# Use Bayes' Theorem to prove that the rate of false positives is accurate (86%) in the following pass

Use Bayes' Theorem to prove that the rate of false positives is accurate (86%) in the following passage.
What I have done so far is list the following
$P\left(A\right)=0.99$
$P\left(B\right)=0.05$
As Bayes' Theorem is as follows
$P\left(A|B\right)=\frac{P\left(B|A\right)P\left(A\right)}{P\left(B\right)}$
I need $P\left(B|A\right)$. I'm stuck on how to find this. Help would be greatly appreciated.
Is it possible to prove this using Bayes'? If not, is there some other sort of mathematics that I can use here?
Roughly 1 per cent of the population suffer from mild cognitive impairment, which might, but doesn’t always, lead to dementia. Suppose that the test is quite a good one, in the sense that 95 per cent of the time it gives the right (negative) answer for people who are free of the condition. That means that 5 per cent of the people who don’t have cognitive impairment will test, falsely, as positive. That doesn’t sound bad. It’s directly analogous to tests of significance which will give 5 per cent of false positives when there is no real effect, if we use a p-value of less than 5 per cent to mean ‘statistically significant’.
But in fact the screening test is not good – it’s actually appallingly bad, because 86 per cent, not 5 per cent, of all positive tests are false positives. So only 14 per cent of positive tests are correct. This happens because most people don’t have the condition, and so the false positives from these people (5 per cent of 99 per cent of the people), outweigh the number of true positives that arise from the much smaller number of people who have the condition (80 per cent of 1 per cent of the people, if we assume 80 per cent of people with the disease are detected successfully).
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assumintdz
Let $A$ be the event that a person randomly chosen doesn't suffer from cognitive impairment and $B$ be the result of the test for dementia which can either be positive $P$ or negative $N$. Then,
$P\left(A\right)=0.99=1-P\left({A}^{C}\right)$
$P\left(B=N|A\right)=0.95$
Also assume that $95\mathrm{%}$ is the accuracy of the test in the sense that if the patient has the disease then $95\mathrm{%}$ of the time the test is positive and if the patient does not suffer from the disease then also $95\mathrm{%}$ of the time the test is negative. So,
$P\left(B=P|{A}^{C}\right)=0.95$ and $P\left(B=P|A\right)=0.05$
So, the probability of false positive is,
$P\left(A|B=P\right)=\frac{P\left(B=P|A\right)P\left(A\right)}{P\left(B=P|A\right)P\left(A\right)+P\left(B=P|{A}^{C}\right)P\left({A}^{C}\right)}=\frac{0.05×0.99}{0.05×0.99+0.95×0.01}=0.838$