Question 1123012: 53% of the residents in a town own a dog. 64% own a cat. 49% of the dog owners also own a cat. If a resident is chosen at random. What are the probabilities for the following?
-Probability(own a dog)
-P(own a cat)
-P(own a cat and a dog)
-P(own a dog given they own a cat)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! probabilities and are related in that, if something has happened 54% of the time in the past, then, unless something changes, it is reasonable to expect that it will happen approximately 54% of the time in the future as well.
that's why, it 54% of the people own dogs, then, it is reasonable to expect that, if you pick a person randomly out of the same population, the probability that the person will own a dog is 54%.
therefore:
given that 54% own dogs and 64% own cats and 49% own both, ........
p(own a dog) = .54
p(own a cat) = .64
p(own a cat and a dog) = .49
p(own a cat given that they own a dog) requires some logic and a formula that was developed to handle this situation.
you have 54% own a dog.
you have 49% own a cat and a dog.
given that a person owns a dog, the probability that the same person also owns a cat is .49 / .54 = .907407407.
the formula developed to handle this situation is:
p(owns a cat given owns a dog) = p(owns a cat and a dog) divided by p(owns a dog).
the generalizerd formula is:
p(A given B) = p(A and B) / p(B)
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