According to the CDC Centers for Disease Control and Prevent

According to the CDC (Centers for Disease Control and Prevention), men who smoke are 23 times more likely to develop lung cancer than men who don\'t smoke. Also according to the CDC, 21.6% of men in the U.S. smoke. What is the probability that a man in the U.S. is a smoker, given that he develops lung cancer?

Solution

let \'x\' be the probebility that a person who don\'t smoke get the lung cancer

probebility that a person who smokes get lung cancer = 23x

probebility that a randomly selected man in US get lung cancer = (0.784*x) + (0.216 * 23x) = 5.752x

probebility that a randomly selected man in US will smoke and got the lung cancer = 0.216*23x = 4.968x

the probability that a man in the U.S. is a smoker, given that he develops lung cancer =

(probebility that a randomly selected man in US will smoke and got the lung cancer / probebility that a randomly selected man in US get lung cancer) = 0.216*23x /{ (0.784*x) + (0.216 * 23x)} = 4.968x / 5.752x = 0.8636