Question

The following lists the joint probabilities associated with smoking and lung disease among 60-to-65 year-old men. Has Lung Disease/smoker 0.1, No Lung Disease/Smoker 0.17, Lung Disease/Nonsmoker 0.03, No Lung Disease/Nonsmoker 0.7. One 60-to-65 year old man is selected at random. What is the probability of the following event: He has lung disease given that he does not smoke?

Answer 1

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

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ 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 $P(A \cap B) = 0.03$

Lung Disease/Nonsmoker 0.03

No Lung Disease/Nonsmoker 0.7

This means that $P(A) = 0.03 + 0.7 = 0.73$

What is the probability of the following event: He has lung disease given that he does not smoke?

$P(B|A) = \frac{0.03}{0.73} = 0.0411$

4.11% probability that he has lung disease given that he does not smoke