1. conditional probability help

Hi, I was wondering if anyone could help me out with where to get started on this prblem

Bad gums may mean a bad heart. Researchers discovered that 78% of people who have suffered a heart attack had 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 11% probability.

A. 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 will have a heart attack?

2. Originally Posted by jason0
Hi, I was wondering if anyone could help me out with where to get started on this prblem

Bad gums may mean a bad heart. Researchers discovered that 78% of people who have suffered a heart attack had 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 11% probability.

A. 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 will have a heart attack?
Draw a tree diagram to see what's happening. The first two branches are Heart attack (H) and no Heart attack (H'). Pr(H) = 0.11, Pr(H') = 0.89.

The two branches that go from H and H' are Dental disease (D) and no Dental disease (D'). Pr(D | H) = 0.78 and so Pr(D' | H) = 0.22. Also Pr(D | H') = 0.3 and so Pr(D' | H') = 0.7.

For part A the tree diagram shuold make it clear how to calculate Pr(H | D).

For part B, you first need to edit the tree diagram because now Pr(H) = 0.4.

3. Originally Posted by mr fantastic
Draw a tree diagram to see what's happening. The first two branches are Heart attack (H) and no Heart attack (H'). Pr(H) = 0.11, Pr(H') = 0.89.

The two branches that go from H and H' are Dental disease (D) and no Dental disease (D'). Pr(D | H) = 0.78 and so Pr(D' | H) = 0.22. Also Pr(D | H') = 0.3 and so Pr(D' | H') = 0.7.

For part A the tree diagram shuold make it clear how to calculate Pr(H | D).

For part B, you first need to edit the tree diagram because now Pr(H) = 0.4.
wouldn't that make the answer for A just .11*.78? because that's not right and i'm still confused

4. Originally Posted by jason0
wouldn't that make the answer for A just .11*.78? because that's not right and i'm still confused
Do you understand conditional probability at all? I suggest you thoroughly review it. $\Pr(H | D) = \frac{\Pr(H \cap D)}{\Pr(D)}$.

From the tree diagram $\Pr(D) = (0.11)(0.78) + (0.89)(0.3) = ....$ and $\Pr(H \cap D) = (0.11)(0.78) = ....$.