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%

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? $13 \%$ $54 \%$ $41 \%$ $59 \%$
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Solution

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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:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the given values:

\[ P(A \cup B) = 0.28 + 0.26 - 0.13 \]

Step 3: Calculate the Probability

Calculating the above expression:

\[ P(A \cup B) = 0.28 + 0.26 - 0.13 = 0.41 \]

Step 4: Convert to Percentage

To express this probability as a percentage:

\[ P(A \cup B) \times 100 = 0.41 \times 100 = 41.0\% \]

Final Answer

The probability that a randomly selected patient has high blood pressure or diabetes is \\(\boxed{41.0\%}\\).

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