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Chapter 2: Q67E (page 85)

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

Expert verified

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

Step by step solution

01

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

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

(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 \({\rm{0}}{\rm{.99 !}}\)

02

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 \({\rm{脡}}_{{\rm{i = 1}}}^{\rm{n}}{{\rm{A}}_{\rm{i}}}{\rm{ = \Omega }}\)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

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

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).

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Most popular questions from this chapter

In October, \({\rm{1994}}\), a flaw in a certain Pentium chip installed in computers was discovered that could result in a wrong answer when performing a division. The manufacturer initially claimed that the chance of any particular division being incorrect was only \({\rm{1}}\) in \({\rm{9}}\) billion, so that it would take thousands of years before a typical user encountered a mistake. However, statisticians are not typical users; some modern statistical techniques are so computationally intensive that a billion divisions over a short time period is not outside the realm of possibility. Assuming that the \({\rm{1}}\) in \({\rm{9}}\) billion figure is correct and that results of different divisions are independent of one another, what is the probability that at least one error occurs in one billion divisions with this chip?

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a. What is the probability that the selected student has at least one of the three types of cards?

b. What is the probability that the selected student has both a Visa card and a MasterCard but not an American Express card?

c. Calculate and interpret \(P\left( {B|A} \right)\)and also \(P\left( {A|B} \right)\)

d. If we learn that the selected student has an American Express card, what is the probability that she or he also has both a Visa card and a MasterCard?

e. Given that the selected student has an American Express card, what is the probability that she or he has at least one of the other two types of cards?

For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are

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a. If he wants to serve \({\rm{3}}\) bottles of zinfandel and serving order is important, how many ways are there to do this?

b. If \({\rm{6}}\) bottles of wine are to be randomly selected from the \({\rm{30}}\) for serving, how many ways are there to do this?

c. If \({\rm{6}}\) bottles are randomly selected, how many ways are there to obtain two bottles of each variety?

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