3.7. Independent Events#
Two events are considered independent if the occurrence of one event does not influence the occurrence of the other event. For independent events, the probability of both events occurring is the product of their individual probabilities. This concept is crucial in determining the likelihood of multiple events happening together when they are independent of each other.
Definition - Independent Events
Two events \(A\) and \(B\) are considered independent if the occurrence of one event does not influence or impact the probability of the other event occurring. In other words, the outcome of event \(A\) has no effect on the likelihood of event \(B\) happening, and vice versa. For clarity, when dealing with several events, we say that they are independent if the occurrence of any single event does not affect the probabilities of the occurrence of the other events. Conversely, if the occurrence of event \(A\) does affect the probability of event \(B\) occurring (or vice versa), then events \(A\) and \(B\) are said to be dependent. In such cases, the outcomes of the events are linked, and the occurrence of one event provides information or alters the likelihood of the other event happening [Devore et al., 2021].
Event \(B\) is said to be independent of event \(A\) if:
3.7.1. Notation \(P(A \text{ and } B)\)#
The notation \(P(A \text{ and } B)\) can have two different interpretations, depending on the context in which it is used.
Multiplication Rule
Addition Rule
3.7.2. Multiplication Rule#
In the context of the multiplication rule, \(P(A \text{ and } B)\) represents the probability that event \(A\) occurs in one trial and event \(B\) occurs in another trial, assuming the trials are independent. This is often used in situations where we have a sequence of events or experiments, and we want to find the overall probability of both events happening. Mathematically, for independent events \(A\) and \(B\) [Devore et al., 2021]:
Definition - Multiplication Rule
To find the probability that event \(A\) occurs in one trial and event \(B\) occurs in another trial, multiply the probability of event \(A\) by the probability of event \(B\), but be sure that the probability of event \(B\) is found by assuming that event \(A\) has already occurred.
Example 3.33
Consider the scenario where you flip a fair coin twice. Define event \( A \) as getting a Head on the first flip, and event \( B \) as getting a Head on the second flip. We want to calculate the probability of getting Heads on both flips.
What is the probability of getting Heads on both flips (Event \( A \) and Event \( B \))?
Solution:
Probability of Event \( A \) (Head on First Flip): The probability of flipping a Head on a fair coin is \( 0.5 \) or 50%.
\[\begin{equation*}P(A) = 0.5\end{equation*}\]Probability of Event \( B \) (Head on Second Flip): Similarly, the probability of flipping a Head on the second flip is also \( 0.5 \) or 50%.
\[\begin{equation*}P(B) = 0.5\end{equation*}\]Probability of Event \( A \) and Event \( B \) (Heads on Both Flips): Since the coin flips are independent, the probability of getting Heads on both flips is the product of the probabilities of each individual event.
\[\begin{equation*}P(A \text{ and } B) = P(A) \times P(B) = 0.5 \times 0.5 = 0.25 \text{ or } 25\%\end{equation*}\]
Therefore, the probability of flipping a Head on both the first and second flip of a fair coin is \( 0.25 \) or 25%.
Definition - Special Multiplication Rule
If events \(A, B, C, \ldots\) are independent:
3.7.3. Addition Rule#
In the context of the addition rule, \(P(A \text{ and } B)\) denotes that events \(A\) and \(B\) both occur in the same trial. This is often used in situations where the events are not mutually exclusive, meaning that they can happen simultaneously. Mathematically, for events \(A\) and \(B\) [Devore et al., 2021]:
Example 3.34
Consider a standard deck of 52 playing cards. You are about to draw a single card from the deck. Event \(A\) is the event of drawing a heart, and event \(B\) is the event of drawing an Ace. We want to calculate the probability of drawing either a heart or an Ace.
What is the probability of drawing either a heart or an Ace (Event \(A\) or Event \(B\))?
