At a certain hospital, 24 patients have lung cancer, 15 patients smoke, and 12 have lung cancer and smoke. Suppose the hospital contains 120 patients. If a patient chosen at random is known to smoke, what is the probability that the patient has lung cancer? If a patient chosen at random is known to smoke, the probability that the patient has lung cancer is (Type an integer or a simplified fraction.)

To find the probability that a patient who smokes has lung cancer, we can use conditional probability. Let: - \( A \) = event that a patient has lung cancer - \( B \) = event that a patient smokes We need to find \( P(A | B) \), which is given by the formula: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] From the information provided: - The number of patients with lung cancer (\( n(A) \)) = 24 - The number of patients who smoke (\( n(B) \)) = 15 - The number of patients who both smoke and have lung cancer (\( n(A \cap B) \)) = 12 First, we determine \( P(B) \): Since there are 120 patients total, the probability that a patient smokes is: \[ P(B) = \frac{n(B)}{120} = \frac{15}{120} = \frac{1}{8} \] Next, we find \( P(A \cap B) \): \[ P(A \cap B) = \frac{n(A \cap B)}{120} = \frac{12}{120} = \frac{1}{10} \] Now, substituting our values into the conditional probability formula: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{10}}{\frac{1}{8}} = \frac{1}{10} \times \frac{8}{1} = \frac{8}{10} = \frac{4}{5} \] Thus, the probability that a patient chosen at random, who is known to smoke, has lung cancer is: \[ \frac{4}{5} \]