Answer:
4.11% probability that he has lung disease given that he does not smoke
Step-by-step explanation:
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Does not smoke
Event B: Lung disease
Lung Disease/Nonsmoker 0.03
This means that [tex]P(A \cap B) = 0.03[/tex]
Lung Disease/Nonsmoker 0.03
No Lung Disease/Nonsmoker 0.7
This means that [tex]P(A) = 0.03 + 0.7 = 0.73[/tex]
What is the probability of the following event: He has lung disease given that he does not smoke?
[tex]P(B|A) = \frac{0.03}{0.73} = 0.0411[/tex]
4.11% probability that he has lung disease given that he does not smoke
Probabilities are used to determine the chances of an event.
The probability that he has lung disease given that he does not smoke is 0.231
The required probability is calculated as:
[tex]\mathbf{P = \frac{P(Lung\ Disease\ and\ Non\ Smoker)}{P(Lung\ Disease)}}[/tex]
From the question, we have:
[tex]\mathbf{P(Lung\ Disease\ and\ Non\ Smoker) = 0.03}[/tex]
[tex]\mathbf{P(Lung\ Disease) = P(Has Lung Disease/smoker) + P(Lung Disease/Nonsmoker)}[/tex]
[tex]\mathbf{P(Lung\ Disease) = 0.1 + 0.03}[/tex]
[tex]\mathbf{P(Lung\ Disease) = 0.13}[/tex]
So, we have:
[tex]\mathbf{P = \frac{P(Lung\ Disease\ and\ Non\ Smoker)}{P(Lung\ Disease)}}[/tex]
[tex]\mathbf{P = \frac{0.03}{0.13}}[/tex]
[tex]\mathbf{P = 0.231}[/tex]
Hence, the probability that he has lung disease given that he does not smoke is 0.231
Read more about probabilities at:
https://brainly.com/question/11234923
