
Sam S. answered 10/11/16
Tutor
5.0
(212)
5/5 AP Calc AB, engineering calc (BS-level), higher calc (MS-level)
By definition of conditional probability,
P (smoke|ectopic) = P (ectopic|smoke) x P (smoke) / P (ectopic)
By law of total probability,
P (ectopic) = P (ectopic|smoke) × P (smoke) + P (ectopic|not smoke) × P (not smoke)
Suppose y = P (ectopic|not smoke). Then from the problem, we know P(ectopic|smoke) = 2 × P (ectopic|not smoke) = 2y. Substitute this into the above two equations:
P(ectopic) = 2y × 0.15 + y × 0.85 = 1.15y
P (smoke|ectopic) = 2y × 0.15 / 1.15y = 6/23