3/2y - 3 + 5/3 z when y= 6 and z = 3.
1 answer:
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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:
In which a is the probability that a household has a cat but not a dog and is the probability that a household has both a cat and a dog.
By the same logic, we have that:
The probability that the household has a cat or a dog is 0.5
The probability that the household has a dog is 0.4
The probability that the household has a cat is 0.2.
So
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
There is a 50% probability that the household has a dog, given that the household has a cat.