A test has false alarm probability of $p$ and miss probability $q$, if the rate of true positives in the population is $r$, what is the probability someone is a positive given he or she tested positive?
If $t/f$ represents the result of the test and $T/F$ represents the actual positivity/negativity of the person. How exactly can I find the probability someone is a positive given he or she tested positive?
I see that $$P(t|F) = p, P(f|T)=q, P(t|T)=r, \text{ and } P(T) = P(T|f)P(f) + P(T|t)P(t)$$
I am trying to find $$P(T|t) = \frac{P(t|T)P(T)}{P(t)}$$
But I'm not seeing any way to find $P(T|t)$ explicitly.
Anyone have any ideas as to how to find this?
Given that $p$ is a false-alarm rate we say that $1 - p$ is the probability the test is a true negative, and likewise $1 - q$ is the rate the test returns positive given that the individual has the condition. $r$ must mean that $100r$% of the population has the condition; I can't think of any other explanation for it (otherwise it would seem that $r$ and $q$ are redundant and we would have no way to describe what proportion of the population has the condition.) Thus, $1 - r$ is the probability of not having the condition.
So a simple application of Bayes' Theorem gives
$$P(\text{positive} | \text{has condition}) = \frac{(1 - q)r}{(1 - q)r + p(1 - r)}$$