Conditional probability question

[jason0]

can anyone help me out with where to get started on this problem?


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?

 

[tkhunny]

1) All those cancer patients will be delighted to know they are "healthy", just because they don't have either periodonal disease or bad gums. What an odd presumption.

2) Did you build a chart?

a) Make Row 1 "Hear Attack"
b) MAke Row 2 "Healthy"
c) Make Column 1 "Periodontal"
d) Make Column 2 "Clean Gums"

Now start filling in the boxes. Let's see what you get.

 

[soroban]

Hello, jason0!

\(\displaystyle \text{We are expected to know Bayes' Theorem: }\(A|B) \:=\:\frac{P(A \cap B)}{P(B)}\)

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. . . . . not sure what this means
Suppose that in a certain community heart attacks occur with 11% probability.

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

\(\displaystyle \text{Let: }\:\begin{array}{ccc} G &=& \text{has gum disease} \\ H &=& \text{has a heart attack} \end{array}\)

\(\displaystyle \text{We are given: }\:\begin{array}{cccc} P(H) &=& 0.11 \\ P(G) &=& 0.30 &\leftarrow \text{ I assume this is true.}\\ P(G|H) &=& 0.78 \end{array}\)

\(\displaystyle \text{Since: }\(G|H} \:=\:\frac{P(G \cap H)}{P(H)} \quad\Rightarrow\quad \frac{P(G \cap H)}{0.11} \:=\:0.78 \quad\Rightarrow\quad P(G \cap H) \:=\:0.0858\)


\(\displaystyle \text{We want: }\(H|G) \:=\:\frac{P(H \cap G)}{P(H)} \;=\;\frac{0.0858}{0.30} \;=\;0.286 \;=\;28.6\%\)

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?

\(\displaystyle \text{We are given: }\(H) \:=\:0.40\)

\(\displaystyle \text{If the other data are the same, there is something wrong with the problem.}\)


\(\displaystyle \text{We have: }\(G|H) \:=\:\frac{P(G \cap H)}{P(H)} \quad\Rightarrow\quad \frac{P(G \cap H)}{0.40} \:=\:0.78 \quad\Rightarrow\quad P(G \cap H) \:=\:0.312\)

\(\displaystyle \text{Then: }\;P(H|G) \:=\:\frac{P(H \cap G)}{P(G)} \:=\: \frac{0.312}{0.3} \;=\;1.04 \;=\;104\%\;\;??\)