Solution:
Probability of Drawing a Heart (Event A): There are 13 hearts in a deck of 52 cards, so the probability of drawing a heart is:
\[\begin{equation*}P(A) = \dfrac{13}{52}\end{equation*}\]Probability of Drawing an Ace (Event B): There are 4 Aces in the deck, so the probability of drawing an Ace is:
\[\begin{equation*}P(B) = \dfrac{4}{52}\end{equation*}\]Probability of Drawing Either a Heart or an Ace (Event A or Event B): To calculate this, we use the addition rule for probabilities, which states that the probability of either event A or event B occurring is the sum of their individual probabilities minus the probability of both events occurring together (since the Ace of Hearts is counted in both probabilities). There is 1 Ace of Hearts in the deck, so:
\[\begin{equation*}P(A \text{ and } B) = \dfrac{1}{52}\end{equation*}\]Using the addition rule:
\[\begin{align*}P(A \text{ or } B) &= P(A) + P(B) - P(A \text{ and } B) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52} \\&= \dfrac{4}{13} \approx 0.308 \text{ or } 30.8\%\end{align*}\]
As shown in the two interpretations, the notation \(P(A \text{ and } B)\) can represent different concepts depending on whether it’s used in the context of the multiplication rule or the addition rule.
3.7.4. Sampling With Replacement and Without Replacement#
Sampling with replacement and without replacement are two different methods of selecting items from a population. They are commonly used in statistics, probability theory, and various fields of research.
Sampling With Replacement: When sampling is done with replacement, it means that after each member of a population is selected to be part of the sample, that member is returned to the population before the next selection is made. Because of this sampling method [Latpate et al., 2021]:
Repetition: Each member of the population has the possibility of being chosen more than once to be included in the sample. Since the items are put back into the population after each selection, they can be picked again in subsequent selections.
Independence: Events in sampling with replacement are independent. The outcome of one pick does not influence the probabilities for subsequent picks. Each selection is treated as a separate and distinct event, unaffected by previous selections.
Example: Consider a bag with five colored balls (red, blue, green, yellow, and orange), and we want to create a sample of three balls using sampling with replacement. After each draw, the ball is returned to the bag before the next draw. Therefore, it’s possible to draw the same color multiple times in our sample. For instance, we could end up with a sample like {red, red, blue}, where “red” appears twice. Since the selections are independent, the probability of drawing any specific color in the second draw remains the same as the probability of drawing that color in the first draw. The result of the first pick does not affect the probabilities for the second pick.
Sampling Without Replacement: When sampling is done without replacement, it means that each member of a population can only be chosen once included in the sample. Because of this sampling method [Latpate et al., 2021]:
Non-repetition: Once a member is selected and included in the sample, it is removed from the population, and it cannot be picked again for subsequent selections. Each member can appear only once in the sample.
Dependence: Events in sampling without replacement are dependent. The outcome of one pick affects the probabilities for subsequent picks. As items are removed from the population after each selection, the composition of the population changes, impacting the likelihood of selecting specific items in subsequent draws.
Example: Consider a deck of cards with four cards (Ace, King, Queen, and Jack), and we want to create a sample of two cards using sampling without replacement. After the first card is drawn, it is not returned to the deck before the second draw. Therefore, it is not possible to draw the same card twice in our sample. The probabilities for the second pick are influenced by the result of the first pick. For instance, if we draw the Ace in the first draw, the probability of drawing the King in the second draw is now higher because there is one less card in the deck.
Example 3.35 (Drawing Marbles)
You have a bag containing 50 marbles: 15 red, 10 blue, and 25 of various other colors. You will draw one marble from the bag.
a. What is the probability of drawing a marble that is neither red nor blue?
b. If you draw one marble, replace it, and then draw another marble, what is the probability that both marbles will be red?
c. If you draw two marbles at the same time without replacement, what is the probability that both are red?
