Problem with a probability question. I'm having a problem with this probability question: 2 robbers disguises themselves as customers in a bank that has 38 customers (non-robbers). Police try to reveal the robbers by using a lie-detector. The detector can identify a robber with 85% probability, however, it has 8% chance to identify a customer as a robber. One person got identified by the detector as a robber. What are the chances that this person is one of the robbers?
I tried solving this with binomial probability but I can't get the right answer, which is supposed to be 0,36.
 A: This is actually a question concerning the Bayes' Theorem.
Denote $D$ be the event that the tested man is the robber, and $E$ the event that the detector said he is the robber.
$$P(D \mid E) = \frac{P(D \cap E)}{P(E)}=\frac {P(E \mid D)P(D)}{P(E \mid D)P(D)+P(E \mid D^c)P(D^c)}$$
If the computation is correct, you should then get $0.3586\approx 0.36$.
A: We interpret the question as asking the following. Suppose that a certain person gets identified by the test as a robber. What is the probability that she is indeed a robber? 
Consider a particular individual, selected at random from the $40$. Let $T$ be the event that the test identifies her as a robber, and let $R$ be the event that she is a robber. We want to find the conditional probability $\Pr(R|T)$. By the usual definition of conditional probability, we have
$$\Pr(R|T)=\frac{\Pr(R\cap T)}{\Pr(T)}.\tag{1}$$
We now proceed to find the two probabilities on the right of (1).
The event $T$ can happen in two disjoint ways: (i) the person is a robber, and is so identified or (ii) the person is a non-robber, but is identified by the test as a robber. 
We first find the probability of (i). The probability someone is a robber is $\frac{2}{40}$. Given that she is a robber, the probability the test identifies her as a robber is $0.85$. It follows that the probability of (i) is $\frac{2}{40}(0.85)$.
A similar argument shows that the probability of (ii) is $\frac{38}{40}(0.08)$.
It follows that
$$\Pr(T)=\frac{2}{40}(0.85)+\frac{38}{40}(0.08).$$
Now we find $\Pr(R\cap T)$. Conveniently, we have already found it, since it is the probability of (i). Now put things together. We find that
$$\Pr(R|T)=\frac{\frac{2}{40}(0.85)}{\frac{2}{40}(0.85)+\frac{38}{40}(0.08)}.$$
