Compute probability from conditional probability

Question

Define $Q$ as the conditional probability given a positive test result, i.e. $Q(A) := P(A|+)$ for event $A$. Compute $Q(S)$ and $Q(S^c)$ where $Q(S)$ is the probability of being sick and $Q(S^c)$ is the probability of being healthy.

The test's sensitivity is 0,9 and the specivity is 0,3. This means that of a population of sick persons there is a 0,9 probability of getting a positive result. In a population of healthy persons there is a 0,3 probability of getting a positive result.

To my disposal I have the following definitions and theorems:

Definition of probability measure

The function $P$ is defined on subsets of the sample space $\Omega$ that satisfies Kolmogorovs axioms is said to be a probability measure

Theorem 1

Let $P$ be a probability measure on $\Omega$ and $A_{1}$ be a given event with $P(A_{1}) > 0$. Then $Q$ is defined for every event $A$ as:

$$Q(A) := P(A|A_{1}) = \frac{P(A\cap A_{1})}{P(A_{1})}$$

Theorem 2

Let $P(A_{1}) > 0$ and let $A$ and $B$ be random events. Then:

$$P(A^c|A_{1}) = 1-P(A|A_{1})$$ $$P(A \cup B|A_{1}) = P(A|A_{1}) + P(B|A_{1})-P(A \cap B|A_{1})$$

Naive attempt

I used the sensitivity and speciality to compute the probability:

$$ Q(S) = \frac{0.9}{0.9+0.3} = 0.75 $$ $$ Q(S_c) = 1 - Q(S) = 0.25 $$

I found an example that then used the $Q(S)$ probability above in the following way to get to the result:

$$ Q(S) = \frac{0.75}{0.75+1} = 0.429$$

I have a hard time understanding what is going on. I have a hunch that Bayes theorem is being used to arrive at the solution. Any hint's in the right direction is greatly appreciated.

Answer 1

To simplify notation, in decimal fractions,

Let $i =$ Incidence,
$v =$ sensitiVity,
$f =$ speciFicity,
$D,D^C =$ Diseased, not diseased
then look at the following chart:

$\begin{array}{|c|c|c|}\hline Status & + & -\\ \hline D & iv & i(1-v)\\ \hline D^C & (1-i)(1-f) & (1-i)f \\ \hline \end{array}$

Then you can find out various things, eg

$D|+ = \dfrac{iv}{iv + (1-i)(1-f)}$

$D^C|- = \dfrac{(1-i)f}{i(1-v) + (1-i)f}$

Note that you can't find these out without knowing $i$

Answer 2

The test's sensitivity is 0,9 and the specivity is 0,3.

This means that of a population of sick persons there is a 0,9 probability of getting a positive result.

$\color{green}\checkmark$Sensitivity is the true positive rate: the probability for a positive result given truly having the trait. $$\mathsf P(+\mid S)=0.9$$

In a population of healthy persons there is a 0,3 probability of getting a positive test result.

$\color{red}\sim$Specificity is the true negative rate: the probability for a negative test result given truly not having the trait. $$\mathsf P(-\mid S^\complement)=0.3$$

But without knowing what the populations of sick and non-sick are (the incident rate), all we can say is:

$$\begin{align}Q(S)&=\mathsf P(S\mid +) \\&= \dfrac{\mathsf P(+\mid S)\mathsf P(S)}{\mathsf P(+\mid S)\mathsf P(S)+(1-\mathsf P(-\mid S^\complement))(1-\mathsf P(S))}\\&=\dfrac{0.9~\mathsf P(S)}{0.9~\mathsf P(S)+0.7(1-\mathsf P(S))}\\&=\dfrac{9\mathsf P(S)}{2\mathsf P(S)+7}\end{align}$$