Question

in a cortain community, 24% of the families own a dog, and 25% of the families that own a dog also own a cat. it is also known that 31% of all the families own a cat. What is the probability that a randomly selected family owns a dog? What is the conditional probability that a randomly solected family owns a dog diven that it doesn't own a cat?

Answer 1

Firstly, let's clarify the probabilities given:

- P(Dog) = 0.24, which is the probability of a family owning a dog.

- P(Cat|Dog) = 0.25, which is the probability of a family owning a cat given that they own a dog.

- P(Cat) = 0.31, which is the probability of a family owning a cat.

The first question asks for the probability that a randomly selected family owns a dog, which we already know is 0.24 or 24%.

Now for the second question, we need to find the probability of a family owning a dog given that they don't own a cat, i.e., P(Dog|~Cat). We know from the Bayes theorem that P(A|B) = P(B|A)*P(A) / P(B). Using this formula with our probabilities, we get:

P(Cat|Dog) = P(Dog|Cat) * P(Cat) / P(Dog)

0.25 = P(Dog|Cat) * 0.31 / 0.24

Solving for P(Dog|Cat), we get:

P(Dog|Cat) = 0.25 * 0.24 / 0.31 ≈ 0.1935

That is, the probability of a family owning a dog given that they own a cat is approximately 0.1935 or 19.35%.

To find the conditional probability P(Dog|~Cat), we should first determine the probability of not owning a cat, which is P(~Cat) = 1 - P(Cat) = 1 - 0.31 = 0.69.

Then, we know that P(Dog) = P(Dog and Cat) + P(Dog and ~Cat), and P(Dog and Cat) = P(Dog|Cat) * P(Cat) = 0.1935 * 0.31 ≈ 0.06.

We can find P(Dog and ~Cat) = P(Dog) - P(Dog and Cat) = 0.24 - 0.06 = 0.18.

Finally, we can find P(Dog|~Cat) = P(Dog and ~Cat) / P(~Cat) = 0.18 / 0.69 ≈ 0.2609 or 26.09%.

Therefore, the probability that a randomly selected family owns a dog is 24%, and the conditional probability that a randomly selected family owns a dog given that it doesn't own a cat is approximately 26.09%.