Solution:
Let’s first visualize the marble distribution:
15 Marbles
10 Marbles
25 Marbles
a. Probability of drawing a marble that is neither red nor blue:
Total marbles = 50
Marbles that are neither red nor blue = 25
b. Probability of drawing two red marbles with replacement:
Probability of drawing a red marble each time = 15/50 = 3/10
Since the marble is replaced, the events are independent
c. Probability of drawing two red marbles without replacement:
Probability of first red marble = 15/50 = 3/10
After drawing one red marble, 14 red marbles remain out of 49 total
Example 3.36 (Study Habits Survey: With and Without Replacement Approach)
A survey was conducted among 2000 college students to understand their study habits. The results are tabulated below:
Studying Habit |
Study With Music |
Study Without Music |
Total |
|---|---|---|---|
Study in Group |
350 |
650 |
1000 |
Study Alone |
150 |
850 |
1000 |
Total |
500 |
1500 |
2000 |
If two college students are randomly selected from the survey participants, find the probability that both study alone and without music using: a) With replacement approach b) Without replacement approach
Solution:
a. With Replacement Approach (Independent Events):
Probability of selecting a student who studies alone and without music:
\[\begin{equation*} P(\text{Alone and No Music}) = \frac{850}{2000} = \frac{17}{40} = 0.425 \end{equation*}\]Probability that both selected students study alone and without music:
\[\begin{equation*} P(\text{Both Alone and No Music}) = \left(\frac{850}{2000}\right)^2 = \left(\frac{17}{40}\right)^2 = \frac{289}{1600} \approx 0.1806 \text{ or } 18.06\% \end{equation*}\]
b. Without Replacement Approach (Dependent Events):
Probability of selecting the first student who studies alone and without music:
\[\begin{equation*} P(\text{First Alone and No Music}) = \frac{850}{2000} = \frac{17}{40} = 0.425 \end{equation*}\]Probability of selecting the second student who studies alone and without music:
\[\begin{equation*} P(\text{Second Alone and No Music}) = \frac{849}{1999} \approx 0.4247 \end{equation*}\]Combined probability:
\[\begin{align*} P(\text{Both Alone and No Music}) &= \frac{850}{2000} \times \frac{849}{1999} \\ & = \frac{721650}{3998000} \approx 0.1805 \text{ or } 18.05\% \end{align*}\]
Remark - Key Difference
The with replacement approach assumes independence between selections, leading to a slightly higher probability. The without replacement approach accounts for the change in the sample space after the first selection, resulting in a marginally lower probability.
Example 3.37 (Library Survey: With and Without Replacement Approach)
A local library surveyed 2000 members about their reading preferences. The results are:
Reading Preference |
Reads Fiction |
Doesn’t Read Fiction |
Total |
|---|---|---|---|
Reads Non-Fiction |
500 |
300 |
800 |
Doesn’t Read Non-Fiction |
200 |
1000 |
1200 |
Total |
700 |
1300 |
2000 |
If two library members are randomly selected, find the probability that both don’t read non-fiction and don’t read fiction using: a) With replacement approach b) Without replacement approach
Solution:
a. With Replacement Approach (Independent Events):
Probability of selecting a member who doesn’t read non-fiction and fiction:
\[\begin{equation*} P(\text{No Non-Fiction and No Fiction}) = \frac{1000}{2000} = \frac{1}{2} = 0.5 \end{equation*}\]Probability that both selected members don’t read non-fiction and fiction:
\[\begin{align*} P(\text{Both No Non-Fiction and No Fiction}) & = \left(\frac{1000}{2000}\right)^2 = \left(\frac{1}{2}\right)^2 \\ & = \frac{1}{4} = 0.25 \text{ or } 25\% \end{align*}\]
b. Without Replacement Approach (Dependent Events):
Probability of selecting the first member who doesn’t read non-fiction and fiction:
\[\begin{equation*} P(\text{First No Non-Fiction and No Fiction}) = \frac{1000}{2000} = \frac{1}{2} = 0.5 \end{equation*}\]Probability of selecting the second member who doesn’t read non-fiction and fiction:
\[\begin{equation*} P(\text{Second No Non-Fiction and No Fiction}) = \frac{999}{1999} \approx 0.4997 \end{equation*}\]Combined probability:
\[\begin{align*} P(\text{Both No Non-Fiction and No Fiction}) & = \frac{1000}{2000} \times \frac{999}{1999} \\ &= \frac{999000}{3998000} \approx 0.2499 \text{ or } 24.99\% \end{align*}\]
Remark - Key Difference
The with replacement approach assumes independence, resulting in a slightly higher probability (25%). The without replacement approach accounts for the change in the sample space after the first selection, leading to a marginally lower probability (24.99%).
