# Why the sum of true positive and false positive does not have to be equal to one?

While reading this question above, I got confused about the sum of true positive and false positive.

If an aircraft is present in a certain area, a radar detects it and generates an alarm signal with probability 99%. If an aircraft is not present, the radar generates a (false) alarm, with probability 10%

I understand that the sum of true positive and false negative should be 1. See question below, since the radar detects the aircraft if it's present only 99 percent of the time(true positive is 99%), that means, it does not detect the aircraft 1 percent of the time and says though it's actually there, which means, false negative is 1%

However, I'm not able to convince myself as to why the sum of true positive and false positive doesn't have to be 100%. I'm confused when I frame the question as follows, "The test result is positive for TP and FP. TP = aircraft present and radar on, FP = aircraft not present and radar on. Now radar being on is an event. Since the sum of P(aircraft present) and P(aircraft not present) should equal 1, so, P(aircraft present, radar on) and P(aircraft not present, radar on) should equal 1. That means, TP + FP should equal 1."

PS : This question is not about precision, recall, accuracy. I would like to know about the sum of TP + FP

• TP and FP probabilities are conditional probabilities, not joint probabilities, so your paragraph in quotes misrepresents what they are. – Glen_b Apr 5 '15 at 0:33

You have four conditional probabilities.

$$\text{tp} = \Pr(\text{detected}\mid\text{aircraft present})$$ $$\text{fn} =\Pr(\text{undetected}\mid\text{aircraft present})$$ $$\text{fp} = \Pr(\text{detected}\mid\text{aircraft not present})$$ $$\text{tn} = \Pr(\text{undetected}\mid\text{aircraft not present})$$

Using the definition of conditional probability, it's easy to prove that $$\text{tp} + \text{fn} = 1$$ and $$\text{fp} + \text{tn} = 1.$$

To prove the first one, since $$\{\text{detected}\}\cap\{\text{undetected}\}=\emptyset,$$ and $$\{\text{detected}\}\cup\{\text{undetected}\}=\Omega,$$ the sure event, then (draw a Venn diagram) $$\{\text{aircraft present}\} = \{\text{detected},\text{aircraft present}\}\cup\{\text{undetected},\text{aircraft present}\},$$ in which I've used the shortcut notation $$\{\text{detected},\text{aircraft present}\} := \{\text{detected}\}\cap\{\text{aircraft present}\}.$$

Hence, $$\Pr(\text{detected}\mid\text{aircraft present}) + \Pr(\text{undetected}\mid\text{aircraft present})$$ $$= \frac{\Pr(\text{detected},\text{aircraft present}) + \Pr(\text{undetected},\text{aircraft present})}{\Pr(\text{aircraft present})}$$ $$= \frac{\Pr(\text{aircraft present})}{\Pr(\text{aircraft present})} = 1.$$

The intuition is that, if you're given the same information, the conditional probabilities of two complementary events must add up to one.

Mathematically, if $\Pr(B)>0$, then $\Pr(\,\cdot\mid B)$ is a probability measure.

But, in general, $$\text{tp} + \text{fp} \neq 1.$$

Consider this: you have a super radar which, given that an aircraft is present, always detect it ($\text{tp}=1$). It never misses a true aircraft. But, sometimes, your super radar is so sensitive that, given that no aircraft is present, it confuses a condor with an aircraft, so that, say, $\text{fp}=0.2$.

• Thanks Zen. Or the opposite, I have a very unintelligent radar which always gives positive irrespective of the aircraft. In that case. TP = 1, and FP = 1. Is this correct? So basically TP + FP doesn't have to 1? – IAMTubby Apr 6 '15 at 2:11