Question 426048: Would you please be able to help me with this question? I am lost!
in a certain small town, 16% of the population developed lung cancer. if 45% of the population are smokers, and 85% of those developing lung cancer are smokers, what is the probability that a smoker in this population will develop lung cancer?
thank you very much for your time and help!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you are given:
.85 of those developing lung cancer are smokers.
.16 of the population developed lung cancer.
this means that .85 * .16 = .136 of the population developed lung cancer and also smoked.
the smokers, however, only make up .45 of the population.
this means that .136 / .45 = .302222222 of the smokers developed lung cancer.
the probability that a smoker in this population will develop lung cancer is equal to .302222222
this is a p(a given b) type problem.
p(a given b) = p(a intersect b) / p(b)
p(b) is the probability that a person is a smoker.
p(b) equals .45
p(a intersect b) is the probability that a person has lung cancer is also a smoker.
since the probability that the person has lung cancer is .16 and the probability that the person who has lung cancer also smokes is .85, then:
p(a intersect b) = .16 * .85 = .136
p(a given b) = p(a intersect b) / p(b) = .136 / .45 = .3022222222
if you are a smoker in this population, then the probability that you will have lung cancer is .302222
this is based on the previous experience that indicates that .30222222 of the persons in the population who smoked also developed lung cancer.
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