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

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

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 = (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.