Kathy Guerra

2022-09-08

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)?
My try
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)

Carson Mueller

You have it almost right. You write $P\left(W|-\right)=P\left(-|W\right)P\left(W\right)/P\left(-\right)$, and you know everything except $P\left(-|W\right)=1-p$ and $P\left(W\right)=m$.
On the other hand, you write the correct expression for $P\left(+\right)$, but then you know $P\left(-\right)=1-P\left(+\right)$. That is, $P\left(-\right)=1-pm-r\left(1-m\right)$.

elisegayezm

Alternate solution:
Compute directly $P\left(-\right)$ instead of $P\left(+\right)$.
The probability that the test is negative is
$P\left(-\right)=P\left(W\right)P\left(-|W\right)+P\left(\sim W\right)P\left(-|\sim W\right)=m\left(1-p\right)+\left(1-m\right)\left(1-r\right)$
The probability that the test is negative and the person has the disease is
$P\left(-\cap W\right)=P\left(W\right)P\left(-|W\right)=m\left(1-p\right)$
Therefore, the probability that a person has the disease, given that the test is negative, is
$P\left(W|-\right)=\frac{P\left(-\cap W\right)}{P\left(-\right)}=\frac{m\left(1-p\right)}{m\left(1-p\right)+\left(1-m\right)\left(1-r\right)}$

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