Questions: Question 10
1 pts
(Section 4.2) Hospital records indicate that 28% of patients have high blood pressure, 26% of patients have diabetes, and 13% have both. What is the probability that a randomly selected patient has high blood pressure or diabetes?
13%
54%
41%
59%
Transcript text: Question 10
1 pts
(Section 4.2) Hospital records indicate that 28\% of patients have high blood pressure, $26 \%$ of patients have diabetes, and $13 \%$ have both. What is the probability that a randomly selected patient has high blood pressure or diabetes?
$13 \%$
$54 \%$
$41 \%$
$59 \%$
Solution
Solution Steps
To find the probability that a randomly selected patient has high blood pressure or diabetes, we can use the principle of inclusion-exclusion for probabilities. The formula is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Where:
\( P(A) \) is the probability of having high blood pressure.
\( P(B) \) is the probability of having diabetes.
\( P(A \cap B) \) is the probability of having both high blood pressure and diabetes.
Given:
\( P(A) = 0.28 \)
\( P(B) = 0.26 \)
\( P(A \cap B) = 0.13 \)
We can plug these values into the formula to find the probability of having either high blood pressure or diabetes.
Step 1: Given Probabilities
We are given the following probabilities:
\( P(A) = 0.28 \) (probability of having high blood pressure)
\( P(B) = 0.26 \) (probability of having diabetes)
\( P(A \cap B) = 0.13 \) (probability of having both high blood pressure and diabetes)
Step 2: Apply Inclusion-Exclusion Principle
To find the probability that a randomly selected patient has high blood pressure or diabetes, we use the inclusion-exclusion principle: