\(P(\text{Lung Cancer} \mid \text{Smoker}) = \dfrac{P(\text{Lung Cancer} \cap \text{Smoker})}{P(\text{Smoker})} = \dfrac{54/500}{180/500} = \dfrac{54}{180} = 0.3\)
So a randomly selected smoker has a 30% chance of having lung cancer.
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In a survey of 500 people, 180 are smokers, 90 have lung cancer, and 54 are both smokers and diagnosed with lung cancer. Determine the probability…
In a survey of 500 people, 180 are smokers, 90 have lung cancer, and 54 are both smokers and diagnosed with lung cancer. Determine the probability that a randomly selected smoker has lung cancer.
Determine the probability, \(P(\text{Lung Cancer} \mid \text{Smoker})\).
From the BUET MSC (CSE) October 2026 – Written exam, question 10 of 14.
Answers (1)
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