In a class of 18 students, 7 have a cat and 12 have a dog. There are 2 students who do not have a cat or a dog. What is the probability that a student who has a cat also has a dog?

Let's solve the problem step by step.

1. Determine the total number of students who have either a cat or a dog or both:

   • Total students in the class = 18
   • Students who do not have a cat or a dog = 2
   • Therefore, students who have either a cat or a dog or both = 18 - 2 = 16

2. Use the principle of inclusion-exclusion to find the number of students who have both a cat and a dog:

   • Let $C$ be the number of students who have a cat.
   • Let $D$ be the number of students who have a dog.
   • Let $B$ be the number of students who have both a cat and a dog.
   • We know $C = 7$, $D = 12$, and $C \cup D = 16$.

   Using the principle of inclusion-exclusion: $C \cup D = C + D - B$ Substituting the known values: $16 = 7 + 12 - B$ Solving for $B$: $16 = 19 - B \implies B = 19 - 16 \implies B = 3$

3. Calculate the probability that a student who has a cat also has a dog:

   • The number of students who have both a cat and a dog = 3
   • The number of students who have a cat = 7

   The probability $P$ that a student who has a cat also has a dog is given by: $P = \frac{\text{Number of students who have both a cat and a dog}}{\text{Number of students who have a cat}} = \frac{3}{7}$

Therefore, the probability that a student who has a cat also has a dog is $\frac{3}{7}$.