Test for Disease
Suppose there is a disease that infects 1 in every 10 people. There is a test for whether a person has the disease, which either says that they do have the disease or don't. However, the test is only 80% accurate, which means that if a person has the disease, there is only 80% probability that the test says they have the disease, and if they don't have the disease, there is 80% probability that the test shows they don't have the disease. If I get tested and it says I do have the disease, the probability I actually do have the disease can be written in the form where and are relatively prime positive integers. Find the value of
The answer is 17.
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1 solution
We know that P ( A ∣ B ) = P ( B ) P ( A ∩ B ) . We need to find the value of P ( I have disease ∣ Test says I have disease ) . We can write this as P ( I have disease ∣ Test says I have disease ) = P ( Test says I have disease ) P ( I have disease ∩ Test says I have disease ) . We know that P ( I have disease ∩ Test says I have disease ) is just 0 . 1 ⋅ 0 . 8 = 0 . 0 8 . Also, P ( Test says I have disease = P ( I have disease ∩ Test says I have disease ) + P ( I don’t have disease ∩ Test says I have disease ) . We therefore know that P ( Test says I have disease ) = 0 . 1 ⋅ 0 . 8 + 0 . 9 ⋅ 0 . 2 = 0 . 2 6 . Therefore, we know that P ( I have disease ∣ Test says I have disease ) = 0 . 2 6 0 . 0 8 = 1 3 4 . Therefore, we have 4 + 1 3 = 1 7 .