Given:
Number of students who has a cat and a dog = 5
Number of students who has a cat but do not have a dog = 11
Number of students who has a dog but do not have a cat = 3
Number of students who neither have a cat nor a dog = 2
To find:
The probability that a student has a cat given that they do not have a dog.
Solution:
Let the following events:
A = Student has a cat
B = Do not have a dog
Total number of outcomes is:
[tex]5+3+11+2=21[/tex]
The probability that a student has a cat but do not have a dog is:
[tex]P(A\cap B)=\dfrac{11}{21}[/tex]
The probability that a student do not have a dog is:
[tex]P(B)=\dfrac{11+2}{21}[/tex]
[tex]P(B)=\dfrac{13}{21}[/tex]
The conditional probability is:
[tex]P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}[/tex]
[tex]P\left(\dfrac{A}{B}\right)=\dfrac{\dfrac{11}{21}}{\dfrac{13}{21}}[/tex]
[tex]P\left(\dfrac{A}{B}\right)=\dfrac{11}{13}[/tex]
Therefore, the probability that a student has a cat given that they do not have a dog is [tex]\dfrac{11}{13}[/tex].
