Respuesta :
Answer:
There is a 50% probability that the household has a dog, given that the household has a cat.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a household has a cat.
B is the probability that a household has a dog.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a household has a cat but not a dog and [tex]A \cap B[/tex] is the probability that a household has both a cat and a dog.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
The probability that the household has a cat or a dog is 0.5
[tex]a + b + (A \cap B) = 0.5[/tex]
The probability that the household has a dog is 0.4
[tex]B = 0.4[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]b = 0.4 - (A \cap B)[/tex]
The probability that the household has a cat is 0.2.
[tex]A = 0.2[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]a = 0.2 - (A \cap B)[/tex]
So
[tex]a + b + (A \cap B) = 0.5[/tex]
[tex]0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5[/tex]
[tex]A \cap B = 0.1[/tex]
What is the probability that the household has a dog, given that the household has a cat?
20% of the households have a cat, and 10% have both a cat and a dog. So
[tex]P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5[/tex]
There is a 50% probability that the household has a dog, given that the household has a cat.