Example 3.38 (Drawing 5 Different Cards)
What is the probability of drawing 5 cards of different values (no 2, 3, or 4 cards of the same value) from a standard deck of 52 cards?
Solution:
Fig. 3.50 A standard deck of playing cards. Source: Open Source Vector Playing Cards#
To solve this problem, we need to calculate the probability of each draw and then multiply these probabilities together. Let’s break it down step by step:
First draw:
Any card can be drawn: \(P_1 = \dfrac{52}{52} = 1\)
Second draw:
Cards left: 51
Cards of different value from the first: 48 (52 - 4 cards of the same value)
Probability: \(P_2 = \dfrac{48}{51}\)
Third draw:
Cards left: 50
Cards of different value from the first two: 44 (52 - 4 - 4 cards of the same values)
Probability: \(P_3 = \dfrac{44}{50}\)
Fourth draw:
Cards left: 49
Cards of different value from the first three: 40 (52 - 4 - 4 - 4 cards of the same values)
Probability: \(P_4 = \dfrac{40}{49}\)
Fifth draw:
Cards left: 48
Cards of different value from the first four: 36 (52 - 4 - 4 - 4 - 4 cards of the same values)
Probability: \(P_5 = \dfrac{36}{48}\)
To find the overall probability of drawing 5 different cards, we multiply these individual probabilities:
Note
This is an example of dependent probability because each draw affects the probability of subsequent draws.
The calculation assumes drawing without replacement, which is why the number of available cards decreases with each draw.
The numerator in each step represents the number of favorable outcomes (cards of different values), while the denominator represents the total remaining cards.
The final probability of about 72.05% indicates that you have a good chance of drawing 5 cards of different values from a standard deck.
Example 3.39
What is the probability of 10 people in a room, nobody sharing a birthday?
Solution: This problem is a classic example of the “birthday paradox,” but in reverse. Instead of finding the probability that two people share a birthday, we’re interested in the probability that no one shares a birthday.
The calculation goes as follows:
The probability that the first person has a birthday on any day of the year is \( \dfrac{365}{365} \) because they can be born on any day of the 365 days in a year.
The probability that the second person has a different birthday from the first person is \( \dfrac{364}{365} \). This is because there are 364 days remaining that are not the first person’s birthday.
Similarly, the probability that the third person doesn’t share a birthday with the first two is \( \dfrac{363}{365} \), and so on.
For 10 people, the probability that none of them share a birthday is the product of each individual probability:
This product is the number of ways to assign different birthdays to each person divided by the total number of ways birthdays can be assigned without restriction:
When calculated, this probability is approximately 0.883, or 88.3%, which means there’s an 88.3% chance that in a room of 10 people, no two will share the same birthday. This assumes a non-leap year and that birthdays are equally likely to occur on any day of the year.
Example 3.40
In a study titled “Factors Influencing the Regular Attendance of Asthma Check-ups in Children,” researchers investigated the impact of various factors on the likelihood of children with asthma attending regular check-ups. The study provided the following joint probability distribution based on the type of healthcare facility and whether the child attended the check-up regularly.
Type of Facility |
Regular Attendance (Yes) |
Regular Attendance (No) |
P(Total) |
|---|---|---|---|
Pediatric Clinic |
0.125 |
0.275 |
0.400 |
General Hospital |
0.095 |
0.305 |
0.400 |
Specialist Clinic |
0.110 |
0.090 |
0.200 |
P(Attendance) |
0.330 |
0.670 |
1.000 |
For a randomly selected child from the study, calculate the following probabilities and interpret the results as percentages:
a. P(Regular Attendance)
b. P(Pediatric Clinic)
c. P(Regular Attendance and Pediatric Clinic)
d. P(Regular Attendance | Pediatric Clinic)
e. P(Pediatric Clinic | General Hospital)
Solution:
a. \(P(\text{Regular Attendance}) = 0.330 \)
This means there is a 33% chance that a child will attend regular check-ups.
b. \(P(\text{Pediatric Clinic}) = 0.400 \)
There is a 40% chance that a child is associated with a pediatric clinic.
c. \(P(\text{Regular Attendance and Pediatric Clinic}) = 0.125 \)
There is a 12.5% chance that a child attends regular check-ups at a pediatric clinic.
d.
