0
$\begingroup$

I know I'm going wrong somewhere just not sure where

I haven't used Bayes' theorem much before so any help would be greatly appreciated.

Suppose you are given following data.

$5\%$ of the population have heart disease.

If you have heart disease, the probability that you have high blood pressure is $90\%$

If you do not have heart disease, the probability that you have high blood pressure is $15\%$

What is the probability that a person chosen at random from the population has high blood pressure?

$$P(B) = P(B|H)P(H) + P(B|H')P(H')$$

$$P(B) = (.9)(.05) + (.15)(.95) = .1875$$

Using Bayes Theorem calculate the probability that the person has heart disease, if they have high blood pressure.

$$P(H|B) = \frac{P(B|H)P(H)}{P(B)}$$

$$(.9)(.05)/.1875 = .24$$

Using Bayes Theorem calculate the probability that the person has heart disease, if they do not have high blood pressure.

$$P(H|B') = \frac{P(B'|H)P(B')}{P(H)}$$

When I sub in for this part I'm getting an invalid answer

$\endgroup$
2
  • 1
    $\begingroup$ There is no reason to use boldfaced characters. $\endgroup$ May 12, 2018 at 13:21
  • $\begingroup$ Welcome to stackexchange. Here's a way to do the problem that may provide more intuition than Bayes Theorem (it is that theorem, in disguise - you can use it to check your work) math.stackexchange.com/questions/2279851/… $\endgroup$ May 12, 2018 at 13:22

3 Answers 3

1
$\begingroup$

The answer is $\frac{3}{16}.$

You divide and conquer, so the probability that someone has heart disease and high blood pressure is $(\frac{1}{20}) (\frac{9}{10})$, which is $\frac{9}{200}$. The probability that someone doesn't have heart disease and has high blood pressure is $(\frac{1}{20}) \frac{3}{20}$, which is $\frac{57}{400}$. $\frac{9}{200}$ is the same as $\frac{18}{400}$. If you add these two fractions, you get $\frac{75}{400}$, or $\frac{3}{16}$.

$\endgroup$
0
$\begingroup$

$$P(H|B) = {P(H \land B) \over P(B)}={P(B|H)P(H) \over P(B)}$$ for the second is fine.

So analogously for the third $$P(H|B') = {P(H \land B')\over P(B')} = {P(B'|H)P(H) \over P(B')}$$

and use that $P(B') = 1-P(B)$ and $P(B'|H) = 1- P(B|H)$. You know all the values already.

$\endgroup$
0
$\begingroup$

The reason why you got an incorrect answer for the 3rd sub-part is because of the incorrect implementation of the Bayes rule.

Your RHS gives $P(B'|H)$ which can be calculated as $1-P(B|H)=1-0.9=0.1$

Here $P(H|B')=\frac{P(H)P(B'|H)}{P(B')} = \frac{P(H).[1-P(B|H)]}{1-P(B)} = \frac{0.05*0.1}{0.8125} = \frac{50}{8125}=0.0061$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .