Question

g according to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers; and of those not having lung disease, 25% are smokers. What is the probability that a smoker has lung disease?

Answer 1

the probability is 0.2132 (21.32%)

Step-by-step explanation:

defining the event S= a person is smoker , then if we choose a person at random , the probability that the person is smoker is

P(S) = probability of having lung disease * probability of being a smoker given that has lung disease + probability of not having lung disease * probability of being a smoker given that has not lung disease = 0.07*0.90 + 0.93*0.25 = 0.2955

then for conditional probability we use the theorem of Bayes . Defining the event L= the person has lung disease then

P(L/S)=P(L∩S)/P(S) = 0.07*0.90 / 0.2955 = 0.2132 (21.32%)

where

P(L∩S) = probability of having lung disease and being a smoker

P(L/S)= probability of having lung disease given that a smoker was chosen