Question
Students were surveyed about their pet preferences. The results are shown below.
Dog Cat Fish Total
Juniors 49 33 19 101
Seniors 56 27 7 90
Total 105 60 26 191
A student from the survey is chosen at random. What is the probability that the student prefers a cat or fish as a pet, given that they are a senior?(1 point)
Responses
1743
17 over 43
930
9 over 30
786
7 over 86
1745
Dog Cat Fish Total
Juniors 49 33 19 101
Seniors 56 27 7 90
Total 105 60 26 191
A student from the survey is chosen at random. What is the probability that the student prefers a cat or fish as a pet, given that they are a senior?(1 point)
Responses
1743
17 over 43
930
9 over 30
786
7 over 86
1745
Answers
GPT-4o mini
To find the probability that a randomly chosen student prefers a cat or fish given that they are a senior, we can use the following formula for conditional probability:
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)}
\]
Where:
- \(P(A \mid B)\) is the probability that event A occurs given that event B has occurred.
- \(P(A \cap B)\) is the probability that both A and B occur.
- \(P(B)\) is the probability that B occurs.
In this context:
- Let event A be "prefers a cat or fish."
- Let event B be "is a senior."
From the table, we have:
- The number of seniors who prefer cats is 27.
- The number of seniors who prefer fish is 7.
- Total seniors = 90.
Now, we find \(P(A \cap B)\), which is the number of seniors who prefer a cat or fish:
\[
P(A \cap B) = \text{Seniors who prefer cats} + \text{Seniors who prefer fish} = 27 + 7 = 34
\]
Next, we find \(P(B)\), which is the total number of seniors:
\[
P(B) = 90
\]
Now we can substitute these values into the conditional probability formula:
\[
P(A \mid B) = \frac{34}{90}
\]
To simplify this fraction:
\[
\frac{34}{90} = \frac{17}{45}
\]
Therefore, the probability that a student prefers a cat or fish as a pet, given that they are a senior, is \( \frac{17}{45} \).
None of the provided options seem to match this exact answer, so let's check our calculations again just to be sure.
- Seniors who prefer cats: 27
- Seniors who prefer fish: 7
- Total seniors: 90
- Total seniors preferring either cats or fish: 27 + 7 = 34
And:
\[
P(A \mid B) = \frac{34}{90} \text{ which simplifies to } \frac{17}{45}
\]
It appears that none of the answers you provided (1743, 17/43, 9/30, 7/86, 1745) match this calculation. It seems that the correct probability is \( \frac{17}{45} \).
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)}
\]
Where:
- \(P(A \mid B)\) is the probability that event A occurs given that event B has occurred.
- \(P(A \cap B)\) is the probability that both A and B occur.
- \(P(B)\) is the probability that B occurs.
In this context:
- Let event A be "prefers a cat or fish."
- Let event B be "is a senior."
From the table, we have:
- The number of seniors who prefer cats is 27.
- The number of seniors who prefer fish is 7.
- Total seniors = 90.
Now, we find \(P(A \cap B)\), which is the number of seniors who prefer a cat or fish:
\[
P(A \cap B) = \text{Seniors who prefer cats} + \text{Seniors who prefer fish} = 27 + 7 = 34
\]
Next, we find \(P(B)\), which is the total number of seniors:
\[
P(B) = 90
\]
Now we can substitute these values into the conditional probability formula:
\[
P(A \mid B) = \frac{34}{90}
\]
To simplify this fraction:
\[
\frac{34}{90} = \frac{17}{45}
\]
Therefore, the probability that a student prefers a cat or fish as a pet, given that they are a senior, is \( \frac{17}{45} \).
None of the provided options seem to match this exact answer, so let's check our calculations again just to be sure.
- Seniors who prefer cats: 27
- Seniors who prefer fish: 7
- Total seniors: 90
- Total seniors preferring either cats or fish: 27 + 7 = 34
And:
\[
P(A \mid B) = \frac{34}{90} \text{ which simplifies to } \frac{17}{45}
\]
It appears that none of the answers you provided (1743, 17/43, 9/30, 7/86, 1745) match this calculation. It seems that the correct probability is \( \frac{17}{45} \).