quiz 2 (id 657919) due: sat sep 13 23:59 2025. until due: 1.4 days. restart = 0, attempt 2. question number…

quiz 2 (id 657919) due: sat sep 13 23:59 2025. until due: 1.4 days. restart = 0, attempt 2. question number 1. hospital records show that 16% of all patients are admitted for heart disease, 26% are admitted for cancer (oncology) treatment, and 4% 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.) 0.70 0.74 0.80 0.84 0.96 none of the above.

quiz 2 (id 657919) due: sat sep 13 23:59 2025. until due: 1.4 days. restart = 0, attempt 2. question number 1. hospital records show that 16% of all patients are admitted for heart disease, 26% are admitted for cancer (oncology) treatment, and 4% 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.) 0.70 0.74 0.80 0.84 0.96 none of the above.

Answer

Explanation:

Step1: Identify relevant probabilities

Let (P(H)) be the probability of being admitted for heart - disease, (P(C)) be the probability of being admitted for cancer. Given (P(H)=0.16), (P(C) = 0.26), and (P(H\cap C)=0.04).

Step2: Use the formula for (P(H\cup C))

The formula for the probability of the union of two events is (P(H\cup C)=P(H)+P(C)-P(H\cap C)). Substitute the values: (P(H\cup C)=0.16 + 0.26- 0.04). [P(H\cup C)=0.38]

Step3: Find the probability of non - coronary care

The probability of being admitted for something other than coronary care is (1 - P(H\cup C)). So (1-0.38 = 0.62). But this is wrong. Let's consider coronary care as heart - disease. The probability of being admitted for heart - disease (P(H)=0.16). The probability of being admitted for something other than coronary care is (1 - 0.16=0.84).

Answer:

0.84