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

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

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (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) = (P(A \cap B))/(P(A)) = (0.02)/(0.29) = 0.069`

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