for work, the nap is not the cause. With this explanation, Tadeo will hopefully understand what Ignacio meant.

b The aim is to determine whether or not the events Ignacio taking a nap after lunch and Ignacio being late for work are independent using the data from the table. Remember, if two events A and B are independent, then P(A) is equal to P(A|B). Therefore, the following equation needs to be checked.

P( Late) ? = P( Late| Nap) To find P(Late), the number of days Ignacio is late must be counted. From the table, Ignacio is late on 5 days. Then, this number will be divided by the total number of days, which is 20. P( Late) = 5/20 = 1/4 To find P( Late| Nap), the number of days that Ignacio takes a nap and is late must be counted. From the table, this happens on 2 days. Next, this number will be divided by the total number of days in which Ignacio takes a nap, which is 8. P( Late| Nap) = 2/8 = 1/4 Since P( Late) and P( Late| Nap) are both equal to 14, the events are independent. Consequently, Ignacio was correct when he said that being late for work has nothing to do with taking a nap after lunch.

Example

Deciding Based on Probabilities

In Maya's new neighborhood, some people have dogs, cats, both, or neither. The following diagram shows the distribution of pets, but Maya has not seen it.

Venn Diagram, 69 only a dog, 45 only a cat, 15 with both, and 61 with no pets
a Some days later, Maya saw her neighbor Ignacio at the mall. Maya believes that Ignacio has exactly one type of pet. What are the chances that Maya is correct? Write the answer as a percentage.
b Later, Maya found two more neighbors, Magdalena and Dylan. She knows that Magdalena does not have a cat and that Dylan does not have a dog. With this information, who is more likely to have a pet, Magdalena or Dylan?
c Maya does not want to be rude by asking them about a pet they do not have. Based on Part B, what should Maya do?

Hint
a How many people only have dogs? How many people only have cats? How many people live in the neighborhood?
b Find the probability that someone has a dog given that they do not have a cat. Compare this probability to the probability that someone has a cat given that they do not have a dog.
c If the probability that a person has a pet is greater than 50 %, Maya can safely ask. Otherwise, she should not.

Solution
a To determine whether Maya is correct in thinking that Ignacio has exactly one type of pet, start by drawing some conclusions from the diagram.
  • There are 69 people that only have dogs.
  • There are 45 people that only have cats.
  • There are 15 people that have both dogs and cats.
  • There are 61 people that have neither a dog nor a cat.

Consequently, there are 69+45=114 people that have exactly one type of pet. Next, the total number of people living in the neighborhood should be found. Number of People 69+45+15+61 = 190 Dividing 114 by 190, the probability that a person chosen at random has exactly one type of pet can be found. P(Exactly one pet type) = 114/190 = 0.6 Therefore, there is a 60 % chance that Maya is correct in thinking that Ignacio has exactly one type of pet.

b In this case, the goal is to find and compare the following probabilities.
  • The probability that Magdalena has a pet.
  • The probability that Dylan has a pet.

Notice that as they are written, these probabilities represent the same situation. However, Maya knows that Magdalena does not have a cat and that Dylan does not have a dog. With this information, the above probabilities can be rewritten.

  • The probability that Magdalena has a pet given that she does not have a cat.
  • The probability that Dylan has a pet given that he does not have a dog.

These are conditional probabilities. Since there are only two types of pets in the survey, the above statement can be written more precisely.

  • The probability that Magdalena has a dog given that she does not have a cat.
  • The probability that Dylan has a cat given that he does not have a dog.

Now that all information is written, the first probability can be found. P(Dog|No Cat) = People who have a dog but not a cat/People who do not have a cat From the diagram, a total of 130 people do not have a cat, and 69 of those people have a dog. P(Dog|No Cat) = 69/130 ≈ 0.53 Therefore, there is about 53 % chance that Magdalena has a dog. The second probability can be found in a similar fashion. P(Cat|No Dog) = People who have a cat but not a dog/People who do not have a dog Using the diagram one more time, a total of 106 people do not have a dog, and 45 of those have a cat. P(Cat|No Dog) = 45/106 ≈ 0.42 Then, there is about 42 % chance that Dylan has a cat. Comparing the two obtained probabilities, the first is greater. Consequently, Magdalena is more likely to have a dog than Dylan is to have a cat.

c If the probability that a person has a pet is more than 50 %, Maya can safely ask about it. Otherwise, she should not do it. The probabilities found in Part B will be rewritten.
  • There is about 53 % chance that Magdalena has a dog.
  • There is about 42 % chance that Dylan has a cat.

