Question

There has been a great deal of controversy over the last several years regarding what types of surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a \({\rm{99\% }}\) chance of correctly identifying a future terrorist and a \({\rm{99}}{\rm{.9\% }}\)chance of correctly identifying someone who is not a future terrorist. If there are \({\rm{1000}}\) future terrorists in a population of \({\rm{300}}\) million, and one of these \({\rm{300}}\) million is randomly selected, scrutinized by the system, and identified as a future terrorist, what is the probability that he/she actually is a future terrorist? Does the value of this probability make you uneasy about using the surveillance system? Explain.

Short answer

The probability that the individual is actually a future terrorist \(P(A\mid B) = 0.003289\)

Step-by-step solution

Determine the value of \({\rm{P(A}}\mid {\rm{B)}}\)

Denote event \({\rm{A = }}\left\{ {{\rm{randomly selected person is a terrorist}}} \right\}{\rm{.}}\)From the exercise we have that 1000 future terrorists in a population of 300 million, which means that

\({\rm{P(A) = }}\frac{{{\rm{1000}}}}{{{\rm{300,000,000}}}}{\rm{ = 0}}{\rm{.0000033}}\)

We are asked to find the probability that the individual is actually a future terrorist.

Denote event \({\rm{B = }}\left\{ {{\rm{identified as a future terorist}}} \right\}{\rm{,}}\)the probability now becomes

\(\begin{array}{l}{\rm{P(A}}\mid {\rm{B)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} \frac{{{\rm{P(A)*P(B}}\mid {\rm{A)}}}}{{{\rm{P(A)*P(B}}\mid {\rm{A) + P}}\left( {{{\rm{A}}^{\rm{¢}}}} \right){\rm{*P}}\left( {{\rm{B}}\mid {{\rm{A}}^{\rm ¢}}} \right)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} \frac{{{\rm{0}}{\rm{.0000033*0}}{\rm{.99}}}}{{{\rm{0}}{\rm{.0000033*0}}{\rm{.99 + (1 - 0}}{\rm{.0000033)*(1 - 0}}{\rm{.999)}}}}\\{\rm{ = 0}}{\rm{.003289}}\end{array}\)

(1): here we use the Bayes' Theorem given below,

(2): all probabilities are given in the exercise. Note: chance of correctly identifying someone who is not a future terrorist is $0.999$ and chance of correctly identifying a future terrorist is \({\rm{0}}{\rm{.99 !}}\)

Explain the Bayes’ theorem

Bayes' Theorem

For \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}\)mutually exclusive events, and \({\rm{É}}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\)the prior probabilities are \(\left. {{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{,i = 1,2, \ldots ,n}}} \right){\rm{.}}\)If \({\rm{B}}\)is event for which \({\rm{P(B) > 0}}\), than the posterior probability of \({{\rm{A}}_{\rm{i}}}\)given that B has occurred

\({\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}\mid {\rm{B}}} \right){\rm{ = }}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}{\rm{ÇB}}} \right)}}{{{\rm{P(B)}}}}{\rm{ = }}\frac{{{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}_{\rm{j}}}} \right){\rm{P}}\left( {{{\rm{A}}_{\rm{j}}}} \right)}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{P}} \left( {{\rm{B}}\mid {{\rm{A}}_{\rm{i}}}} \right){\rm{ \times P}}\left( {{{\rm{A}}_{\rm{i}}}} \right)}}{\rm{,}}\;\;\;{\rm{j = 1,2, \ldots ,n}}\)

About \({\rm{0}}{\rm{.3\% }}\)of all people who "should" be terrorists would actually be a terrorist, therefore the surveillance system does not seem really appropriate (really small percentage).