Question

One household is to be selected at random from a town. ​ ​The probability that ​the household has a cat is 0.20.2 . ​ ​The probability that the household has a dog ​is 0.40.4 . ​ ​The probability that the household has a cat or a dog is 0.50.5 . ​ ​What is ​the probability that the household has a dog, given that the household has a ​cat? ​ ​Show your work on the scratchpad.

Answer 1

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:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog ​is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that ​the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

So

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

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

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a ​cat.