By the Complement Rule, the following pair of conclusions can also be drawn.

  • There is about 47 % chance that Magdalena does not have a dog.
  • There is about 58 % chance that Dylan does not have a cat.

From the four statements, Maya could safely ask Magdalena about her dog, but she should not ask Dylan about his cat. Keep in mind that Magdalena might not have a dog despite the probabilities and conclusions. Similarly, Dylan might have a cat.

Example

Age vs. Music Preference

Tearrik wants to determine if there is a connection between age and music preference. To figure it out, he surveyed 120 people at the mall, asking their age and whether they prefer pop or classical music. After analyzing the data collected, he concluded that there is no connection at all.

Pop Classical Total
35 Years Old or Younger
Older Than 35 40
Total 45 120

Based on the conclusion made by Tearrik, complete the missing information in the two-way frequency table.

Answer
Pop Classical Total
35 Years Old or Younger 30 50 80
Older Than 35 15 25 40
Total 45 75 120

Hint

Since there is no connection between age and music preference, the probability that someone older than 35 likes pop music is the same as the probability that any person likes this type of music. In other words, the events A person likes pop music and A person is older than 35 years old are independent.

Solution

For simplicity, some variables will be assigned to the missing data.

Pop Classical Total
35 Years Old or Younger a b c
Older Than 35 d e 40
Total 45 f 120

In the table, the grand total and two marginal frequencies are given. Knowing that the marginal frequencies in the total row and column add to the grand total 120, the missing marginal frequencies can be calculated. 45+f = 120 c+40 = 120 ⇔ f = 75 c = 80 The obtained values can be added to the table.

Pop Classical Total
35 Years Old or Younger a b 80
Older Than 35 d e 40
Total 45 75 120

To find the joint frequencies, the conclusion made by Tearrik will be used instead of a system of equations.

There is no connection between age and music preference.

One conclusion that can be drawn from the above statement is that the probability that someone older than 35 likes pop music is the same as the probability that any person likes pop. Consequently, the following equation can be written. P(Pop| >35) = P(Pop) By the definition of conditional probability, the right-hand side of this equation can be rewritten. P(Pop|> 35) = A person likes pop and is over35/A person is older than35 From the table, 40 people are older than 35 and d of those like pop music. P(Pop| > 35) = d/40 On the other hand, the probability that someone likes pop music is the number of people who preferred pop music divided by the total number of people surveyed. P(Pop) = 45/120 = 3/8 The value of d can be found by substituting P(Pop|>35)= d40 and P(Pop)= 38 in the equation P(Pop|>35)=P(Pop).
P(Pop| > 35) = P(Pop)
SubstituteII

P(Pop| > 35)= d/40, P(Pop)= 3/8

d/40 = 3/8
▼
Solve for d
MultEqn

LHS * 40=RHS* 40

d = 3/8(40)
MoveRightFacToNum

a/c* b = a* b/c

d = 120/8
CalcQuot

Calculate quotient

d = 15
Therefore, there were 15 people older than 35 years old who preferred pop music.
Pop Classical Total
35 Years Old or Younger a b 80
Older Than 35 15 e 40
Total 45 75 120

The sum of the joint frequencies in a row equals the marginal frequency of the row. Similarly, the sum of the joint frequencies in a column equals the marginal frequency of the column. a+15 = 45 15+e= 40 ⇔ a = 30 e = 25 The obtained values can be added to the table.

Pop Classical Total
35 Years Old or Younger 30 b 80
Older Than 35 15 25 40
Total 45 75 120

The value of b can be found in a similar way. 30+b = 80 ⇔ b = 50 The table can be now completed!

Pop Classical Total
35 Years Old or Younger 30 50 80
Older Than 35 15 25 40
Total 45 75 120

Note that the conclusion made by Tearrik implies that the following pairs of events are independent.

  • A person likes pop music and A person is older than 35 years old.
  • A person likes pop music and A person is 35 years old or younger.
  • A person likes classical music and A person is older than 35 years old.
  • A person likes classical music and A person is 35 years old or younger.
Example

Probabilities and Treats

Diego wants to throw a party at the end of the school year. To determine what kind of treats he should buy, he asked his 80 classmates whether they prefer cupcakes, cookies, donuts, or chocolate.

Frequency Table: Boys/Girls and Cupcake/Cookie/Donuts/Chocolate

On the day of the party, Diego puts the treats on a table.

