Is it possible to calculate this conditional probability?

Question

I have this problem.

Suppose that 20% of the population is made of terrorists.

A company developed an alarm such that, when a terrorist goes through the alarm, the alarm always beeps. The problem is that sometimes the alarm beeps also when the person is not a terrorist.

We can improve the alarm in such a way that it still always beeps when faced with terrorist, but when faced with non-terrorists it commits less mistakes.

Let's call alpha the probability that the alarm beeps when faced with a non-terrorist. What is the value of alpha if P(terrorist|alarm_beeped) = 99,9%?

(P(terrorist|alarm_beeped) meaning the probability that the person is a terrorist given that the alarm beeped).

I'm working with Bayes Theorem, but can't find a solution.

Any help?

Thanks

Answer 1

With Bayes theorem would be enough. First, suppose net events:

$T$ Person is a terrorist

$A$ Alarm beeps

From your statement, we have:

$$ P(T) = 0.2 $$ $$ P(A \vert T) = 1$$ $$ P(A \vert \bar{T}) = \alpha $$

Applying Bayes:

$$ P(T \vert A) = \frac{P(A \vert T) P(T)}{P(A)} = \frac{0.2}{0.2+0.8 \alpha}$$

Where $P(A)$ can be computed from total probability, as follow:

$$P(A) = P(A \vert T) P(T) + P(A \vert \bar{T}) P(\bar{T}) = 0.2 + 0.8 \alpha$$

So, to have $P(T \vert A) = 0.999$, we need:

$$\frac{0.2}{0.2+0.8 \alpha} = 0.999 $$

Where you can resolve for alpha.

Answer 2

For simplicity, let P(T) be the probability a person is a terrorist and P(A) be the probability that the alarm beeped.

alpha is the probability that the alarm beeps when faced with a non-terrorist. This is the same as the probability the alarm beeps given the person is a non-terrorist, i.e. $P(A|\neg T)$.

Since we are provided the probability $P(T|A)$, we can use this to obtain the following equation using Baye's theorem $$P(T|A) = \frac{P(A|T)P(T)}{P(A)}$$ However, we do not know the value of P(A) and this does not help us find P(A|N), so we use the alternative form of Baye's theorem $$P(T|A) = \frac{P(A|T)P(T)}{P(A|T)P(T) + P(A|\neg T)P(\neg T)}$$

It was given that P(T|A)=0.999 and that P(T)=0.20. From here we can note that a person is either a terrorist or not, so P(\neg T)=0.80. It was said that the alarm will always beep for a terrorist, so P(A|T)=1.0. We can now plug in these values into the above equation to obtain $$0.999 = \frac{1\times0.2}{(1\times0.2) + 0.8P(A|\neg T)}$$ $$0.999 = \frac{0.2}{0.2 + 0.8P(A|\neg T)}$$ We can then isolate the desired probability $$\left(0.2 + 0.8P(A|\neg T)\right)0.999 = 0.2$$ $$0.2 + 0.8P(A|\neg T) = 0.2002$$ $$0.8P(A|\neg T) = 0.0002$$ $$P(A|\neg T) = 0.00025$$ Thus, alpha = 0.00025 or 0.025%