Question
In a class of 24 students, 5 have a cat and 13 have a dog. There are 9 students who do not have a cat or a dog. What is the probability that a student chosen randomly from the class does not have a cat? Type your answer as a fraction OR a decimal rounded to 3 decimal places(1 point)
The probabilty that a student does NOT have a cat is:
The probabilty that a student does NOT have a cat is:
Answers
Answered by
GPT-4o mini
To find the probability that a randomly chosen student does not have a cat, we first need to determine the number of students who do not have a cat.
Given:
- Total students = 24
- Students with a cat = 5
- Students with a dog = 13
- Students with neither (no cat, no dog) = 9
To find the number of students with at least one pet (either a cat or a dog), we can subtract the number of students with neither from the total number of students:
\[
\text{Students with at least one pet} = \text{Total students} - \text{Students with neither} = 24 - 9 = 15
\]
Next, to find the number of students who have at least one of the pets (cat or dog), we can use the principle of inclusion-exclusion. Let \( C \) be the set of students with cats and \( D \) be the set of students with dogs.
The number of students with at least one pet is given as:
\[
|C \cup D| = |C| + |D| - |C \cap D|
\]
From the information provided (but not about the overlap, as we do not have it), we know:
\[
15 = 5 + 13 - |C \cap D|
\]
Rearranging gives:
\[
|C \cap D| = 5 + 13 - 15 = 3
\]
This means that:
- Students with only a cat = \( |C| - |C \cap D| = 5 - 3 = 2 \)
- Students with only a dog = \( |D| - |C \cap D| = 13 - 3 = 10 \)
Now, we can find the number of students who do not have a cat:
\[
\text{Students without a cat} = \text{Total students} - \text{Students with a cat} = 24 - 5 = 19
\]
Finally, the probability \( P \) that a randomly chosen student does not have a cat is:
\[
P(\text{not having a cat}) = \frac{\text{Number of students without a cat}}{\text{Total number of students}} = \frac{19}{24}
\]
To express this as a decimal rounded to three decimal places:
\[
\frac{19}{24} \approx 0.79167
\]
Rounded to three decimal places, this is:
\[
0.792
\]
Thus, the probability that a student does NOT have a cat is:
\[
\frac{19}{24} \quad \text{or} \quad 0.792
\]
Given:
- Total students = 24
- Students with a cat = 5
- Students with a dog = 13
- Students with neither (no cat, no dog) = 9
To find the number of students with at least one pet (either a cat or a dog), we can subtract the number of students with neither from the total number of students:
\[
\text{Students with at least one pet} = \text{Total students} - \text{Students with neither} = 24 - 9 = 15
\]
Next, to find the number of students who have at least one of the pets (cat or dog), we can use the principle of inclusion-exclusion. Let \( C \) be the set of students with cats and \( D \) be the set of students with dogs.
The number of students with at least one pet is given as:
\[
|C \cup D| = |C| + |D| - |C \cap D|
\]
From the information provided (but not about the overlap, as we do not have it), we know:
\[
15 = 5 + 13 - |C \cap D|
\]
Rearranging gives:
\[
|C \cap D| = 5 + 13 - 15 = 3
\]
This means that:
- Students with only a cat = \( |C| - |C \cap D| = 5 - 3 = 2 \)
- Students with only a dog = \( |D| - |C \cap D| = 13 - 3 = 10 \)
Now, we can find the number of students who do not have a cat:
\[
\text{Students without a cat} = \text{Total students} - \text{Students with a cat} = 24 - 5 = 19
\]
Finally, the probability \( P \) that a randomly chosen student does not have a cat is:
\[
P(\text{not having a cat}) = \frac{\text{Number of students without a cat}}{\text{Total number of students}} = \frac{19}{24}
\]
To express this as a decimal rounded to three decimal places:
\[
\frac{19}{24} \approx 0.79167
\]
Rounded to three decimal places, this is:
\[
0.792
\]
Thus, the probability that a student does NOT have a cat is:
\[
\frac{19}{24} \quad \text{or} \quad 0.792
\]