Table with cupcakes, cookies, donuts, and chocolates
a Diego sees Mark approach the table. What is the probability that Mark will choose a donut? Write the answer as a percentage rounded to the nearest integer.
b If LaShay approaches the table, what is the probability that she will pick a cupcake? Write the answer as a percentage rounded to the nearest integer.
c Diego asked a friend to bring him something from the table and was given a cupcake. Knowing this, who is Diego more likely to have asked for the treat, Dylan or Emily?
d Write in words what P(Chocolate|Boy) means and find its value. Write the answer as a percentage rounded to the nearest integer.
e Diego wants to find the probability that a person at the party prefers cookies over any other treat. Diego and Mark have the following conversation.
Diego: & I need to know the & gender of the person. [0.2cm] Mark: & No. The probability that a & person prefers cookies does & not depend on their gender. Who is correct?

Hint
a Since Mark is a boy, the conditional probability that someone prefers donuts given that he is a boy must be found.
b Since LaShay is a girl, the conditional probability that someone prefers cupcakes given that she is a girl must be found.
c Find the probability that a person is a boy knowing that he prefers cupcakes. Then, find the probability that a person is a girl knowing that she prefers cupcakes. Which of these probabilities is greater?
d Remember the way a conditional probability is read.
e Find the probability that a random person prefers cookies. Find the probability that a person prefers cookies knowing that they are a boy. Finally, find the probability that a person prefers cookies knowing that they are a girl. Compare these three probabilities.

Solution
a The probability of Mark choosing a donut is not the same as the probability of a random person choosing a donut. The difference is that in the first case, the gender of the person is known.
P(Mark& 🍩) ≠ P(🍩) Since Mark is a boy, the probability that he chooses a donut is the same as the probability of a person choosing a donut knowing that they are a boy. P(Mark& 🍩) = P(🍩|Boy) By the definition of conditional probability, the right-hand side of the equation can be rewritten as follows. P(Mark& 🍩) = #(🍩 and Boy)/#Boys From the table, the number of boys who prefer donuts is 13 and the total number of boys is 45. With this information, the probability that Mark chooses a donut can be determined.
P(Mark& 🍩) = #(🍩 and Boy)/#Boys
SubstituteValues

Substitute values

P(Mark& 🍩) = 13/45
▼
Evaluate right-hand side
CalcQuot

Calculate quotient

P(Mark& 🍩) = 0.288888...
WritePercent

Convert to percent

P(Mark& 🍩) = 28.888888... %
RoundInt

Round to nearest integer

P(Mark& 🍩) ≈ 29 %
Consequently, the probability that Mark will choose a donut is about 29 %.
b As in the previous case, determining the probability that LaShay will choose a cupcake is the same as finding the probability that a person will choose a cupcake given that they are a girl.
P(LaShay& 🧁) = P(🧁|Girl) The right-hand side of the previous equation equals the number of girls that prefer cupcakes divided by the total number of girls. P(LaShay& 🧁) = #(🧁and Girl)/#Girls From the table, the number of girls that prefer cupcakes is 8, and the total number of girls is 35. These values can be substituted into the previous equation.
P(LaShay& 🧁) = #(🧁and Girl)/#Girls
SubstituteValues

Substitute values

P(LaShay& 🧁) = 8/35
▼
Evaluate right-hand side
CalcQuot

Calculate quotient

P(LaShay& 🧁) = 0.228571...
WritePercent

Convert to percent

P(LaShay& 🧁) = 22.857142... %
RoundInt

Round to nearest integer

P(LaShay& 🧁) ≈ 23 %
Consequently, there is 23 % of chance that LaShay chooses a cupcake from the table.
c The only information known to determine who Diego is more likely to have asked for the treat is that they gave him a cupcake. Therefore, the following pair of probabilities need to be found and compared.
P(Dylan | 🧁) and P(Emily | 🧁) Since Dylan is a boy and Emily is a girl, the previous probabilities can be written in terms of gender. This way, the data from the table can be used. P(Boy | 🧁) and P(Girl | 🧁) Now, find the conditional probability that Diego asked a boy for the treat, knowing that he was given a cupcake. P(Boy | 🧁) = #(Boys and🧁)/#🧁 From the table, 15 people prefer cupcakes and 7 of them are boys.
P(Boy | 🧁) = #(Boys and🧁)/#🧁
SubstituteValues

Substitute values

P(Boy | 🧁) = 7/15
▼
Evaluate right-hand side
CalcQuot

Calculate quotient

P(Boy | 🧁) = 0.466666...
WritePercent

Convert to percent

P(Boy | 🧁) = 46.666666... %
RoundInt

Round to nearest integer

P(Boy | 🧁) ≈ 47 %
Next, find the conditional probability that Diego asked a girl for the treat, knowing that he was given a cupcake. P(Girl | 🧁) = #(Girls and🧁)/#🧁 From the table, there are 8 girls that prefer cupcakes.
P(Girl | 🧁) = #(Girls and🧁)/#🧁
SubstituteValues

