Suppose that 7 out of the 19 doctors in a small hospital are General Practitioners, 6 out of the 19 are under the age of 40, and 5 are both General Practitioners and under the age of 40. What is the probability that you are randomly assigned a General Practitioner or a doctor under the age of 40? Enter a fraction or round your answer to 4 decimal places, if necessary. Answer Tables Keypad Keyboard Shortcuts
Suppose that 7 out of the 19 doctors in a small hospital are General Practitioners, 6 out of the 19 are under the age of 40, and 5 are both General Practitioners and under the age of 40. What is the probability that you are randomly assigned a General Practitioner or a doctor under the age of 40? Enter a fraction or round your answer to 4 decimal places, if necessary. Answer Tables Keypad Keyboard Shortcuts
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Probability Problem Example**
Suppose that 7 out of the 19 doctors in a small hospital are General Practitioners, 6 out of the 19 are under the age of 40, and 5 are both General Practitioners and under the age of 40. What is the probability that you are randomly assigned a General Practitioner or a doctor under the age of 40? Enter a fraction or round your answer to 4 decimal places, if necessary.
---
**Problem Explanation:**
To find the probability of being assigned a General Practitioner or a doctor under the age of 40, use the principle of inclusion-exclusion.
- Let \( A \) be the event that the doctor is a General Practitioner.
- Let \( B \) be the event that the doctor is under the age of 40.
Given:
- \( P(A) = \frac{7}{19} \)
- \( P(B) = \frac{6}{19} \)
- \( P(A \cap B) = \frac{5}{19} \) (doctors who are both General Practitioners and under 40)
The probability that a doctor is either a General Practitioner or under the age of 40 is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{7}{19} + \frac{6}{19} - \frac{5}{19} = \frac{8}{19} \]
Therefore, the probability of being assigned a General Practitioner or a doctor under the age of 40 is \(\frac{8}{19}\).
Note: There were no graphs or diagrams present in this problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3ff7712-c3af-4451-90e7-4db287f048cf%2F7d82db8a-f077-4d45-a527-f24b9b372c74%2F5n1cr7c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Probability Problem Example**
Suppose that 7 out of the 19 doctors in a small hospital are General Practitioners, 6 out of the 19 are under the age of 40, and 5 are both General Practitioners and under the age of 40. What is the probability that you are randomly assigned a General Practitioner or a doctor under the age of 40? Enter a fraction or round your answer to 4 decimal places, if necessary.
---
**Problem Explanation:**
To find the probability of being assigned a General Practitioner or a doctor under the age of 40, use the principle of inclusion-exclusion.
- Let \( A \) be the event that the doctor is a General Practitioner.
- Let \( B \) be the event that the doctor is under the age of 40.
Given:
- \( P(A) = \frac{7}{19} \)
- \( P(B) = \frac{6}{19} \)
- \( P(A \cap B) = \frac{5}{19} \) (doctors who are both General Practitioners and under 40)
The probability that a doctor is either a General Practitioner or under the age of 40 is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{7}{19} + \frac{6}{19} - \frac{5}{19} = \frac{8}{19} \]
Therefore, the probability of being assigned a General Practitioner or a doctor under the age of 40 is \(\frac{8}{19}\).
Note: There were no graphs or diagrams present in this problem.
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