Given that a child is associated with a pediatric clinic, there is a 31.25% chance that they attend regular check-ups.
e. \(P(\text{Pediatric Clinic | General Hospital}) \) cannot be directly calculated from the given information because it requires the probability of being in a pediatric clinic given that the child is already in a general hospital, which is not provided in the joint probability distribution. This probability would require additional information about the transition or referral rates between general hospitals and pediatric clinics.
Example 3.41
In a survey conducted to understand the reading habits of high school students, researchers categorized the students based on their preferred reading genre and whether they read at least one book per month. The following table summarizes the joint probability distribution for the survey results:
Preferred Genre |
Reads at Least One Book per Month (Yes) |
Reads at Least One Book per Month (No) |
P(Total) |
|---|---|---|---|
Fiction |
0.18 |
0.32 |
0.50 |
Non-Fiction |
0.12 |
0.28 |
0.40 |
Other Genres |
0.05 |
0.05 |
0.10 |
P(Reading Habit) |
0.35 |
0.65 |
1.00 |
For a randomly selected student from the survey, calculate the following probabilities and interpret the results as percentages:
a. P(Reads at Least One Book per Month)
b. P(Fiction)
c. P(Reads at Least One Book per Month and Fiction)
d. P(Reads at Least One Book per Month | Fiction)
e. P(Fiction | Reads at Least One Book per Month)
Solution:
a. \(P(\text{Reads at Least One Book per Month}) = 0.35 \)
This means there is a 35% chance that a student reads at least one book per month.
b. \(P(\text{Fiction}) = 0.50 \)
There is a 50% chance that a student’s preferred reading genre is fiction.
c. \(P(\text{Reads at Least One Book per Month and Fiction}) = 0.18 \)
There is an 18% chance that a student reads at least one book per month and prefers fiction.
d.
Given that a student prefers fiction, there is a 36% chance that they read at least one book per month.
e.
Given that a student reads at least one book per month, there is approximately a 51.43% chance that their preferred genre is fiction.
Example 3.42
The Bureau of Labor Statistics collects data on the employment status of the population. Here is a joint probability distribution for the number of months individuals have been unemployed, by age group, for U.S. adults during one year.
Age Group |
Under 6 Months |
6-12 Months |
12-18 Months |
Over 18 Months |
P(Total) |
|---|---|---|---|---|---|
18 - 24 |
0.125 |
0.075 |
0.050 |
0.025 |
0.275 |
25 - 34 |
0.150 |
0.100 |
0.050 |
0.025 |
0.325 |
35 - 44 |
0.100 |
0.075 |
0.025 |
0.015 |
0.215 |
45 - 54 |
0.080 |
0.060 |
0.020 |
0.010 |
0.170 |
55+ |
0.050 |
0.030 |
0.015 |
0.005 |
0.100 |
P(Unemployment Duration) |
0.505 |
0.340 |
0.160 |
0.080 |
1.000 |
For an individual selected at random from this distribution, calculate the following probabilities and interpret the results as percentages:
a. P(Unemployed for Under 6 Months)
b. P(Age 25 - 34)
c. P(Unemployed for Under 6 Months and Age 25 - 34)
d. P(Unemployed for Under 6 Months | Age 25 - 34)
e. P(Age 25 - 34 | Unemployed for 6 - 12 Months)
Solution:
a. \(P(\text{Unemployed for Under 6 Months}) = 0.505\)
This means there is a 50.5% chance that an individual has been unemployed for under 6 months.
b. \(P(\text{Age 25 - 34}) = 0.325\)
There is a 32.5% chance that an individual is in the age group of 25 to 34 years.
c. \(P(\text{Unemployed for Under 6 Months and Age 25 - 34}) = 0.150\)
There is a 15% chance that an individual is in the age group of 25 to 34 years and has been unemployed for under 6 months.
d.
Given that an individual is in the age group of 25 to 34 years, there is approximately a 46.15% chance that they have been unemployed for under 6 months.
e.
Given that an individual has been unemployed for 6 to 12 months, there is approximately a 29.41% chance that they are in the age group of 25 to 34 years.
Example 3.43
A study on the use of public transportation in a city revealed the following joint probability distribution for the choice of transportation and the time of day.
Time of Day |
Bus |
Train |
Bicycle |
P(Total) |
|---|---|---|---|---|
Morning |
0.20 |
0.15 |
0.10 |
0.45 |
Afternoon |
0.10 |
0.20 |
0.05 |
0.35 |
Evening |
0.05 |
0.10 |
0.05 |
0.20 |
P(Transportation Choice) |
0.35 |
0.45 |
0.20 |
1.00 |
For a randomly selected individual from the study, calculate the following probabilities and interpret the results as percentages:
a. P(Choosing Bus)
b. P(Traveling in the Morning)
c. P(Choosing Bus and Traveling in the Morning)
d. P(Choosing Bus | Traveling in the Morning)
e. P(Traveling in the Morning | Choosing Train)
Solution:
a. \(P(\text{Choosing Bus}) = 0.35\)
This means there is a 35% chance that an individual chooses the bus as their mode of transportation.
b. \(P(\text{Traveling in the Morning}) = 0.45\)
There is a 45% chance that an individual travels in the morning.
c. \(P(\text{Choosing Bus and Traveling in the Morning}) = 0.20\)
There is a 20% chance that an individual chooses the bus as their mode of transportation and travels in the morning.
d.
Given that an individual travels in the morning, there is approximately a 44.44% chance that they choose the bus.
e.
Given that an individual chooses the train, there is approximately a 33.33% chance that they travel in the morning.
Example 3.44
A local health clinic conducted a survey to understand the frequency of exercise among different age groups. The survey results are summarized in the joint probability table below, based on the frequency of exercise per week and the age group of the participants.
Age Group |
1-2 Times/Week |
3-4 Times/Week |
5+ Times/Week |
P(Total) |
|---|---|---|---|---|
18-29 |
0.15 |
0.20 |
0.10 |
0.45 |
30-39 |
0.10 |
0.15 |
0.05 |
0.30 |
40-49 |
0.05 |
0.10 |
0.05 |
0.20 |
50+ |
0.02 |
0.03 |
0.00 |
0.05 |
P(Exercise Frequency) |
0.32 |
0.48 |
0.20 |
1.00 |
For a randomly selected participant from the survey, calculate the following probabilities and interpret the results as percentages:
a. P(Exercising 1-2 Times per Week)
b. P(Age Group 30-39)
c. P(Exercising 1-2 Times per Week and Age Group 30-39)
d. P(Exercising 1-2 Times per Week | Age Group 30-39)
e. P(Age Group 30-39 | Exercising 3-4 Times per Week)
Solution:
a. \(P(\text{Exercising 1-2 Times per Week}) = 0.32\)
This means there is a 32% chance that a participant exercises 1-2 times per week.
b. \(P(\text{Age Group 30-39}) = 0.30\)
There is a 30% chance that a participant is in the age group of 30-39 years.
c. \(P(\text{Exercising 1-2 Times per Week and Age Group 30-39}) = 0.10\)
There is a 10% chance that a participant is in the age group of 30-39 years and exercises 1-2 times per week.
d.
Given that a participant is in the age group of 30-39 years, there is approximately a 33.33% chance that they exercise 1-2 times per week.
e.
Given that a participant exercises 3-4 times per week, there is approximately a 31.25% chance that they are in the age group of 30-39 years.
Example 3.45
A survey on smartphone usage patterns was conducted among different age groups to understand the preferred operating system (OS). The survey results are summarized in the joint probability table below, based on the preferred OS and the age group of the respondents.