Substitute values

P(Girl | 🧁) = 8/15
▼
Evaluate right-hand side
CalcQuot

Calculate quotient

P(Girl | 🧁) = 0.53333...
WritePercent

Convert to percent

P(Girl | 🧁) = 53.333333... %
RoundInt

Round to nearest integer

P(Girl | 🧁) ≈ 53 %
Comparing the two probabilities, the second is greater. Therefore, knowing that Diego was given a cupcake, he is more likely to have asked a girl for the treat. Consequently, Diego is more likely to have asked Emily for the treat.
d The given expression represents a conditional probability.
P(Chocolate|Boy) ⇕ Probability that a person prefers chocolate given they are a boy. In other words, the given expression represents the chance of a boy preferring chocolate. To find it, the number of boys that like chocolate will be divided by the total number of boys. P(🍫|Boy) = #(🍫 and Boy)/#Boys From the table, there are 16 boys that like chocolate, and the total number of boys is 45.
P(🍫|Boy) = #(🍫 and Boy)/#Boys
SubstituteValues

Substitute values

P(🍫|Boy) = 16/45
▼
Evaluate right-hand side
CalcQuot

Calculate quotient

P(🍫|Boy) = 0.355555...
WritePercent

Convert to percent

P(🍫|Boy) = 35.555555... %
RoundInt

Round to nearest integer

P(🍫|Boy) ≈ 36 %
The probability that a person prefers chocolate given that they are a boy is about 36 %.
e Diego was asked to find the probability that a person chosen at random prefers cookies.
P(🍪) = #(People that prefer🍪)/# People However, Diego thinks that the gender of a person affects this probability. He thinks that the probability of a person preferring cookies varies depending on whether they are a boy or a girl. That is, Diego is considering the following conditional probabilities. P(🍪|Boy) and P(🍪|Girl) In contrast, Mark says that gender has no influence. To determine who is correct, find and compare the three probabilities. Start by finding P(🍪). From the table, the total number of people is 80 and 16 of them prefer cookies.
P(🍪) = #(People that like🍪)/#People
SubstituteValues

Substitute values

P(🍪) = 16/80
ReduceFrac

a/b=.a /16./.b /16.

P(🍪) = 1/5
Now, the probability that a person prefers cookies given that they are a boy is obtained by dividing the number of boys that prefer cookies by the total number of boys. P(🍪|Boy) = #(Boys and🍪)/# Boys From the table, there are 9 boys that like cookies and the total number of boys is 45.
P(🍪|Boy) = #(Boys and🍪)/# Boys
SubstituteValues

Substitute values

P(🍪|Boy) = 9/45
ReduceFrac

a/b=.a /9./.b /9.

P(🍪|Boy) = 1/5
Similarly, the probability that a person prefers cookies given that they are a girl is calculated by dividing the number of girls who prefer cookies 7 by the total number of girls 35.
P(🍪|Girl) = #(Girl and🍪)/#Girls
SubstituteValues

Substitute values

P(🍪|Girl) = 7/35
ReduceFrac

a/b=.a /7./.b /7.

P(🍪|Girl) = 1/5
As can be seen, the three probabilities found are equal. P(🍪) = P(🍪|Boy) = P(🍪|Girl) This implies that the probability that a person prefers cookies is the same, no matter whether they are a boy or a girl. Consequently, Mark is correct in saying that Diego does not need to know the gender of the person.
Closure

Answering a Test at Random

Davontay took a multiple-choice test where each question had two choices. He randomly guessed the answers to all the five questions in the test.

A paper with 5 questions and each question has two options

Let A be the event of guessing the answer to the second question correctly. Let B be the event of guessing correctly on the fifth question.

a Are the events independent?
b In the affirmative case, what does it mean for the events to be independent? Write the answer using everyday language.

Answer
a Yes, the events are independent.
b The events being independent means that having guessed the second question correctly does not influence having guessed the fifth question correctly and vice versa.

Hint
a Find the sample space, P(A), P(B), and P(A⋂ B). Then, compare P(A⋂ B) to P(A)* P(B).
b Independence means that the occurrence of one event does not affect the probability of the other.

