# What is the probability that at least three guilty parties are caught at the same time and at least four of the innocent are released?

A lie detector will be used by police to investigate 10 suspects of involvement in a particular crime. Admit that among them, five are guilty (but will plead innocence) and the other five are really innocent. It is also known that:

even when a person tells the truth, the detector has a 5% chance of failing, indicating that she lied;

even when she lies, the detector has a 30% chance of not being able to detect the lie.

What is the probability that at least three guilty parties are caught at the same time and at least four of the innocent are released?

Let C be the event "is a criminal" and L be "lie detector worked OK":

$$P(C \cap L) = P(C) \cdot P(L|C) = 0.35$$

$$P(\lnot C \cap L) = P(\lnot C) \cdot P(L|\lnot C) = 0.475$$

Let X be the number of criminals caught, Y the number of released innocents and E the event that we are interested in:

$$X:Bin(10, 0.35)$$

$$Y:Bin(10, 0.475)$$

$$P(E) = \sum_{i=3}^{6} \sum_{j=4}^{10 - i} P(X=i, Y=j)$$

$$P(E) = \sum_{i=3}^{6} \sum_{j=4}^{10 - i} P(X=i) \cdot P(Y=j|X=i)$$

$$P(E) = \sum_{i=3}^{6} \sum_{j=4}^{10 - i} \binom{10}{i} \cdot 0.35^i \cdot 0.65^{10-i} \cdot \binom{10-i}{j} \cdot 0.475^j \cdot 0.525^{10-i-j}$$

Question:

Are the steps of the solution correct? If so, is there a smarter way to solve this problem?

If the police do not know the number of actual criminals, i.e. it is possible that they could release all or none of the suspects, then the basic approach appears to be correct. But, since there are 5 innocents and 5 criminals and what happens to a criminal is independent of what happens to an innocent then the definitions of X and Y and the summation indexing needs to reflect that. Specifically, the range of possible values for both X and Y are the integers between 1 and 5. Recommend this revision:

The probability of the i'th criminal being caught is:

$$P(X_i = 1) = P(L|C) = 0.7$$

The probability of the i'th innocent being released is:

$$P(Y_i = 1) = P(L|\lnot C) = 0.95$$

Therefore,

$$X: Bin(5,0.7)$$

$$Y: Bin(5,0.95)$$

$$P(E) = P(X>3, Y>4)$$

by independence,

$$P(E) = P(X>3) P(Y>4)$$

$$P(E) = \sum_{i=3}^{5} \sum_{j=4}^{5} P(X=i)P(Y=j)$$

$$P(E) = \sum_{i=3}^{5} \sum_{j=4}^{5} \binom{5}{i} \cdot 0.7^i \cdot 0.3^{5-i} \cdot \binom{5}{j} \cdot 0.95^j \cdot 0.05^{5-j}$$

If, on the other hand, the police will release 5 individuals, no matter what the results of the lie detector are, the problem becomes significantly more complicated and the information provided in the question is not sufficient to answer the question. As an example of this, it is possible that every single individual could be flagged by the lie detector (although highly unlikely, it is still possible). In this situation, how would the police proceed?

• Thanks for your comments! Please, consider that police knows nothing about the suspects so it is possible that they could release (or not) them all. – David Duarte Jan 11 at 13:12
• David, my pleasure! In that case, I would follow the basic outline in my answer. The key is leveraging the independence of the two groups to make the calculations a bit easier. – Ezra Jan 11 at 17:01