Question

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.

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