Question

Bad gums may mean a bad heart. Researchers discovered that 81% of people who have suffered a heart attack has periodontal disease, an inflammation 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 15% probability.

If someone has periodontal disease, what is the probability that he or she will have a heart attack?

Answer 1

Using Bayes' theorem, the probability that someone with periodontal disease will have a heart attack is approximately 40.5%.

Step-by-step explanation:

To calculate the probability that someone with periodontal disease will have a heart attack, we use Bayes' theorem. Given:

The probability of having a heart attack (A) is 15% or 0.15.

The probability of having periodontal disease (P) given that someone has had a heart attack is 81% or 0.81.

The probability of having periodontal disease in the healthy population is 30% or 0.3.

Bayes' theorem states:

P(A|P) = (P(P|A) * P(A)) / P(P)

Substituting in the given values:

P(A|P) = (0.81 * 0.15) / 0.3

P(A|P) = 0.1215 / 0.3

P(A|P) ≈ 0.405

So, the probability that someone with periodontal disease will have a heart attack is approximately 40.5%.

Answer 2

0.405 or 40.5%

Step-by-step explanation:

Let event A=having a periodontal disease

event B=having a heart attack

we are given

P(A)=P(having a periodontal disease)=0.30

P(B)=P(having a heart attack)=0.15

P(A/B)=P(have a periodontal disease/have a heart attack)=0.81

P(B/A)=P(have a heart attack/have a periodontal disease)=?

P(A/B)=P(A∩B)/P(B)

P(A∩B)=P(B)*P(A/B)=0.15*0.81=0.1215

P(B/A)=P(A∩B)/P(A)=0.1215/0.3=0.405

There is 40.5% probability that if someone has periodontal disease will have a heart attack.