Here are a few recommended readings before getting started with this lesson.
In a multiple-choice test, Davontay randomly selected the answers to all five questions. Each question had two options to choose from.
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?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 $(18−29years old)$ | $55$ | $36$ |
Adults $(30−44years old)$ | $51$ | $41$ |
Middle-Aged Adults $(45−64years old)$ | $44$ | $52$ |
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.
Descriptions:
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.
Descriptions:
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 $64.$
Descriptions:
Descriptions:
Clinton | Trump | Total | |
---|---|---|---|
Young Adults $(18−29years old)$ | $55$ | $36$ | $91$ |
Adults $(30−44years old)$ | $51$ | $41$ | $92$ |
Middle-Aged Adults $(45−64years old)$ | $44$ | $52$ | $96$ |
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)P(Trump∣Young Adult) =9155 ≈0.60=9136 ≈0.40 $ Consequently, the following two conclusions can be made.
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.
$P(Adult∣Clinton)P(Middle-Aged Adult∣Clinton) =19551 ≈0.26=19544 ≈0.23 $ Consequently, the following two conclusions can be made.
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 $64.$
$P(Clinton∣Middle-Aged Adult) =9644 ≈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) =18152 ≈0.29 $ In conclusion, knowing that someone voted for Trump, there is a $29%$ probability that they are between $45$ and $64$ years old.
$P(Senior∣Clinton) =19545 ≈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) =9752 ≈0.54 $ In conclusion, knowing that someone is older than $64$ years old, there is a $54%$ chance that they voted for Trump.
Last month, Ignacio got a part-time job working from $4:00P.M.$ to $8:00P.M.$ 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.
Late | On Time | Total | |
---|---|---|---|
Nap | $2$ | $6$ | $8$ |
No Nap | $3$ | $9$ | $12$ |
Total | $5$ | $15$ | $20$ |
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.
Ignacio taking a nap after lunchand
Ignacio being late for workare 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)=205 =41 $ 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)=82 =41 $ Since $P(Late)$ and $P(Late∣Nap)$ are both equal to $41 ,$ 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.
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.
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 People69+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)=190114 =0.6 $ Therefore, there is a $60%$ chance that Maya is correct in thinking that Ignacio has exactly one type of 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.
These are conditional probabilities. Since there are only two types of pets in the survey, the above statement can be written more precisely.
Now that all information is written, the first probability can be found. $P(Dog∣No Cat)=People who do not have a catPeople who have a dog but not 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)=13069 ≈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 do not have a dogPeople who have a cat but not 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)=10645 ≈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.
By the Complement Rule, the following pair of conclusions can also be drawn.
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.
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.
Pop | Classical | Total | |
---|---|---|---|
$35$ Years Old or Younger | $30$ | $50$ | $80$ |
Older Than $35$ | $15$ | $25$ | $40$ |
Total | $45$ | $75$ | $120$ |
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.
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=120c+40=120 ⇔{f=75c=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.
$P(Pop∣>35)=40d $, $P(Pop)=83 $
$LHS⋅40=RHS⋅40$
$ca ⋅b=ca⋅b $
Calculate quotient
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=4515+e=40 ⇔{a=30e=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 musicand
A person is older than $35$ years old.
A person likes pop musicand
A person is $35$ years old or younger.
A person likes classical musicand
A person is older than $35$ years old.
A person likes classical musicand
A person is $35$ years old or younger.
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.
On the day of the party, Diego puts the treats on a table.
Substitute values
Calculate quotient
Convert to percent
Round to nearest integer
Substitute values
Calculate quotient
Convert to percent
Round to nearest integer
Substitute values
Calculate quotient
Convert to percent
Round to nearest integer
Substitute values
Calculate quotient
Convert to percent
Round to nearest integer
Substitute values
Calculate quotient
Convert to percent
Round to nearest integer
Substitute values
$ba =b/16a/16 $
Substitute values
$ba =b/9a/9 $
Substitute values
$ba =b/7a/7 $
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.
Substitute values
Multiply fractions
The events being independent means that having guessed the second question correctly does not influence having guessed the fifth question correctly and vice versa.