Question
A survey asked students that are athletes and not athletes whether they prefer to watch golf or tennis on TV. The results are shown below.
Prefers Watching Golf Prefers Watching Tennis Total
Athlete 60 30 90
Not an Athlete 96 48 144
Total 156 78 234
Let A be the event "student is not an athlete" and B be the event "student prefers watching tennis." Determine if A and B are independent events.
(1 point)
Responses
P(A∣B)=4878≈0.62
and P(A)=48144≈0.33
The events are not independent.
cap p times open paren cap A divides cap b close paren is equal to 48 over 78 almost equal to 0 point 6 2 and cap p times cap A is equal to 48 over 144 almost equal to 0 point 3 3 The events are not independent.
P(A∣B)=48144≈0.33
and P(A)=144234≈0.62
The events are not independent.
cap p times open paren cap A divides cap b close paren is equal to 48 over 144 almost equal to 0 point 3 3 and cap p times cap A is equal to 144 over 234 almost equal to 0 point 6 2 The events are not independent.
P(A∣B)=4878≈0.62
and P(A)=48234≈0.21
The events are not independent.
cap p times open paren cap A divides cap b close paren is equal to 48 over 78 almost equal to 0 point 6 2 and cap p times cap A is equal to 48 over 234 almost equal to 0 point 2 1 The events are not independent.
P(A∣B)=4878≈0.62
and P(A)=144234≈0.62
The events are independent.
Prefers Watching Golf Prefers Watching Tennis Total
Athlete 60 30 90
Not an Athlete 96 48 144
Total 156 78 234
Let A be the event "student is not an athlete" and B be the event "student prefers watching tennis." Determine if A and B are independent events.
(1 point)
Responses
P(A∣B)=4878≈0.62
and P(A)=48144≈0.33
The events are not independent.
cap p times open paren cap A divides cap b close paren is equal to 48 over 78 almost equal to 0 point 6 2 and cap p times cap A is equal to 48 over 144 almost equal to 0 point 3 3 The events are not independent.
P(A∣B)=48144≈0.33
and P(A)=144234≈0.62
The events are not independent.
cap p times open paren cap A divides cap b close paren is equal to 48 over 144 almost equal to 0 point 3 3 and cap p times cap A is equal to 144 over 234 almost equal to 0 point 6 2 The events are not independent.
P(A∣B)=4878≈0.62
and P(A)=48234≈0.21
The events are not independent.
cap p times open paren cap A divides cap b close paren is equal to 48 over 78 almost equal to 0 point 6 2 and cap p times cap A is equal to 48 over 234 almost equal to 0 point 2 1 The events are not independent.
P(A∣B)=4878≈0.62
and P(A)=144234≈0.62
The events are independent.
Answers
GPT-4o mini
To determine whether events A and B are independent, we need to compare P(A|B) with P(A). If P(A|B) = P(A), then A and B are independent; otherwise, they are dependent.
We can calculate the necessary probabilities using the data provided.
1. **Calculate P(A)**:
\[
P(A) = \frac{\text{Total number of students who are not athletes}}{\text{Total number of students}}
= \frac{144}{234} \approx 0.6154
\]
2. **Calculate P(B)**:
P(B) is the probability that a student prefers watching tennis:
\[
P(B) = \frac{\text{Total number of students who prefer watching tennis}}{\text{Total number of students}}
= \frac{78}{234} \approx 0.3333
\]
3. **Calculate P(A ∩ B)**:
This is the probability that a student is not an athlete and prefers watching tennis. From the data, we see that there are 48 students who are not athletes and prefer watching tennis.
\[
P(A ∩ B) = \frac{\text{Number of non-athletes who prefer tennis}}{\text{Total number of students}}
= \frac{48}{234} \approx 0.2051
\]
4. **Calculate P(A|B)**:
This is the probability that a student is not an athlete given that they prefer watching tennis:
\[
P(A|B) = \frac{P(A ∩ B)}{P(B)} = \frac{48/234}{78/234} = \frac{48}{78} \approx 0.6154
\]
5. **Compare P(A) and P(A|B)**:
Now, we check if \( P(A|B) \) is equal to \( P(A) \):
- \( P(A) \approx 0.6154 \)
- \( P(A|B) \approx 0.6154 \)
Since these probabilities are approximately equal, events A and B are independent.
Thus, the correct conclusion is that the events are independent, and the proper option is:
**P(A|B) = P(A) = 0.6154; the events are independent.**
We can calculate the necessary probabilities using the data provided.
1. **Calculate P(A)**:
\[
P(A) = \frac{\text{Total number of students who are not athletes}}{\text{Total number of students}}
= \frac{144}{234} \approx 0.6154
\]
2. **Calculate P(B)**:
P(B) is the probability that a student prefers watching tennis:
\[
P(B) = \frac{\text{Total number of students who prefer watching tennis}}{\text{Total number of students}}
= \frac{78}{234} \approx 0.3333
\]
3. **Calculate P(A ∩ B)**:
This is the probability that a student is not an athlete and prefers watching tennis. From the data, we see that there are 48 students who are not athletes and prefer watching tennis.
\[
P(A ∩ B) = \frac{\text{Number of non-athletes who prefer tennis}}{\text{Total number of students}}
= \frac{48}{234} \approx 0.2051
\]
4. **Calculate P(A|B)**:
This is the probability that a student is not an athlete given that they prefer watching tennis:
\[
P(A|B) = \frac{P(A ∩ B)}{P(B)} = \frac{48/234}{78/234} = \frac{48}{78} \approx 0.6154
\]
5. **Compare P(A) and P(A|B)**:
Now, we check if \( P(A|B) \) is equal to \( P(A) \):
- \( P(A) \approx 0.6154 \)
- \( P(A|B) \approx 0.6154 \)
Since these probabilities are approximately equal, events A and B are independent.
Thus, the correct conclusion is that the events are independent, and the proper option is:
**P(A|B) = P(A) = 0.6154; the events are independent.**