Question

In a region, 12% of the adult population are smokers, 0.8% are smokers with emphysema, and 0.2% are non-smokers with emphysema.

a. What is the probability that a person selected at random has emphysema?

b. Given that the selected person is a smoker, what is the probability that the person has emphysema?

c. Given that the selected person is not a smoker, what is the probability that this person has emphysema?

Answer 1

P(smokers) = 0.12 and therefore, P(non - smokers) = 1 - P(smokers) = 1 - 0.12 = 0.88

P(smokers and emphysema) = 0.008

P(non - smokers and emphysema) = 0.002

a. Answer :

P(emphysema) = P(smokers and emphysema) + P(non - smokers and emphysema)

                         = 0.008 + 0.002

                         = 0.01

Therefore, the probability that a person selected at random has emphysema is 0.01

b. Answer :

P(emphysema | smoker) = P(emphysema and smoker) / P(smoker)   ...........(conditional probability)

                                        = 0.008 / 0.12

                                        = 0.0667

Therefore, given that the selected person is a smoker, the probability that the person has emphysema is 0.0667

c. Answer :

P(emphysema | non - smoker) = P(emphysema and non - smoker) / P(non - smoker)     ..........(conditional probability)

                                                  = 0.002 / 0.88

                                                  = 0.0023

Therefore, given that the selected person is not a smoker, the probability that this person has emphysema is 0.0023