Probability of a false negative

Question

A population includes a fraction m of individuals carrying a disease exists and has the following characteristics:

P(positive test | individual with the disease) = p

P(positive test | individual without the disease) = r

What is the probability of a false negative : P(individual with the disease | negative test) ?

Here is what I have so far:

Let W = with the disease, ~W = without the disease;

Let + = positive test, - = negative test

From the given, we have: P(W) = m, P(~W) = 1-m

We want P(W | -) = (P(- |W) P(W))/(P(-)) = (P(- | W)P(W)/(P(- | W)P(W) + P(- ~W)P(~W))

We have: P (+) = P(+|W)P(W) + P(+|~W)P(~W) = p * P(W) + r * P(~W)

However, I do not know how to proceed from here. Could anyone please help me? Thank you very much!

Answer 1

You have it almost right. You write $P(W|-)=P(-|W)P(W)/P(-)$, and you know everything except $P(-)$: $P(-|W)=1-p$ and $P(W)=m$.

On the other hand, you write the correct expression for $P(+)$, but then you know $P(-)=1-P(+)$. That is, $P(-)=1-pm-r(1-m)$.

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Here is an alternate solution, where I compute directly $P(-)$ instead of $P(+)$.

The probability that the test is negative is

$$P(-)=P(W)P(-|W)+P(\sim W)P(-|\sim W)=m(1-p)+(1-m)(1-r)$$

The probability that the test is negative and the person has the disease is

$$P(-\cap W)=P(W)P(-|W)=m(1-p)$$

Therefore, the probability that a person has the disease, given that the test is negative, is

$$P(W|-)=\frac{P(-\cap W)}{P(-)}=\frac{m(1-p)}{m(1-p)+(1-m)(1-r)}$$