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% are admitted for 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. question number 2. p(e)=0.37, p(f)=0.53, and p(e∪f)=0.72

Explanation:

Step1: Identify given percentages

The percentage of patients admitted for heart - disease is 16%, for cancer is 26%, and 4% for both coronary and oncology care. Heart - disease is considered a coronary care issue.

Step2: Use the principle of inclusion - exclusion for probabilities

The probability of being admitted for coronary or oncology care $P(C\cup O)$ is given by $P(C)+P(O)-P(C\cap O)$. Let $P(C)$ be the probability of coronary care (including heart - disease) and $P(O)$ be the probability of oncology care. We want to find the probability of being admitted for something other than coronary care. First, find $P(C\cup O)=0.16 + 0.26-0.04=0.38$.

Step3: Calculate the complementary probability

The probability of being admitted for something other than coronary care is $1 - P(C\cup O)$. So, $1 - 0.38 = 0.62$. But this seems to be wrong approach. Let's consider another way.
The probability of being admitted for coronary care (heart - disease which is a type of coronary care) is 16% and for oncology is 26% with 4% overlap. The probability of being admitted for coronary care is 16%. The probability of being admitted for non - coronary care is $1 - 0.16=0.84$.

Answer:

0.84