Question

Bad gums may mean a bad heart. Researchers discovered that 80% of people who have su ered a heart attack had periodontal disease, an inammation of the gums. Only 30% of healthy people have this disease. Suppose that in a certain community heart attacks are quite rare, occurring with only 10% probability. If someone does not have periodontal disease, what is the probability that he or she will have a heart attack?

Answer 1

Answer:

0.069 = 6.9% probability that he or she will have a heart attack

Step-by-step explanation:

Conditional Probability

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 have periodontal disease

Event B: Has a heart attack.

Probability of not having a periodontal disease:

100 - 80 = 20% of 10%(had a heart attack).

30% of 100-10 = 90%(did not have a heart attack). So

$P(A) = 0.2*0.1 + 0.3*0.9 = 0.29$

Intersection of A and B:

Not having the disease, suffering a heart attack, so 20% of 10%.

$P(A cap B) = 0.2*0.1 = 0.02$

What is the probability that he or she will have a heart attack?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.29} = 0.069$

0.069 = 6.9% probability that he or she will have a heart attack