I stumble upon this probability problem. I would like to know if there's anything I haven't thought of?
Problem: The probability of a pet being a dog is 30%. The probability that a pet dog weighs more than 3 kg is 60%. If a pet is not a dog, the probability of it weighing less than or equal to 3 kg is 70%.
Question: What is the probability of a pet being a dog if it weighs more than 3 kg?
Solution:
(1) P(Pet = Dog) = 0.3
(2) P(Weight > 2 | Pet = Dog) = 0.6
(3) P(Weight ≤ 2 | Pet != Dog) = 0.7
One can obtain the answer by applying Bayes' Rule and the Total Probability Theorem, that is:
P(Pet = Dog | Weight > 3) = P(Pet = Dog, Weight > 3) / P(Weight > 3) = 0.18 / 0.46 = 0.3913
Where we get P(Pet = Dog, Weight > 3) = P(Pet = Dog) * P(Weight > 3 | Pet = Dog) = 0.3 * 0.6 = 0.18 from (1) and (2)
Where we get P(Weight > 3) = P(Pet = Dog) * P(Weight > 3 | Pet = Dog) + P(Pet != Dog) * P(Weight > 3 | Pet != Dog) = P(Pet = Dog) * P(Weight > 3 | Pet = Dog) + P(Pet != Dog) * (1 - P(Weight ≤ 3 | Pet != Dog)) = 0.3 * 0.6 + 0.7 * 0.4 = 0.46 from (2) and (3) and applying the Total Probability Theorem.
Thanks for your help!