- #1
sessomw5098
- 8
- 0
Hey,
A virus test is 98% accurate and 1 in 10000 people have the virus. Given that the test is positive, what are the chances that you have the virus?
This is what I've got:
Since keyword “given,” I’m assuming Bayes Theorem. So, let A be the event that you have the virus. Let B be the event that the test is positive. So, I have P(A|B). But, my problem is how do I find P(B)? In other words, P(B) is the probability of a virus test 98% accurate and 1 in 10000 people have the virus. Would I have to break event B into two separate events? For example, let C = virus test is 98% accurate and D = 1 in 10000 people have the virus. Then, P(B) = P(C AND D). So, P(A | (C AND D))? Would this be the right approach? Any leads would be greatly appreciated.
Thanks
A virus test is 98% accurate and 1 in 10000 people have the virus. Given that the test is positive, what are the chances that you have the virus?
This is what I've got:
Since keyword “given,” I’m assuming Bayes Theorem. So, let A be the event that you have the virus. Let B be the event that the test is positive. So, I have P(A|B). But, my problem is how do I find P(B)? In other words, P(B) is the probability of a virus test 98% accurate and 1 in 10000 people have the virus. Would I have to break event B into two separate events? For example, let C = virus test is 98% accurate and D = 1 in 10000 people have the virus. Then, P(B) = P(C AND D). So, P(A | (C AND D))? Would this be the right approach? Any leads would be greatly appreciated.
Thanks