Understanding Independence in Everyday Decisions: Conditional Probability Explored

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.

Probability of an Event
Independent and Dependent Events
Conditional Probability
Two-way Tables, Marginal Frequency, and Joint Frequency
Marginal Relative Frequency and Joint Relative Frequency

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

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.

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) = \frac{5}{2 0} = \frac{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) = \frac{2}{8} = \frac{1}{4}

Since

P (Late)

and

P (Late ∣ Nap)

are both equal to

\frac{1}{4},

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

190,

the probability that a person chosen at random has exactly one type of pet can be found.

P (Exactly one pet type) = \frac{1 1 4}{1 9 0} = 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) = \frac{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) = \frac{6 9}{1 3 0} \approx 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) = \frac{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) = \frac{4 5}{1 0 6} \approx 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	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 \Leftrightarrow {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) = \frac{A person likes pop and is over 3 5}{A person is older than 3 5}

From the table,

40

people are older than

35

and

d

of those like pop music.

P (Pop ∣ > 35) = \frac{d}{4 0}

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) = \frac{4 5}{1 2 0} = \frac{3}{8}

The value of

d

can be found by substituting

P (Pop ∣ > 35) = \frac{d}{4 0}

and

P (Pop) = \frac{3}{8}

in the equation

P (Pop ∣ > 35) = P (Pop) .

P (Pop ∣ > 35) = P (Pop)

SubstituteII

$P (Pop ∣ > 35) = \frac{d}{4 0}$ , $P (Pop) = \frac{3}{8}$

\frac{d}{4 0} = \frac{3}{8}

Solve for

d

MultEqn

$LHS \cdot 40 = RHS \cdot 40$

d = \frac{3}{8} (40)

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

d = \frac{1 2 0}{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 \Leftrightarrow {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 \Leftrightarrow 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 : Mark : I need to know the gender of the person . 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 & 🍩) \neq = 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 & 🍩) = \frac{# ( 🍩 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 & 🍩) = \frac{# ( 🍩 and Boy )}{# Boys}

SubstituteValues

Substitute values

P (Mark & 🍩) = \frac{1 3}{4 5}

Evaluate right-hand side

CalcQuot

Calculate quotient

P (Mark & 🍩) = 0.288888 \dots

WritePercent

Convert to percent

P (Mark & 🍩) = 28.888888 \dots %

RoundInt

Round to nearest integer

P (Mark & 🍩) \approx 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 & 🧁) = \frac{# ( 🧁 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 & 🧁) = \frac{# ( 🧁 and Girl )}{# Girls}

SubstituteValues

Substitute values

P (LaShay & 🧁) = \frac{8}{3 5}

Evaluate right-hand side

CalcQuot

Calculate quotient

P (LaShay & 🧁) = 0.228571 \dots

WritePercent

Convert to percent

P (LaShay & 🧁) = 22.857142 \dots %

RoundInt

Round to nearest integer

P (LaShay & 🧁) \approx 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 ∣ 🧁) = \frac{# ( Boys and 🧁 )}{# 🧁}

From the table,

15

people prefer cupcakes and

7

of them are boys.

P (Boy ∣ 🧁) = \frac{# ( Boys and 🧁 )}{# 🧁}

SubstituteValues

Substitute values

P (Boy ∣ 🧁) = \frac{7}{1 5}

Evaluate right-hand side

CalcQuot

Calculate quotient

P (Boy ∣ 🧁) = 0.466666 \dots

WritePercent

Convert to percent

P (Boy ∣ 🧁) = 46.666666 \dots %

RoundInt

Round to nearest integer

P (Boy ∣ 🧁) \approx 47 %

Next, find the conditional probability that Diego asked a girl for the treat, knowing that he was given a cupcake.

P (Girl ∣ 🧁) = \frac{# ( Girls and 🧁 )}{# 🧁}

From the table, there are

8

girls that prefer cupcakes.

P (Girl ∣ 🧁) = \frac{# ( Girls and 🧁 )}{# 🧁}

SubstituteValues

Substitute values

P (Girl ∣ 🧁) = \frac{8}{1 5}

Evaluate right-hand side

CalcQuot

Calculate quotient

P (Girl ∣ 🧁) = 0.53333 \dots

WritePercent

Convert to percent

P (Girl ∣ 🧁) = 53.333333 \dots %

RoundInt

Round to nearest integer

P (Girl ∣ 🧁) \approx 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) = \frac{# ( 🍫 and Boy )}{# Boys}

From the table, there are

16

boys that like chocolate, and the total number of boys is

45 .

P (🍫 ∣ Boy) = \frac{# ( 🍫 and Boy )}{# Boys}

SubstituteValues

Substitute values

P (🍫 ∣ Boy) = \frac{1 6}{4 5}

Evaluate right-hand side

CalcQuot

Calculate quotient

P (🍫 ∣ Boy) = 0.355555 \dots

WritePercent

Convert to percent

P (🍫 ∣ Boy) = 35.555555 \dots %

RoundInt

Round to nearest integer

P (🍫 ∣ Boy) \approx 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 (🍪) = \frac{# ( 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 (🍪) = \frac{# ( People that like 🍪 )}{# People}

SubstituteValues

Substitute values

P (🍪) = \frac{1 6}{8 0}

ReduceFrac

$\frac{a}{b} = \frac{a / 1 6}{b / 1 6}$

P (🍪) = \frac{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) = \frac{# ( Boys and 🍪 )}{# Boys}

From the table, there are

9

boys that like cookies and the total number of boys is

45 .

P (🍪 ∣ Boy) = \frac{# ( Boys and 🍪 )}{# Boys}

SubstituteValues

Substitute values

P (🍪 ∣ Boy) = \frac{9}{4 5}

ReduceFrac

$\frac{a}{b} = \frac{a / 9}{b / 9}$

P (🍪 ∣ Boy) = \frac{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) = \frac{# ( Girl and 🍪 )}{# Girls}

SubstituteValues

Substitute values

P (🍪 ∣ Girl) = \frac{7}{3 5}

ReduceFrac

$\frac{a}{b} = \frac{a / 7}{b / 7}$

P (🍪 ∣ Girl) = \frac{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.

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 \cap B) .

Then, compare

P (A \cap B)

P (A) \cdot 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 \cap B) = P (A) \cdot 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.

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 .

With this data, the probability of events

A

and

B

can be found.

P (A) P (B) = \frac{1 6}{3 2} = \frac{1}{2} = \frac{1 6}{3 2} = \frac{1}{2}

Next, to find

P (A \cap B),

the number of outcomes that are common for both events will be counted.

There are

8

outcomes that satisfy both events.

P (A \cap B) = \frac{8}{3 2} = \frac{1}{4}

Finally, substitute

P (A),

P (B),

and

P (A \cap B)

into the initial equation to see if a true statement is obtained.

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

SubstituteValues

Substitute values

\frac{1}{4} = ? \frac{1}{2} \cdot \frac{1}{2}

MultFrac

Multiply fractions

\frac{1}{4} = \frac{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.

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Catch-Up and Review

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