Question

It is estimated that 5.9% of Americans have diabetes. Suppose a medic lab uses a test for dieabetes that is 98% accurate for people who have the disease and 95% acc who do not have it. Find the conditional probabill=ity that a randomly selected person actually has diabetes given taht the lab test says they have it.

Answer 1

To solve this, we can use Bayes' Theorem, which states that the probability of A given B is equal to the probability of B given A times the probability of A divided by the probability of B. More concisely, \[(A|B)=(B|A)*\frac{A}{B}\] We want to know the probability a person has diabetes, given that they have a positive result. We don't know this yet. What we DO know is that the probability of B given A (that is, the probability they get a positive result when they have diabetes), which is 98%. The probability of A is 5.9%, and the probability of B is (fraction who do have diabetes and get positive) + (fraction who don't and get positive) = 0.059*0.98+0.941*0.05. So, we get our answer to be:\[0.98*\frac{ 0.059 }{ 0.059*0.98+0.941*0.05}\] \[=0.98*\frac{0.059}{0.10487}\] \[≈0.55135\] So we get a probability of a little over half.

Answer 2

i need to make a tree how would i do that?

Answer 3

As you can see from the probability tree the required probability is 0.05782