Solution
a Two events A and B are independent when the following equation is satisfied.
P(A⋂ B) = P(A)* P(B) To find the probabilities involved, the sample space will be first written. Let C represent a correct answer and I represent an incorrect answer. Using these variables, all the possible outcomes in the sample space can be listed.
Interactive table showing all the possible outcomes
By adding the number of possible outcomes of each case, there are 32 possible outcomes in the sample space. Of the 32 outcomes, there are 16 that satisfy event A and 16 that satisfy event B.
Interactive table showing the outcomes satisfying event A or event B
With this data, the probability of events A and B can be found. P(A) &= 16/32 = 1/2 [0.3cm] P(B) &= 16/32 = 1/2 Next, to find P(A⋂ B), the number of outcomes that are common for both events will be counted.
Interactive table showing the outcomes satisfying event 'A and B'
There are 8 outcomes that satisfy both events. P(A⋂ B) = 8/32 = 1/4 Finally, substitute P(A), P(B), and P(A⋂ B) into the initial equation to see if a true statement is obtained.
P(A⋂ B) = P(A)* P(B)
SubstituteValues

Substitute values

1/4 ? = 1/2* 1/2
MultFrac

Multiply fractions

1/4 = 1/4 ✓
Since a true statement was obtained, A and B are independent events.
b The events being independent means that the occurrence of one of them does not affect the probability of the occurrence of the other. Therefore, the independence of the given events can be written using everyday language as follows.

The events being independent means that having guessed the second question correctly does not influence having guessed the fifth question correctly and vice versa.


Level 1
Level 2
Level 3
Recognizing Independence and Making Decisions
Exercise 2.1

When penalties are called in basketball, a player may be given one to three free throw attempts. During a basketball season, a statistician working for a pro team's analytics department made the following observations for instances where two free throws were awarded.

Statistics on the number of penalties made in basketball
According to this data, is the probability of making the second free throw affected by whether the player makes the first free throw? Justify the answer.

Let's make a two-way frequency table using the statistician's data set.

To determine if the probability of making the second free throw is dependent on whether the player made the first free throw, we will calculate the conditional probability of making the second free throw given that the player made the first free throw and given that the player missed the first free throw. $P(make second|made first)$ [0.4em] $P(make second|missed first)$ If these probabilities are different, we can argue that there may be a dependency.

Knowing the total number of observations when a player made or missed the first free throw allows us to calculate the conditional probabilities of making the second free throw. P (make second|made first )& = 51/71 ≈ 72 % [1em] P (make second|missed first )& = 25/39 ≈ 64 % As we can see, according to this data, a basketball player is more likely to make the second free throw if they made their first attempt (72 %) than if they missed it (64 %). Perhaps the hot-hand fallacy is not a fallacy after all? (All kidding aside, it is a fallacy.)

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Hint & Answer
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Solution

In the lot of a car dealership, 80 % of all cars have five seats, 25 % are sports cars, and 10 % are both sports cars and have five seats.

a group of cars
 What is the probability that a randomly selected car in the lot is not a sports car and does not have five seats?
What is the probability that a sports car does not have five seats?
Are a car not having five seats and the car being a sports car associated? Explain?

We know that 25 % of the cars are sports cars. Therefore, 75 % must not be sports cars. Similarly, if 80 % have five seats, then 20 % of the cars must not have five seats. Let's make a two-way table and include all of these statistics.

Using the statistics in the two-way table, we can calculate some conditional probabilities. P(not five seats|sports car) [-1em] 25 %-10 %=15 % P(not sports car|five seats) [-1em] 80 %-10 %=70 % Let's add these probabilities to the diagram.

Now we can calculate the probability that a randomly selected car does not have five seats and is not a sports car. 20 % -15 % = 5 % ✓ We can also calculate the percentage of cars that are neither five seaters nor sports cars as the difference between the total percentage of non sports cars and the percentage of non sports cars that are five seaters. 75 % -70 % = 5 % Now we can complete our table.


We want to know the probability that a sports car does not have five seats. Therefore, we must divide the percentage of sports cars that do not have five seats by the total percentage of sports cars.

Now we can determine the conditional probability of a car not having five seats given that it is a sports car. P(not five seats|sports car)&=0.15/0.25 [0.15cm] &=60 %

If sports cars are associated with not having five seats, the probability of a sports car not having five seats must be greater than the general probability that any given car does not have five seats. P(not five seats for all cars)&=20 % [0.5em] P(not five seats|sports car)&=60 % As we can see, the events are associated because the probability that a sports car does not have five seats is greater than the overall probability of a car not having five seats.

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Hint & Answer
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Solution

Recognizing Independence and Making Decisions

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