If 58% of the people at a certain conference are doctors, 38% are men, and 18% are male doctors, what is the probability that a person selected at random at this conference is a doctor or man (or both)? Write your answer as a decimal (not as a percentage). 0 ?
If 58% of the people at a certain conference are doctors, 38% are men, and 18% are male doctors, what is the probability that a person selected at random at this conference is a doctor or man (or both)? Write your answer as a decimal (not as a percentage). 0 ?
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![**Problem Statement:**
If 58% of the people at a certain conference are doctors, 38% are men, and 18% are male doctors, what is the probability that a person selected at random at this conference is a doctor or man (or both)?
Write your answer as a decimal (not as a percentage).
-----
**Explanation:**
To solve this problem, we use the principle of inclusion and exclusion for probabilities.
Let \( P(D) \) = probability of selecting a doctor = 0.58\
Let \( P(M) \) = probability of selecting a man = 0.38\
Let \( P(D \cap M) \) = probability of selecting a male doctor = 0.18
We need to find the probability of selecting either a doctor or a man (or both), represented as \( P(D \cup M) \).
The formula for the probability of the union of two events is:
\[ P(D \cup M) = P(D) + P(M) - P(D \cap M) \]
Substitute the values:
\[ P(D \cup M) = 0.58 + 0.38 - 0.18 \]
\[ P(D \cup M) = 0.78 \]
Therefore, the probability that a person selected at random at this conference is a doctor, or a man, or both is **0.78**.
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Transcribed Image Text:**Problem Statement:**
If 58% of the people at a certain conference are doctors, 38% are men, and 18% are male doctors, what is the probability that a person selected at random at this conference is a doctor or man (or both)?
Write your answer as a decimal (not as a percentage).
-----
**Explanation:**
To solve this problem, we use the principle of inclusion and exclusion for probabilities.
Let \( P(D) \) = probability of selecting a doctor = 0.58\
Let \( P(M) \) = probability of selecting a man = 0.38\
Let \( P(D \cap M) \) = probability of selecting a male doctor = 0.18
We need to find the probability of selecting either a doctor or a man (or both), represented as \( P(D \cup M) \).
The formula for the probability of the union of two events is:
\[ P(D \cup M) = P(D) + P(M) - P(D \cap M) \]
Substitute the values:
\[ P(D \cup M) = 0.58 + 0.38 - 0.18 \]
\[ P(D \cup M) = 0.78 \]
Therefore, the probability that a person selected at random at this conference is a doctor, or a man, or both is **0.78**.
------
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This box allows you to enter your calculated answer.
**Answer Verification Buttons:**
- **✓:** Verifies your answer.
- **↻:** Resets the question.
- **❓:** Provides a hint or additional information about the problem.
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