Answer to Question #166115 in Statistics and Probability for Sunera Shakthi

Question

Bad gums may be evidence of a bad heart. Researchers have discovered that 85% of people who have suffered a heart attack had periodontal disease, an inflammation of the gums. Only 29% of healthy people have this disease.

a) Suppose that in a certain community, heart attacks are quite rare, occurring with only 10% probability. If someone has periodontal disease, what is the probability that he or she will have a heart attack?

b) If 40% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease in this community will have a heart attack?

Answer 1

Solution:

a) We will assume that the statistics above part a) apply to this community. Then let us denote H - heart attack, D - periodontal disease. We are given:

$P(D)=0.29, P(D|H)=0.85, P(H) = 0.1$ ,

We need to calculate the probability of Heart Attack, given that person has Periodontal Disease.

Using the law of conditional probability:

$P(A)*P(B|A)=P(A|B)*P(B)$

We get:

$P(H|D)*P(D)=P(H)*P(D|H),$ P(H|D) is what we need to find.

$P(H|D) = \frac{P(H)*P(D|H)}{P(D)} = \frac{0.1*0.85}{0.29} \approx 0.293$ or $29.3\%$

b) Assuming the statistics above the a) are correct for this case. We are given the following:

$P(D) = 0.29, P(D|H)=0.85, P(H)=0.4$ and we basically need to calculate the same probability as in a) but with different given values:

$P(H|D) = \frac{P(H)*P(D|H)}{P(D)}=\frac{0.4*0.85}{0.29}=1.17$

It can mean either that all people who have Periodontal Disease will have a heart attack or that statistics presented above a) is incorrect.

Answer:

a) $0.293$, or $29.3\%$

b) 1 or statistics are incorrect