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:
\[ \frac{19}{24} \quad \text{or} \quad 0.792 \]