Question

In a hospital, 10% of patients have liver disease and 5% are alcoholics. Among those diagnosed with liver disease, 7% are alcoholics. If the patient is an alcoholic, what is their chances of having liver disease

Answer 1

Let A be event that patient has a liver disease, so
P(A) = 10% = 0.1
Let B be event that patient is an alcoholic, so
P(B) = 5% = 0.05

Among patients with liver disease, 7% are alcoholic; this means that probability that a patient is alcoholic, given that they have liver disease, is 7%
So, this is our 
P(B|A) = 7% = 0.07
Now, we need to find probability that patient has liver disease given, he is alcoholic, so
this will be P(A|B)
So, using Bayes Theorem, we get
P(A|B) = (0.07 * 0.1) / (0.05)
P(A|B) =0.14
Therefore, if the patient is an alcoholic, his chances of having liver disease is 0.14 or 14%