conditional probability - 2 different tests done

Question

Question:

A medical research team wanted to evaluate a screening method proposed for 
Alzheimer's disease. The method was evaluated on 450 patients randomly selected 
patients with Alzheimers (in 436 cases, the screening was positive). Moreover,
the method was also done on 500 patients without the disease and returned 
positive for 5 out of the 500. 7.7% of all Canadians over the age of 65 have
Alzhiemers. Find the probability that a Canadian over the age of 65 has Alzhiemers
given the test was positive.

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It seems that the test returns a 96.8% chance of success given that the patient has alzhiemers and a .01% chance of success given the patient DOESN'T have alzhiemers.

So I used bayes theorem for this one, so that I would find

  A = Alzhiemers
  Positive = Positive test result given the patient actually had alzhiemers 
  Positive2 = Positive test result given the patient doesn't have alzhiemers

$$P(A | Positive) = \frac{P(Positive|A)(A)}{P(Positive|A)(A) + P(Positive2|A)(A)}$$

$$= \frac{(.968)(.077)}{(.968)(.077) + (.01)(.077)}$$

$$= approx. .98$$

Did i do this correctly? As in was i supposed to even use bayes theorem for this problem?

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ahh seems like i multiplied the probability of positive test result given the person doesn't have alzhiemers with the probability of all canadians who do have alzheimers. Thanks guys.

Answer 1

Let $A$ denote the event that the person has Alzheimer's and $A^c$ denote the event that the person does not have Alzheimer's.

Let $P$ denote the event that the person tested positive for the disease and $P^c$ denote that the person tested negative for the disease.

According to the problem statement, our experimental results from having tested 450 people who are sick with Alzheimer's are that $\mathscr{P}(P|A) \approxeq \frac{436}{450} \approxeq .969$, $\mathscr{P}(P^c|A) \approxeq \frac{14}{450} \approxeq .031$

From having tested 500 people who are not sick with Alzheimer's, we have our experimental results as $\mathscr{P}(P^c|A^c)\approxeq \frac{495}{500} = .99$ and $\mathscr{P}(P|A^c)\approxeq \frac{5}{500} = .01~~~$ (note, $.01 = 1\%\neq .01\% = .0001$. be careful about putting % signs where they don't belong)

So, the question asks us, using our experimental results and the fact that $\mathscr{P}(A) = .077$ (and assuming that the test's accuracy has nothing to do with age of the patient), what is the chance that someone who tested positive infact has Alzheimer's, I.e: $\mathscr{P}(A|P)$

By Baye's Theorem, $\mathscr{P}(A|P) = \dfrac{\mathscr{P}(P|A)\mathscr{P}(A)}{\mathscr{P}(P)}$

The top of the fraction is given by interpretation of the problem statement. We do have to think a bit longer as to how to calculate $\mathscr{P}(P)$ however. Note that $$\mathscr{P}(P) = \mathscr{P}(P\cap(A\cup A^c)) = \mathscr{P}(P\cap A) + \mathscr{P}(P\cap A^c)$$

$$=\mathscr{P}(A)\cdot\mathscr{P}(P|A) + \mathscr{P}(A^c)\cdot\mathscr{P}(P|A^c)$$

So, we have, $\mathscr{P}(P) = .077\cdot.969 + .923\cdot .01 = .083$

Plugging back into the fraction from before, we get $\mathscr{P}(A|P) = \dfrac{\mathscr{P}(P|A)\mathscr{P}(A)}{\mathscr{P}(P)} = \dfrac{.969\cdot .077}{.969\cdot .077 + .923\cdot .01} = \dfrac{.074}{.083}\approxeq .89$

Answer 2

Where you have $$\frac{P(Positive|A)(A)}{P(Positive|A)(A) + P(Positive2|A)(A)}$$ you should probably have $$\frac{P(Positive|A)P(A)}{P(Positive|A)P(A) + P(Positive2|\text{not }A)P(\text{not }A)}$$ which would give a rather smaller result. Yes you should use Bayes' theorem, and it is this smaller result which is apparently particularly counter-intuitive to medical students (and law students).