Age Group |
iOS |
Android |
Other |
P(Total) |
|---|---|---|---|---|
16-25 |
0.18 |
0.22 |
0.05 |
0.45 |
26-35 |
0.14 |
0.16 |
0.04 |
0.34 |
36-45 |
0.06 |
0.10 |
0.03 |
0.19 |
46+ |
0.02 |
0.04 |
0.01 |
0.07 |
P(OS Preference) |
0.40 |
0.52 |
0.13 |
1.00 |
For a randomly selected respondent from the survey, calculate the following probabilities and interpret the results as percentages:
a. P(Prefer iOS)
b. P(Age Group 26-35)
c. P(Prefer iOS and Age Group 26-35)
d. P(Prefer iOS | Age Group 26-35)
e. P(Age Group 26-35 | Prefer Android)
Solution:
a. \(P(\text{Prefer iOS}) = 0.40\)
This means there is a 40% chance that a respondent prefers the iOS operating system.
b. \(P(\text{Age Group 26-35}) = 0.34\)
There is a 34% chance that a respondent is in the age group of 26-35 years.
c. \(P(\text{Prefer iOS and Age Group 26-35}) = 0.14\)
There is a 14% chance that a respondent is in the age group of 26-35 years and prefers the iOS operating system.
d.
Given that a respondent is in the age group of 26-35 years, there is approximately a 41.18% chance that they prefer the iOS operating system.
e.
Given that a respondent prefers the Android operating system, there is approximately a 30.77% chance that they are in the age group of 26-35 years.
Example 3.46
A survey was conducted in a neighborhood to understand pet ownership and preferences. The survey results are summarized in the table below, based on the type of pet owned and the type of housing of the pet owner.
Type of Housing |
Dog |
Cat |
Bird |
None |
P(Total) |
|---|---|---|---|---|---|
Apartment |
320 |
280 |
50 |
100 |
750 |
House |
480 |
220 |
30 |
120 |
850 |
P(Pet Type) |
800 |
500 |
80 |
220 |
1600 |
For a randomly selected individual from the survey, calculate the following probabilities and interpret the results as percentages:
a) Apartment: Find the probability of randomly selecting an individual and getting someone who lives in an apartment.
b) Cat Given Apartment: Find the probability of randomly selecting an individual and getting someone who owns a cat given that the selected person lives in an apartment.
c) Apartment Given Cat: Find the probability of randomly selecting an individual and getting someone who lives in an apartment given that the selected person owns a cat.
d) Cat or Apartment: Find the probability of randomly selecting an individual and getting someone who owns a cat or lives in an apartment.
e) Cat or House: Find the probability of randomly selecting an individual and getting someone who owns a cat or lives in a house.
f) Bird: If two of the survey subjects are randomly selected without replacement, find the probability that they both own birds.
g) Bird: If two of the survey subjects are randomly selected with replacement, find the probability that they both own birds.
h) Complement: If C represents the event of randomly selecting an individual and getting someone who owns a cat, what does \( C^c \) represent? Find the value of \( P(C^c) \).
i) Complement: If A represents the event of randomly selecting an individual and getting someone who lives in an apartment, what does \( A^c \) represent? Find the value of \( P(A^c) \).
j) All Three Dogs: If three of the survey subjects are randomly selected without replacement, find the probability that they all own dogs.
Solution:
a) \(P(\text{Apartment}) = \dfrac{750}{1600} = 0.46875\)
There is a 46.875% chance that a randomly selected individual lives in an apartment.
b) \(P(\text{Cat | Apartment}) = \dfrac{280}{750} = 0.37333\)
Given that the selected individual lives in an apartment, there is a 37.333% chance that they own a cat.
c) \(P(\text{Apartment | Cat}) = \dfrac{280}{500} = 0.56\)
Given that the selected individual owns a cat, there is a 56% chance that they live in an apartment.
d) \(P(\text{Cat or Apartment}) \) is calculated by adding the probabilities of owning a cat and living in an apartment, then subtracting the probability of both owning a cat and living in an apartment to avoid double-counting:
There is a 59.375% chance that a randomly selected individual either owns a cat or lives in an apartment.
e) \(P(\text{Cat or House}) \) is calculated similarly to part d):
There is a 70.625% chance that a randomly selected individual either owns a cat or lives in a house.
f) For two subjects selected without replacement, the probability that both own birds is:
There is approximately a 0.39% chance that both selected individuals own birds.
g) For two subjects selected with replacement, the probability that both own birds is:
There is a 0.25% chance that both selected individuals own birds.
h) \( C^c \) represents the event of randomly selecting an individual and getting someone who does not own a cat. The value of \( P(C^c) \) is:
There is a 68.75% chance that a randomly selected individual does not own a cat.
i) \( A^c \) represents the event of randomly selecting an individual and getting someone who does not live in an apartment. The value of \( P(A^c) \) is:
There is a 53.125% chance that a randomly selected individual does not live in an apartment.
j) For three subjects selected without replacement, the probability that all own dogs is:
There is approximately a 7.9% chance that all three selected individuals own dogs.
Example 3.47
A transportation survey was conducted in a city to understand the commuting preferences of its residents. The survey results are summarized in the table below, based on the preferred mode of commuting and the area of residence.
Area of Residence |
Car |
Bus |
Bike |
Walk |
P(Total) |
|---|---|---|---|---|---|
Downtown |
150 |
200 |
100 |
50 |
500 |
Suburban |
300 |
150 |
50 |
0 |
500 |
Rural |
200 |
50 |
25 |
25 |
300 |
P(Commuting Preference) |
650 |
400 |
175 |
75 |
1300 |
For a randomly selected resident from the survey, calculate the following probabilities and interpret the results as percentages:
a) Downtown: Find the probability of randomly selecting a resident and getting someone who lives downtown.
b) Bus Given Downtown: Find the probability of randomly selecting a resident and getting someone who prefers the bus given that the selected person lives downtown.
c) Downtown Given Bus: Find the probability of randomly selecting a resident and getting someone who lives downtown given that the selected person prefers the bus.
d) Bus or Downtown: Find the probability of randomly selecting a resident and getting someone who prefers the bus or lives downtown.
e) Bike or Suburban: Find the probability of randomly selecting a resident and getting someone who prefers the bike or lives in the suburban area.
f) Walk: If two of the survey subjects are randomly selected without replacement, find the probability that they both prefer to walk.
g) Walk: If two of the survey subjects are randomly selected with replacement, find the probability that they both prefer to walk.
h) Complement: If B represents the event of randomly selecting a resident and getting someone who prefers the bus, what does \( B^c \) represent? Find the value of \( P(B^c) \).
i) Complement: If D represents the event of randomly selecting a resident and getting someone who lives downtown, what does \( D^c \) represent? Find the value of \( P(D^c) \).
j) All Three Car: If three of the survey subjects are randomly selected without replacement, find the probability that they all prefer the car.
Solution:
a) \(P(\text{Downtown}) = \dfrac{500}{1300} \approx 0.3846\)
There is approximately a 38.46% chance that a randomly selected resident lives downtown.
b) \(P(\text{Bus | Downtown}) = \dfrac{200}{500} = 0.4\)
Given that the selected resident lives downtown, there is a 40% chance that they prefer the bus.
c) \(P(\text{Downtown | Bus}) = \dfrac{200}{400} = 0.5\)
Given that the selected resident prefers the bus, there is a 50% chance that they live downtown.
d) \(P(\text{Bus or Downtown}) \) is calculated by adding the probabilities of preferring the bus and living downtown, then subtracting the probability of both preferring the bus and living downtown to avoid double-counting:
There is approximately a 53.85% chance that a randomly selected resident either prefers the bus or lives downtown.
e) \(P(\text{Bike or Suburban}) \) is calculated similarly to part d):
There is approximately a 48.08% chance that a randomly selected resident either prefers the bike or lives in the suburban area.
f) For two subjects selected without replacement, the probability that both prefer to walk is:
There is approximately a 0.43% chance that both selected residents prefer to walk.
g) For two subjects selected with replacement, the probability that both prefer to walk is:
There is approximately a 0.35% chance that both selected residents prefer to walk.
h) \( B^c \) represents the event of randomly selecting a resident and getting someone who does not prefer the bus. The value of \( P(B^c) \) is:
There is approximately a 69.23% chance that a randomly selected resident does not prefer the bus.
i) \( D^c \) represents the event of randomly selecting a resident and getting someone who does not live downtown. The value of \( P(D^c) \) is:
There is approximately a 61.54% chance that a randomly selected resident does not live downtown.
j) For three subjects selected without replacement, the probability that all prefer the car is:
There is approximately an 8.12% chance that all three selected residents prefer the car.