Question

popper 2 1) hospital records show that 14% of all patients are admitted for heart disease, 16% are admitted for cancer (oncology) treatment, and 8% receive both coronary and oncology care. what is the probability that a randomly selected patient is admitted for something other than coronary care? (note that heart disease is a coronary care issue.) a) 0.84 b) .92 c) .86 d) .78 e) .76

Explanation:

Step1: Define Events

Let \( A \) be the event of being admitted for heart disease (coronary care), \( B \) be the event of being admitted for cancer (oncology). We know \( P(A) = 0.14 \), \( P(B) = 0.16 \), \( P(A \cap B) = 0.08 \). We need to find the probability of not being in \( A \), i.e., \( P(\text{not } A) \), but wait, no—wait, the problem is "admitted for something other than coronary care", so we can also use the principle of inclusion - exclusion for the union, then find the complement of \( A \) or use the formula for the probability of the complement. Wait, actually, the total probability of being in \( A \) or \( B \) or both is \( P(A \cup B)=P(A)+P(B)-P(A \cap B) \). Then the probability of being in neither (but wait, no—the problem is "admitted for something other than coronary care", so it's \( 1 - P(A) \)? Wait, no, wait the problem says "admitted for something other than coronary care", so it's the probability that the patient is not in \( A \). Wait, but let's check the numbers. Wait, maybe I misread. Wait, the problem: "14% for heart (coronary), 16% for cancer (oncology), 8% both. What is the probability a randomly selected patient is admitted for something other than coronary care?" So "other than coronary care" means not \( A \). So \( P(\text{not } A)=1 - P(A) \)? But \( P(A) = 0.14 \), so \( 1 - 0.14 = 0.86 \)? Wait, but wait, maybe the problem is about being in \( B \) or other, but no—"other than coronary care" is all patients not in \( A \). Wait, let's confirm. The formula for the complement: \( P(\text{not } A)=1 - P(A) \). But wait, is there a mistake here? Wait, no—because "other than coronary care" means the patient is not admitted for coronary care, regardless of other admissions. So \( P(\text{not } A)=1 - P(A)=1 - 0.14 = 0.86 \)? Wait, but let's check with inclusion - exclusion. The total probability of \( A \cup B \) is \( 0.14 + 0.16 - 0.08 = 0.22 \). Then the probability of being in neither \( A \) nor \( B \) is \( 1 - 0.22 = 0.78 \), but that's not what the problem is asking. The problem is "admitted for something other than coronary care", which is "not \( A \)", so it's \( 1 - P(A)=1 - 0.14 = 0.86 \). Wait, that matches option c. Let's re - express:

Let \( A \) be the event of coronary care admission. We need \( P(\overline{A}) \). By the complement rule, \( P(\overline{A}) = 1 - P(A) \).

Step2: Substitute Values

We know \( P(A)=0.14 \). So \( P(\overline{A}) = 1 - 0.14 = 0.86 \).

Answer:

c. 0.86